Calculus And Linear Algebra

CT 1, 2, 3 Compilation PYQs | Detailed Step-by-Step Solutions

Paper 1 - Question 1 (Marks: 8)
Find the Eigenvalues and Eigenvectors of \( \begin{pmatrix} 6 & -6 & 5 \\ 14 & -13 & 10 \\ 7 & -6 & 4 \end{pmatrix} \)
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Correct Answer: Eigenvalues: \( \lambda = -1, -1, -1 \). Eigenvectors: \( X_1 = \begin{pmatrix} 6 \\ 7 \\ 0 \end{pmatrix} \) and \( X_2 = \begin{pmatrix} -5 \\ 0 \\ 7 \end{pmatrix} \)
Detailed Solution:
Step 1: Write the characteristic equation
The characteristic equation is given by \( |A - \lambda I| = 0 \).
For a \( 3 \times 3 \) matrix, it can be written as:
\( \lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0 \)
Step 2: Find \( S_1, S_2, S_3 \)
\( S_1 = \text{Trace}(A) = 6 + (-13) + 4 = -3 \)
\( S_2 = \text{Sum of minors of principal diagonal elements} \)
\( M_{11} = \begin{vmatrix} -13 & 10 \\ -6 & 4 \end{vmatrix} = (-13)(4) - (10)(-6) = -52 + 60 = 8 \)
\( M_{22} = \begin{vmatrix} 6 & 5 \\ 7 & 4 \end{vmatrix} = 24 - 35 = -11 \)
\( M_{33} = \begin{vmatrix} 6 & -6 \\ 14 & -13 \end{vmatrix} = -78 - (-84) = 6 \)
\( S_2 = 8 - 11 + 6 = 3 \)
\( S_3 = |A| = 6(-52 + 60) - (-6)(56 - 70) + 5(-84 + 91) \)
\( S_3 = 6(8) + 6(-14) + 5(7) = 48 - 84 + 35 = -1 \)
Step 3: Solve the characteristic equation
Substitute \( S_1, S_2, S_3 \) into the equation:
\( \lambda^3 - (-3)\lambda^2 + 3\lambda - (-1) = 0 \)
\( \lambda^3 + 3\lambda^2 + 3\lambda + 1 = 0 \)
This is the algebraic expansion of \( (\lambda + 1)^3 = 0 \).
Thus, the eigenvalues are \( \lambda = -1, -1, -1 \).
Step 4: Find Eigenvectors for \( \lambda = -1 \)
We solve \( (A - \lambda I)X = 0 \):
\( \begin{pmatrix} 6 - (-1) & -6 & 5 \\ 14 & -13 - (-1) & 10 \\ 7 & -6 & 4 - (-1) \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
\( \begin{pmatrix} 7 & -6 & 5 \\ 14 & -12 & 10 \\ 7 & -6 & 5 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
Since all three rows are proportional (R2 = 2*R1, R3 = R1), we only have one independent equation:
\( 7x - 6y + 5z = 0 \)
Since there are 3 variables and 1 equation, we have \( 3 - 1 = 2 \) free variables. We can find two linearly independent eigenvectors.
Case A: Let \( z = 0 \), then \( 7x - 6y = 0 \implies x = 6, y = 7 \).
Eigenvector \( X_1 = \begin{pmatrix} 6 \\ 7 \\ 0 \end{pmatrix} \)
Case B: Let \( y = 0 \), then \( 7x + 5z = 0 \implies x = -5, z = 7 \).
Eigenvector \( X_2 = \begin{pmatrix} -5 \\ 0 \\ 7 \end{pmatrix} \)
Paper 1 - Question 2a (Marks: 4)
Find the nature of the quadratic form without reducing to its canonical form: \( x^2 + 2y^2 + 3z^2 + 2xy + 2yz - 2xz \).
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Correct Answer: The quadratic form is Indefinite.
Detailed Solution:
Step 1: Write the matrix of the quadratic form (A)
The general form is \( ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy \).
Comparing this to \( x^2 + 2y^2 + 3z^2 + 2xy + 2yz - 2xz \), we have:
\( a = 1, b = 2, c = 3 \)
\( 2f = 2 \implies f = 1 \)
\( 2g = -2 \implies g = -1 \)
\( 2h = 2 \implies h = 1 \)
The symmetric matrix \( A \) is \( \begin{pmatrix} a & h & g \\ h & b & f \\ g & f & c \end{pmatrix} = \begin{pmatrix} 1 & 1 & -1 \\ 1 & 2 & 1 \\ -1 & 1 & 3 \end{pmatrix} \)
Step 2: Find the principal minors
\( D_1 = |1| = 1 \quad (\text{which is } > 0) \)

\( D_2 = \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = (1)(2) - (1)(1) = 1 \quad (\text{which is } > 0) \)

\( D_3 = |A| = \begin{vmatrix} 1 & 1 & -1 \\ 1 & 2 & 1 \\ -1 & 1 & 3 \end{vmatrix} \)
Expanding along the first row:
\( D_3 = 1(6 - 1) - 1(3 - (-1)) - 1(1 - (-2)) \)
\( D_3 = 1(5) - 1(4) - 1(3) = 5 - 4 - 3 = -2 \quad (\text{which is } < 0) \)
Step 3: Determine the nature
Since the principal minors are \( D_1 > 0 \), \( D_2 > 0 \), and \( D_3 < 0 \), the signs do not alternate completely (for negative definite), nor are they all positive (for positive definite).
By Sylvester's criterion, the quadratic form is Indefinite.
Paper 1 - Question 2b (Marks: 4)
Two of the Eigenvalues of \( A = \begin{pmatrix} 1 & 1 & -2 \\ -1 & 2 & 1 \\ 0 & 1 & -1 \end{pmatrix} \) are \( 1 \) and \( -1 \). Find the third Eigenvalue and hence find the Eigenvalues of \( A^{-1} \).
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Correct Answer: The third eigenvalue of \( A \) is \( 2 \). The eigenvalues of \( A^{-1} \) are \( 1, -1, \frac{1}{2} \).
Detailed Solution:
Step 1: Find the sum of the eigenvalues of A
A fundamental property of matrices is that the sum of the eigenvalues equals the Trace of the matrix (sum of principal diagonal elements).
Trace(A) = \( 1 + 2 + (-1) = 2 \).
Step 2: Find the third eigenvalue (\(\lambda_3\))
Let the eigenvalues be \( \lambda_1, \lambda_2, \lambda_3 \).
We are given \( \lambda_1 = 1 \) and \( \lambda_2 = -1 \).
\( \lambda_1 + \lambda_2 + \lambda_3 = \text{Trace}(A) \)
\( 1 + (-1) + \lambda_3 = 2 \implies \lambda_3 = 2 \).
So, the eigenvalues of A are \( 1, -1, 2 \).
Step 3: Find the eigenvalues of \( A^{-1} \)
By property, if \( \lambda \) is an eigenvalue of \( A \), then \( \frac{1}{\lambda} \) is an eigenvalue of \( A^{-1} \) (provided \( |A| \neq 0 \)).
The eigenvalues of \( A^{-1} \) are \( \frac{1}{1}, \frac{1}{-1}, \frac{1}{2} \).
Thus, the eigenvalues of \( A^{-1} \) are \( 1, -1, \frac{1}{2} \).
Paper 1 - Question 3 (Marks: 8)
Diagonalize the matrix \( A = \begin{pmatrix} 3 & 1 & 1 \\ 1 & 3 & -1 \\ 1 & -1 & 3 \end{pmatrix} \) through orthogonal transformation, where the Eigenvalues are 1, 4, 4.
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Correct Answer: The orthogonal matrix is \( N = \begin{pmatrix} -1/\sqrt{3} & 1/\sqrt{2} & -1/\sqrt{6} \\ 1/\sqrt{3} & 1/\sqrt{2} & 1/\sqrt{6} \\ 1/\sqrt{3} & 0 & -2/\sqrt{6} \end{pmatrix} \) and the diagonalized matrix is \( D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} \).
Detailed Solution:
Step 1: Find the Eigenvectors for \( \lambda = 1 \)
Substitute \( \lambda = 1 \) into \( (A - \lambda I)X = 0 \):
\( \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
Solving the first two equations using cross-multiplication:
\( 2x + y + z = 0 \) and \( x + 2y - z = 0 \)
\( \frac{x}{(1)(-1) - (1)(2)} = \frac{y}{(1)(1) - (2)(-1)} = \frac{z}{(2)(2) - (1)(1)} \)
\( \frac{x}{-3} = \frac{y}{3} = \frac{z}{3} \implies \frac{x}{-1} = \frac{y}{1} = \frac{z}{1} \)
Eigenvector \( X_1 = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \)
Step 2: Find the Eigenvectors for \( \lambda = 4 \)
Substitute \( \lambda = 4 \) into \( (A - \lambda I)X = 0 \):
\( \begin{pmatrix} -1 & 1 & 1 \\ 1 & -1 & -1 \\ 1 & -1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
This yields the single independent equation: \( -x + y + z = 0 \implies x = y + z \).
Let \( y = 1, z = 0 \implies x = 1 \). So \( X_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \).
Since A is symmetric, eigenvectors corresponding to distinct eigenvalues are orthogonal. Check \( X_1 \cdot X_2 = (-1)(1) + (1)(1) + (1)(0) = 0 \).
For the third eigenvector \( X_3 = \begin{pmatrix} l \\ m \\ n \end{pmatrix} \), it must be orthogonal to both \( X_1 \) and \( X_2 \):
\( -l + m + n = 0 \) and \( l + m = 0 \).
Using cross-multiplication on \( -l + m + n = 0 \) and \( l + m + 0n = 0 \):
\( \frac{l}{(1)(0) - (1)(1)} = \frac{m}{(1)(1) - (-1)(0)} = \frac{n}{(-1)(1) - (1)(1)} \)
\( \frac{l}{-1} = \frac{m}{1} = \frac{n}{-2} \implies X_3 = \begin{pmatrix} -1 \\ 1 \\ -2 \end{pmatrix} \)
Step 3: Normalize the Eigenvectors
Length of \( X_1 = \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3} \implies e_1 = \begin{pmatrix} -1/\sqrt{3} \\ 1/\sqrt{3} \\ 1/\sqrt{3} \end{pmatrix} \)
Length of \( X_2 = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2} \implies e_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \)
Length of \( X_3 = \sqrt{(-1)^2 + 1^2 + (-2)^2} = \sqrt{6} \implies e_3 = \begin{pmatrix} -1/\sqrt{6} \\ 1/\sqrt{6} \\ -2/\sqrt{6} \end{pmatrix} \)
Step 4: Form Orthogonal Matrix N and Diagonalize
The orthogonal matrix \( N = [e_1 \quad e_2 \quad e_3] = \begin{pmatrix} -1/\sqrt{3} & 1/\sqrt{2} & -1/\sqrt{6} \\ 1/\sqrt{3} & 1/\sqrt{2} & 1/\sqrt{6} \\ 1/\sqrt{3} & 0 & -2/\sqrt{6} \end{pmatrix} \).
The diagonal matrix is \( D = N^T A N = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} \).
Paper 1 - Question 4 (Marks: 14)
Verify the Cayley Hamilton theorem for \( A = \begin{pmatrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix} \), hence find \( A^{-1} \).
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Correct Answer: The Cayley Hamilton theorem is verified. \( A^{-1} = \frac{1}{35} \begin{pmatrix} -4 & 11 & -5 \\ -1 & -6 & 25 \\ 6 & 1 & -10 \end{pmatrix} \).
Detailed Solution:
Step 1: Find the Characteristic Equation
The equation is \( \lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0 \).
\( S_1 = \text{Trace}(A) = 1 + 2 + 1 = 4 \).
\( S_2 = M_{11} + M_{22} + M_{33} = (2 - 6) + (1 - 7) + (2 - 12) = -4 - 6 - 10 = -20 \).
\( S_3 = |A| = 1(2 - 6) - 3(4 - 3) + 7(8 - 2) = -4 - 3 + 42 = 35 \).
The characteristic equation is \( \lambda^3 - 4\lambda^2 - 20\lambda - 35 = 0 \).
Step 2: State Cayley Hamilton Theorem
By the theorem, every square matrix satisfies its own characteristic equation:
\( A^3 - 4A^2 - 20A - 35I = 0 \).
Step 3: Compute \( A^2 \) and \( A^3 \)
\( A^2 = A \cdot A = \begin{pmatrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 20 & 23 & 23 \\ 15 & 22 & 37 \\ 10 & 9 & 14 \end{pmatrix} \)
\( A^3 = A^2 \cdot A = \begin{pmatrix} 20 & 23 & 23 \\ 15 & 22 & 37 \\ 10 & 9 & 14 \end{pmatrix} \begin{pmatrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 135 & 152 & 232 \\ 140 & 163 & 208 \\ 60 & 76 & 111 \end{pmatrix} \)
Step 4: Verify the Theorem
Substitute the matrices into \( A^3 - 4A^2 - 20A - 35I \):
\( \begin{pmatrix} 135 & 152 & 232 \\ 140 & 163 & 208 \\ 60 & 76 & 111 \end{pmatrix} - \begin{pmatrix} 80 & 92 & 92 \\ 60 & 88 & 148 \\ 40 & 36 & 56 \end{pmatrix} - \begin{pmatrix} 20 & 60 & 140 \\ 80 & 40 & 60 \\ 20 & 40 & 20 \end{pmatrix} - \begin{pmatrix} 35 & 0 & 0 \\ 0 & 35 & 0 \\ 0 & 0 & 35 \end{pmatrix} \)
Element (1,1): \( 135 - 80 - 20 - 35 = 0 \)
Element (2,2): \( 163 - 88 - 40 - 35 = 0 \)
Element (3,3): \( 111 - 56 - 20 - 35 = 0 \)
All elements resolve to 0. Hence, \( A^3 - 4A^2 - 20A - 35I = 0 \). Verified.
Step 5: Find \( A^{-1} \)
Multiply the equation by \( A^{-1} \):
\( A^2 - 4A - 20I - 35A^{-1} = 0 \implies 35A^{-1} = A^2 - 4A - 20I \)
\( 35A^{-1} = \begin{pmatrix} 20 & 23 & 23 \\ 15 & 22 & 37 \\ 10 & 9 & 14 \end{pmatrix} - \begin{pmatrix} 4 & 12 & 28 \\ 16 & 8 & 12 \\ 4 & 8 & 4 \end{pmatrix} - \begin{pmatrix} 20 & 0 & 0 \\ 0 & 20 & 0 \\ 0 & 0 & 20 \end{pmatrix} \)
\( 35A^{-1} = \begin{pmatrix} -4 & 11 & -5 \\ -1 & -6 & 25 \\ 6 & 1 & -10 \end{pmatrix} \)
\( A^{-1} = \frac{1}{35} \begin{pmatrix} -4 & 11 & -5 \\ -1 & -6 & 25 \\ 6 & 1 & -10 \end{pmatrix} \)
Paper 1 - Question 5 (Marks: 14)
Reduce the quadratic form \( x^2 + 2yz \) into a canonical form by means of orthogonal transformation and hence determine its nature, index, rank and signature.
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Correct Answer: Canonical form is \( y_1^2 + y_2^2 - y_3^2 \).
Nature: Indefinite, Index: 2, Rank: 3, Signature: 1.
Detailed Solution:
Step 1: Write the matrix of the quadratic form (A)
Comparing \( x^2 + 2yz \) with \( ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy \):
\( a=1, b=0, c=0, 2f=2 \implies f=1, g=0, h=0 \).
\( A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \)
Step 2: Find the Characteristic Equation and Eigenvalues
\( |A - \lambda I| = \begin{vmatrix} 1-\lambda & 0 & 0 \\ 0 & -\lambda & 1 \\ 0 & 1 & -\lambda \end{vmatrix} = 0 \)
\( (1-\lambda)( \lambda^2 - 1 ) = 0 \implies (1-\lambda)(\lambda-1)(\lambda+1) = 0 \)
The eigenvalues are \( \lambda = 1, 1, -1 \).
Step 3: Find the Eigenvectors
For \( \lambda = -1 \): \( \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
\( 2x = 0 \implies x = 0 \) and \( y + z = 0 \implies y = -z \).
Let \( z = 1 \implies y = -1 \). So \( X_1 = \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix} \).

For \( \lambda = 1 \): \( \begin{pmatrix} 0 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
This gives \( -y + z = 0 \implies y = z \). \( x \) is a free variable.
Let \( x = 1, z = 0 \implies y = 0 \). So \( X_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \).
Let \( x = 0, z = 1 \implies y = 1 \). So \( X_3 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \).
Note that \( X_2 \cdot X_1 = 0 \), \( X_3 \cdot X_1 = 0 \), and \( X_2 \cdot X_3 = 0 \). They are mutually orthogonal.
Step 4: Normalize Eigenvectors to Form Matrix N
\( e_1 = \begin{pmatrix} 0 \\ -1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix}, e_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, e_3 = \begin{pmatrix} 0 \\ 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix} \).
Orthogonal Matrix \( N = [e_2 \quad e_3 \quad e_1] = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1/\sqrt{2} & -1/\sqrt{2} \\ 0 & 1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix} \). (Ordered for eigenvalues 1, 1, -1).
Step 5: Write the Canonical Form and Determine Nature
The diagonal matrix is \( D = \text{diag}(1, 1, -1) \).
The canonical form is \( \lambda_1 y_1^2 + \lambda_2 y_2^2 + \lambda_3 y_3^2 = 1 \cdot y_1^2 + 1 \cdot y_2^2 + (-1) \cdot y_3^2 \).
Canonical Form: \( y_1^2 + y_2^2 - y_3^2 \).
Nature: Indefinite (Contains both positive and negative terms).
Rank: Total non-zero terms in canonical form = 3.
Index: Total positive terms in canonical form = 2.
Signature: (Index) - (Number of negative terms) = 2 - 1 = 1.
Paper 2 - Question 1 (Marks: 8)
Find the Eigenvalues and Eigenvectors of \( A = \begin{pmatrix} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{pmatrix} \)
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Correct Answer: Eigenvalues: \( \lambda = 1, 1, 5 \). Eigenvectors: \( X_1 = \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix}, X_2 = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \) and \( X_3 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \)
Detailed Solution:
Step 1: Write the characteristic equation
\( |A - \lambda I| = 0 \implies \lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0 \)
Step 2: Find \( S_1, S_2, S_3 \)
\( S_1 = \text{Trace}(A) = 2 + 3 + 2 = 7 \)
\( S_2 = \begin{vmatrix} 3 & 1 \\ 2 & 2 \end{vmatrix} + \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} + \begin{vmatrix} 2 & 2 \\ 1 & 3 \end{vmatrix} = (6 - 2) + (4 - 1) + (6 - 2) = 4 + 3 + 4 = 11 \)
\( S_3 = |A| = 2(6 - 2) - 2(2 - 1) + 1(2 - 3) = 2(4) - 2(1) - 1 = 8 - 2 - 1 = 5 \)
Step 3: Solve the characteristic equation
\( \lambda^3 - 7\lambda^2 + 11\lambda - 5 = 0 \)
By trial, if \( \lambda = 1 \): \( 1 - 7 + 11 - 5 = 0 \), so \( \lambda = 1 \) is a root.
Factor out \( (\lambda - 1) \): \( (\lambda - 1)(\lambda^2 - 6\lambda + 5) = 0 \)
\( (\lambda - 1)(\lambda - 1)(\lambda - 5) = 0 \)
Eigenvalues are \( \lambda = 1, 1, 5 \).
Step 4: Find Eigenvectors for \( \lambda = 5 \)
\( (A - 5I)X = 0 \implies \begin{pmatrix} -3 & 2 & 1 \\ 1 & -2 & 1 \\ 1 & 2 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
Taking the first two equations:
\( -3x + 2y + z = 0 \)
\( x - 2y + z = 0 \)
Adding them: \( -2x + 2z = 0 \implies x = z \).
Substitute \( x = z \) into the second equation: \( z - 2y + z = 0 \implies 2y = 2z \implies y = z \).
Let \( x = 1 \implies y = 1, z = 1 \). So \( X_3 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \).
Step 5: Find Eigenvectors for \( \lambda = 1 \)
\( (A - I)X = 0 \implies \begin{pmatrix} 1 & 2 & 1 \\ 1 & 2 & 1 \\ 1 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
This gives one independent equation: \( x + 2y + z = 0 \).
Since there are 2 free variables, we get 2 independent eigenvectors.
Let \( z = 0, y = 1 \implies x = -2 \). So \( X_1 = \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix} \).
Let \( y = 0, z = 1 \implies x = -1 \). So \( X_2 = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix} \).
Paper 2 - Question 2a (Marks: 4)
Find the nature of the quadratic form without reducing to its canonical form: \( x^2 + 5y^2 + z^2 + 2xy + 2yz + 6xz \).
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Correct Answer: The quadratic form is Indefinite.
Detailed Solution:
Step 1: Form the symmetric matrix A
Comparing to \( ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy \):
\( a=1, b=5, c=1 \)
\( 2f=2 \implies f=1 \)
\( 2g=6 \implies g=3 \)
\( 2h=2 \implies h=1 \)
\( A = \begin{pmatrix} 1 & 1 & 3 \\ 1 & 5 & 1 \\ 3 & 1 & 1 \end{pmatrix} \)
Step 2: Find the principal minors
\( D_1 = |1| = 1 \quad (> 0) \)
\( D_2 = \begin{vmatrix} 1 & 1 \\ 1 & 5 \end{vmatrix} = 5 - 1 = 4 \quad (> 0) \)
\( D_3 = |A| = 1(5 - 1) - 1(1 - 3) + 3(1 - 15) \)
\( D_3 = 1(4) - 1(-2) + 3(-14) = 4 + 2 - 42 = -36 \quad (< 0) \)
Step 3: Determine the nature
Since \( D_1 > 0 \), \( D_2 > 0 \), and \( D_3 < 0 \), the sequence of signs is (+, +, -). This means it is neither positive definite (all +) nor negative definite (-, +, -).
Therefore, the quadratic form is Indefinite.
Paper 2 - Question 2b (Marks: 4)
Two of the Eigenvalues of \( A = \begin{pmatrix} 2 & -2 & 2 \\ 1 & 1 & 1 \\ 1 & 3 & -1 \end{pmatrix} \) are \( -2 \) and \( 2 \). Find the third Eigenvalue. Hence find the Eigenvalues of \( A^{-1} \) and \( A^4 \).
Click to Reveal Answer & Solution
Correct Answer:
Third eigenvalue is \( 2 \).
Eigenvalues of \( A^{-1} \) are \( -1/2, 1/2, 1/2 \).
Eigenvalues of \( A^4 \) are \( 16, 16, 16 \).
Detailed Solution:
Step 1: Find the sum of eigenvalues
Sum of eigenvalues = \( \text{Trace}(A) = 2 + 1 + (-1) = 2 \).
Step 2: Find the third eigenvalue (\(\lambda_3\))
Let the eigenvalues be \( \lambda_1 = -2, \lambda_2 = 2 \), and \( \lambda_3 \).
\( -2 + 2 + \lambda_3 = 2 \implies \lambda_3 = 2 \).
Eigenvalues of A are \( -2, 2, 2 \).
Step 3: Find eigenvalues of \( A^{-1} \)
Eigenvalues of \( A^{-1} \) are \( 1/\lambda \):
\( \frac{1}{-2}, \frac{1}{2}, \frac{1}{2} \implies -0.5, 0.5, 0.5 \).
Step 4: Find eigenvalues of \( A^4 \)
Eigenvalues of \( A^k \) are \( \lambda^k \):
\( (-2)^4, 2^4, 2^4 \implies 16, 16, 16 \).
Paper 2 - Question 3 (Marks: 8)
Diagonalize the matrix \( A = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 3 & 0 \\ 1 & 0 & 2 \end{pmatrix} \) through orthogonal transformation, where the Eigenvalues are 1, 3, 3.
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Correct Answer: The orthogonal matrix is \( N = \begin{pmatrix} 1/\sqrt{2} & 0 & 1/\sqrt{2} \\ 0 & 1 & 0 \\ -1/\sqrt{2} & 0 & 1/\sqrt{2} \end{pmatrix} \) and the diagonalized matrix is \( D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix} \).
Detailed Solution:
Step 1: Find the Eigenvectors for \( \lambda = 1 \)
Substitute \( \lambda = 1 \) into \( (A - \lambda I)X = 0 \):
\( \begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
This gives equations: \( x + z = 0 \implies x = -z \) and \( 2y = 0 \implies y = 0 \).
Let \( z = -1 \implies x = 1 \). So \( X_1 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \).
Step 2: Find the Eigenvectors for \( \lambda = 3 \)
Substitute \( \lambda = 3 \) into \( (A - \lambda I)X = 0 \):
\( \begin{pmatrix} -1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
This yields the single independent equation: \( -x + z = 0 \implies x = z \).
The variable \( y \) is free. We need two eigenvectors \( X_2 \) and \( X_3 \) that satisfy \( x = z \) and are orthogonal to \( X_1 \).
Let \( x = 0, z = 0, y = 1 \). So \( X_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \).
Let \( x = 1, z = 1, y = 0 \). So \( X_3 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \).
Check orthogonality: \( X_1 \cdot X_2 = 0 \), \( X_2 \cdot X_3 = 0 \), \( X_1 \cdot X_3 = 0 \). They are mutually orthogonal.
Step 3: Normalize the Eigenvectors
Length of \( X_1 = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2} \implies e_1 = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ -1/\sqrt{2} \end{pmatrix} \)
Length of \( X_2 = \sqrt{0^2 + 1^2 + 0^2} = 1 \implies e_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \)
Length of \( X_3 = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2} \implies e_3 = \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ 1/\sqrt{2} \end{pmatrix} \)
Step 4: Form Orthogonal Matrix N and Diagonalize
The orthogonal matrix \( N = [e_1 \quad e_2 \quad e_3] = \begin{pmatrix} 1/\sqrt{2} & 0 & 1/\sqrt{2} \\ 0 & 1 & 0 \\ -1/\sqrt{2} & 0 & 1/\sqrt{2} \end{pmatrix} \).
The diagonal matrix is \( D = N^T A N = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix} \).
Paper 2 - Question 4 (Marks: 14)
Verify the Cayley Hamilton theorem for \( A = \begin{pmatrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix} \), hence find \( A^{-1} \).
Click to Reveal Answer & Solution
Correct Answer: The Cayley Hamilton theorem is verified. \( A^{-1} = \frac{1}{35} \begin{pmatrix} -4 & 11 & -5 \\ -1 & -6 & 25 \\ 6 & 1 & -10 \end{pmatrix} \).
Detailed Solution:
(This question is identical to Paper 1, Question 4.)
Step 1: Find the Characteristic Equation
The equation is \( \lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0 \).
\( S_1 = \text{Trace}(A) = 1 + 2 + 1 = 4 \).
\( S_2 = M_{11} + M_{22} + M_{33} = (2 - 6) + (1 - 7) + (2 - 12) = -4 - 6 - 10 = -20 \).
\( S_3 = |A| = 1(2 - 6) - 3(4 - 3) + 7(8 - 2) = -4 - 3 + 42 = 35 \).
The characteristic equation is \( \lambda^3 - 4\lambda^2 - 20\lambda - 35 = 0 \).
Step 2: State Cayley Hamilton Theorem
By the theorem, every square matrix satisfies its own characteristic equation:
\( A^3 - 4A^2 - 20A - 35I = 0 \).
Step 3: Compute \( A^2 \) and \( A^3 \)
\( A^2 = A \cdot A = \begin{pmatrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 20 & 23 & 23 \\ 15 & 22 & 37 \\ 10 & 9 & 14 \end{pmatrix} \)
\( A^3 = A^2 \cdot A = \begin{pmatrix} 20 & 23 & 23 \\ 15 & 22 & 37 \\ 10 & 9 & 14 \end{pmatrix} \begin{pmatrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 135 & 152 & 232 \\ 140 & 163 & 208 \\ 60 & 76 & 111 \end{pmatrix} \)
Step 4: Verify the Theorem
Substitute the matrices into \( A^3 - 4A^2 - 20A - 35I \):
\( \begin{pmatrix} 135 & 152 & 232 \\ 140 & 163 & 208 \\ 60 & 76 & 111 \end{pmatrix} - \begin{pmatrix} 80 & 92 & 92 \\ 60 & 88 & 148 \\ 40 & 36 & 56 \end{pmatrix} - \begin{pmatrix} 20 & 60 & 140 \\ 80 & 40 & 60 \\ 20 & 40 & 20 \end{pmatrix} - \begin{pmatrix} 35 & 0 & 0 \\ 0 & 35 & 0 \\ 0 & 0 & 35 \end{pmatrix} = 0 \). Verified.
Step 5: Find \( A^{-1} \)
Multiply the equation by \( A^{-1} \):
\( 35A^{-1} = A^2 - 4A - 20I = \begin{pmatrix} -4 & 11 & -5 \\ -1 & -6 & 25 \\ 6 & 1 & -10 \end{pmatrix} \implies A^{-1} = \frac{1}{35} \begin{pmatrix} -4 & 11 & -5 \\ -1 & -6 & 25 \\ 6 & 1 & -10 \end{pmatrix} \)
Paper 2 - Question 5 (Marks: 14)
Reduce the quadratic form \( x^2 + 2yz \) into a canonical form by means of orthogonal transformation and hence determine its nature, index, rank and signature.
Click to Reveal Answer & Solution
Correct Answer: Canonical form is \( y_1^2 + y_2^2 - y_3^2 \).
Nature: Indefinite, Index: 2, Rank: 3, Signature: 1.
Detailed Solution:
(This question is identical to Paper 1, Question 5.)
Step 1: Write the matrix of the quadratic form (A)
\( A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \)
Step 2: Find the Characteristic Equation and Eigenvalues
\( |A - \lambda I| = \begin{vmatrix} 1-\lambda & 0 & 0 \\ 0 & -\lambda & 1 \\ 0 & 1 & -\lambda \end{vmatrix} = 0 \)
\( (1-\lambda)( \lambda^2 - 1 ) = 0 \implies (1-\lambda)(\lambda-1)(\lambda+1) = 0 \)
The eigenvalues are \( \lambda = 1, 1, -1 \).
Step 3: Find the Eigenvectors
For \( \lambda = -1 \): \( \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies x = 0, y = -z \implies X_1 = \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix} \).

For \( \lambda = 1 \): \( \begin{pmatrix} 0 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \implies y = z \). \( x \) is a free variable.
Let \( x = 1, z = 0 \implies y = 0 \implies X_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \).
Let \( x = 0, z = 1 \implies y = 1 \implies X_3 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \).
Step 4: Normalize Eigenvectors to Form Matrix N
Orthogonal Matrix \( N = [e_2 \quad e_3 \quad e_1] = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1/\sqrt{2} & -1/\sqrt{2} \\ 0 & 1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix} \).
Step 5: Write the Canonical Form and Determine Nature
The canonical form is \( 1 \cdot y_1^2 + 1 \cdot y_2^2 + (-1) \cdot y_3^2 \).
Canonical Form: \( y_1^2 + y_2^2 - y_3^2 \).
Nature: Indefinite.
Rank: 3.
Index: 2.
Signature: 2 - 1 = 1.
Paper 3 - Question 1 (Marks: 8)
Find the Eigen values and Eigen vectors of the matrix \( A = \begin{pmatrix} 1 & 0 & -1 \\ 1 & 2 & 1 \\ 2 & 2 & 3 \end{pmatrix} \).
Click to Reveal Answer & Solution
Correct Answer: Eigenvalues are \( \lambda = 1, 2, 3 \). Eigenvectors are \( X_1 = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}, X_2 = \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}, X_3 = \begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix} \).
Detailed Solution:
Step 1: Find the Characteristic Equation
\( \lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0 \)
\( S_1 = \text{Trace}(A) = 1 + 2 + 3 = 6 \)
\( S_2 = \begin{vmatrix} 2 & 1 \\ 2 & 3 \end{vmatrix} + \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} + \begin{vmatrix} 1 & 0 \\ 1 & 2 \end{vmatrix} = (6 - 2) + (3 - (-2)) + (2 - 0) = 4 + 5 + 2 = 11 \)
\( S_3 = |A| = 1(6 - 2) - 0 + (-1)(2 - 4) = 4 + 2 = 6 \)
Characteristic Equation: \( \lambda^3 - 6\lambda^2 + 11\lambda - 6 = 0 \).
Step 2: Find the Eigenvalues
By trial, if \( \lambda = 1 \): \( 1 - 6 + 11 - 6 = 0 \), so 1 is a root.
Factoring gives \( (\lambda - 1)(\lambda^2 - 5\lambda + 6) = 0 \implies (\lambda - 1)(\lambda - 2)(\lambda - 3) = 0 \).
Eigenvalues: \( \lambda = 1, 2, 3 \).
Step 3: Find Eigenvector for \( \lambda = 1 \)
\( (A - I)X = 0 \implies \begin{pmatrix} 0 & 0 & -1 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
From R1: \( -z = 0 \implies z = 0 \).
From R2: \( x + y + z = 0 \implies x + y = 0 \implies x = -y \).
Let \( y = -1 \implies x = 1 \). So \( X_1 = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \).
Step 4: Find Eigenvector for \( \lambda = 2 \)
\( (A - 2I)X = 0 \implies \begin{pmatrix} -1 & 0 & -1 \\ 1 & 0 & 1 \\ 2 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
From R1: \( -x - z = 0 \implies x = -z \).
From R3: \( 2x + 2y + z = 0 \implies 2(-z) + 2y + z = 0 \implies 2y - z = 0 \implies 2y = z \).
Let \( y = 1 \implies z = 2 \implies x = -2 \). So \( X_2 = \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix} \).
Step 5: Find Eigenvector for \( \lambda = 3 \)
\( (A - 3I)X = 0 \implies \begin{pmatrix} -2 & 0 & -1 \\ 1 & -1 & 1 \\ 2 & 2 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
From R1: \( -2x - z = 0 \implies z = -2x \).
From R3: \( 2x + 2y = 0 \implies y = -x \).
Let \( x = 1 \implies y = -1 \implies z = -2 \). So \( X_3 = \begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix} \).
Paper 3 - Question 2(i) (Marks: 2)
Find the nature of the quadratic form \( x_1^2 + 2x_2^2 \) without reducing to canonical form.
Click to Reveal Answer & Solution
Correct Answer: The quadratic form is Positive Definite.
Detailed Solution:
Step 1: Write the symmetric matrix A
The variables are \( x_1, x_2 \). There are no cross terms, so the matrix is diagonal.
\( A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \).
Step 2: Find the principal minors
\( D_1 = |1| = 1 \quad (> 0) \)
\( D_2 = |A| = \begin{vmatrix} 1 & 0 \\ 0 & 2 \end{vmatrix} = 2 - 0 = 2 \quad (> 0) \)
Step 3: Determine the nature
Since \( D_1 > 0 \) and \( D_2 > 0 \), all principal minors are positive.
Therefore, the quadratic form is Positive Definite.
Paper 3 - Question 2(ii) (Marks: 2)
If 2 and 3 are the Eigen values of \( A = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 2 & 0 \\ a & 0 & 2 \end{pmatrix} \), Find the value of a.
Click to Reveal Answer & Solution
Correct Answer: \( a = 1 \).
Detailed Solution:
Step 1: Use the property of the Trace
Sum of eigenvalues = Trace of matrix A.
Trace(A) = Sum of diagonal elements = \( 2 + 2 + 2 = 6 \).
Let the third eigenvalue be \( \lambda_3 \).
\( 2 + 3 + \lambda_3 = 6 \implies \lambda_3 = 1 \).
The eigenvalues are \( 1, 2, 3 \).
Step 2: Use the property of the Determinant
Product of eigenvalues = Determinant of A.
Product = \( 1 \cdot 2 \cdot 3 = 6 \).
Determinant \( |A| = 2(4 - 0) - 0 + 1(0 - 2a) = 8 - 2a \).
Equate the two: \( 8 - 2a = 6 \implies 2a = 2 \implies a = 1 \).
Paper 3 - Question 3 (Marks: 8)
Diagonalize the matrix \( A = \begin{pmatrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{pmatrix} \) by an orthogonal transformation, given the Eigen values are 0, 3 and 15.
Click to Reveal Answer & Solution
Correct Answer: The orthogonal matrix is \( N = \begin{pmatrix} 1/3 & 2/3 & 2/3 \\ 2/3 & 1/3 & -2/3 \\ 2/3 & -2/3 & 1/3 \end{pmatrix} \) and the diagonal matrix is \( D = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 15 \end{pmatrix} \).
Detailed Solution:
Step 1: Find Eigenvector for \( \lambda = 0 \)
\( (A - 0I)X = 0 \implies \begin{pmatrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
Take R1 and R2: \( 8x - 6y + 2z = 0 \) and \( -6x + 7y - 4z = 0 \).
Cross-multiplication:
\( \frac{x}{(-6)(-4) - (2)(7)} = \frac{y}{(2)(-6) - (8)(-4)} = \frac{z}{(8)(7) - (-6)(-6)} \)
\( \frac{x}{24 - 14} = \frac{y}{-12 + 32} = \frac{z}{56 - 36} \implies \frac{x}{10} = \frac{y}{20} = \frac{z}{20} \implies \frac{x}{1} = \frac{y}{2} = \frac{z}{2} \).
\( X_1 = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} \).
Step 2: Find Eigenvector for \( \lambda = 3 \)
\( (A - 3I)X = 0 \implies \begin{pmatrix} 5 & -6 & 2 \\ -6 & 4 & -4 \\ 2 & -4 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
Take R1 and R2: \( 5x - 6y + 2z = 0 \) and \( -6x + 4y - 4z = 0 \).
Cross-multiplication:
\( \frac{x}{24 - 8} = \frac{y}{-12 + 20} = \frac{z}{20 - 36} \implies \frac{x}{16} = \frac{y}{8} = \frac{z}{-16} \implies \frac{x}{2} = \frac{y}{1} = \frac{z}{-2} \).
\( X_2 = \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix} \).
Step 3: Find Eigenvector for \( \lambda = 15 \)
\( (A - 15I)X = 0 \implies \begin{pmatrix} -7 & -6 & 2 \\ -6 & -8 & -4 \\ 2 & -4 & -12 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \)
Take R1 and R2: \( -7x - 6y + 2z = 0 \) and \( -6x - 8y - 4z = 0 \).
Cross-multiplication:
\( \frac{x}{24 + 16} = \frac{y}{-12 - 28} = \frac{z}{56 - 36} \implies \frac{x}{40} = \frac{y}{-40} = \frac{z}{20} \implies \frac{x}{2} = \frac{y}{-2} = \frac{z}{1} \).
\( X_3 = \begin{pmatrix} 2 \\ -2 \\ 1 \end{pmatrix} \).
Step 4: Normalize vectors and form N
Length of each vector is \( \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3 \).
\( e_1 = \begin{pmatrix} 1/3 \\ 2/3 \\ 2/3 \end{pmatrix}, e_2 = \begin{pmatrix} 2/3 \\ 1/3 \\ -2/3 \end{pmatrix}, e_3 = \begin{pmatrix} 2/3 \\ -2/3 \\ 1/3 \end{pmatrix} \).
\( N = [e_1 \quad e_2 \quad e_3] = \begin{pmatrix} 1/3 & 2/3 & 2/3 \\ 2/3 & 1/3 & -2/3 \\ 2/3 & -2/3 & 1/3 \end{pmatrix} \).
\( D = N^T A N = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 15 \end{pmatrix} \).
Paper 3 - Question 4 (Marks: 14)
Verify Cayley Hamilton theorem and hence find \( A^{-1} \) for \( A = \begin{pmatrix} 2 & -1 & 2 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix} \).
Click to Reveal Answer & Solution
Correct Answer: The Cayley Hamilton theorem is verified. \( A^{-1} = \frac{1}{3} \begin{pmatrix} 3 & 0 & -3 \\ 1 & 2 & 0 \\ -1 & 1 & 3 \end{pmatrix} \).
Detailed Solution:
Step 1: Find the Characteristic Equation
\( \lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0 \)
\( S_1 = \text{Trace}(A) = 2 + 2 + 2 = 6 \).
\( S_2 = (4 - 1) + (4 - 2) + (4 - 1) = 3 + 2 + 3 = 8 \).
\( S_3 = |A| = 2(4 - 1) - (-1)(-2 + 1) + 2(1 - 2) = 6 - 1 - 2 = 3 \).
Equation: \( \lambda^3 - 6\lambda^2 + 8\lambda - 3 = 0 \).
Step 2: State Cayley Hamilton Theorem
\( A^3 - 6A^2 + 8A - 3I = 0 \).
Step 3: Compute \( A^2 \) and \( A^3 \)
\( A^2 = A \cdot A = \begin{pmatrix} 2 & -1 & 2 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix} \begin{pmatrix} 2 & -1 & 2 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix} = \begin{pmatrix} 7 & -6 & 9 \\ -5 & 6 & -6 \\ 5 & -5 & 7 \end{pmatrix} \)

\( A^3 = A^2 \cdot A = \begin{pmatrix} 7 & -6 & 9 \\ -5 & 6 & -6 \\ 5 & -5 & 7 \end{pmatrix} \begin{pmatrix} 2 & -1 & 2 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix} = \begin{pmatrix} 29 & -28 & 38 \\ -22 & 23 & -28 \\ 22 & -22 & 29 \end{pmatrix} \)
Step 4: Verify the Theorem
\( A^3 - 6A^2 + 8A - 3I = \)
\( \begin{pmatrix} 29 & -28 & 38 \\ -22 & 23 & -28 \\ 22 & -22 & 29 \end{pmatrix} - \begin{pmatrix} 42 & -36 & 54 \\ -30 & 36 & -36 \\ 30 & -30 & 42 \end{pmatrix} + \begin{pmatrix} 16 & -8 & 16 \\ -8 & 16 & -8 \\ 8 & -8 & 16 \end{pmatrix} - \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix} \)
\( = \begin{pmatrix} 29-42+16-3 & -28+36-8 & 38-54+16 \\ -22+30-8 & 23-36+16-3 & -28+36-8 \\ 22-30+8 & -22+30-8 & 29-42+16-3 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \). Verified.
Step 5: Find \( A^{-1} \)
Multiply the equation by \( A^{-1} \): \( A^2 - 6A + 8I - 3A^{-1} = 0 \implies 3A^{-1} = A^2 - 6A + 8I \).
\( 3A^{-1} = \begin{pmatrix} 7 & -6 & 9 \\ -5 & 6 & -6 \\ 5 & -5 & 7 \end{pmatrix} - \begin{pmatrix} 12 & -6 & 12 \\ -6 & 12 & -6 \\ 6 & -6 & 12 \end{pmatrix} + \begin{pmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{pmatrix} \)
\( 3A^{-1} = \begin{pmatrix} 3 & 0 & -3 \\ 1 & 2 & 0 \\ -1 & 1 & 3 \end{pmatrix} \implies A^{-1} = \frac{1}{3} \begin{pmatrix} 3 & 0 & -3 \\ 1 & 2 & 0 \\ -1 & 1 & 3 \end{pmatrix} \)
Paper 3 - Question 5 (Marks: 14)
Reduce the quadratic form \( x^2 + y^2 + z^2 - 2xy - 2yz - 2zx \) to canonical form through an orthogonal transformation and hence find its nature, Rank, Index and Signature.
Click to Reveal Answer & Solution
Correct Answer: Canonical form is \( 2y_1^2 + 2y_2^2 - y_3^2 \).
Nature: Indefinite, Rank: 3, Index: 2, Signature: 1.
Detailed Solution:
Step 1: Write the symmetric matrix A
\( A = \begin{pmatrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1 \end{pmatrix} \)
Step 2: Find the Characteristic Equation and Eigenvalues
\( \lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0 \)
\( S_1 = 3 \).
\( S_2 = (1 - 1) + (1 - 1) + (1 - 1) = 0 \).
\( S_3 = |A| = 1(1 - 1) - (-1)(-1 - 1) + (-1)(1 + 1) = 0 - 2 - 2 = -4 \).
Char Eq: \( \lambda^3 - 3\lambda^2 + 4 = 0 \).
Roots: \( \lambda = 2 \) works. Factoring gives \( (\lambda - 2)(\lambda^2 - \lambda - 2) = 0 \implies (\lambda - 2)(\lambda - 2)(\lambda + 1) = 0 \).
Eigenvalues are \( \lambda = -1, 2, 2 \).
Step 3: Find the Eigenvectors
For \( \lambda = -1 \): \( \begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = 0 \). Solving gives \( x = y = z \implies X_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \).

For \( \lambda = 2 \): \( \begin{pmatrix} -1 & -1 & -1 \\ -1 & -1 & -1 \\ -1 & -1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = 0 \). This gives \( x + y + z = 0 \).
Let \( y = -1, z = 0 \implies x = 1 \implies X_2 = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \).
For \( X_3 \), we need a vector satisfying \( x + y + z = 0 \) that is orthogonal to \( X_2 \) (so \( x - y = 0 \implies x = y \)).
Substitute \( x = y \): \( 2x + z = 0 \implies z = -2x \). Let \( x = 1 \implies X_3 = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} \).
Step 4: Normalize Eigenvectors to Form Matrix N
\( e_1 = \begin{pmatrix} 1/\sqrt{3} \\ 1/\sqrt{3} \\ 1/\sqrt{3} \end{pmatrix} \), \( e_2 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} \), \( e_3 = \begin{pmatrix} 1/\sqrt{6} \\ 1/\sqrt{6} \\ -2/\sqrt{6} \end{pmatrix} \).
\( N = \begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{6} & 1/\sqrt{3} \\ -1/\sqrt{2} & 1/\sqrt{6} & 1/\sqrt{3} \\ 0 & -2/\sqrt{6} & 1/\sqrt{3} \end{pmatrix} \)
Step 5: Write the Canonical Form and Determine Nature
The canonical form is \( \lambda_2 y_1^2 + \lambda_3 y_2^2 + \lambda_1 y_3^2 \).
Canonical Form: \( 2y_1^2 + 2y_2^2 - y_3^2 \).
Nature: Indefinite (both + and - signs present).
Rank: 3 (number of non-zero terms).
Index: 2 (number of positive terms).
Signature: Index - Negative terms = 2 - 1 = 1.
Paper 4 - Question 1 (Marks: 1)
If \( x = u - 2v \), \( y = 2u - v \), then \( \frac{\partial(x,y)}{\partial(u,v)} = \) ?
Click to Reveal Answer & Solution
Correct Answer: (A) 3
Detailed Solution:
Step 1: Write down the definition of the Jacobian
The Jacobian is defined as the determinant of the partial derivatives:
\( J = \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \)
Step 2: Calculate the partial derivatives
Given \( x = u - 2v \):
\( \frac{\partial x}{\partial u} = 1 \) and \( \frac{\partial x}{\partial v} = -2 \)
Given \( y = 2u - v \):
\( \frac{\partial y}{\partial u} = 2 \) and \( \frac{\partial y}{\partial v} = -1 \)
Step 3: Evaluate the determinant
\( J = \begin{vmatrix} 1 & -2 \ 2 & -1 \end{vmatrix} \)
\( J = (1)(-1) - (-2)(2) \)
\( J = -1 - (-4) = -1 + 4 = 3 \)
Paper 4 - Question 2 (Marks: 1)
If \( f(x,y) = 0 \) is an implicit function, then \( \frac{dy}{dx} = \) ?
Click to Reveal Answer & Solution
Correct Answer: (B) \( -\frac{\partial f/\partial x}{\partial f/\partial y} \)
Detailed Solution:
Step 1: Differentiate the implicit function with respect to \(x\)
Given \( f(x,y) = 0 \), we apply the chain rule considering \(y\) as a function of \(x\).
\( \frac{d}{dx} [f(x,y)] = \frac{d}{dx} [0] \)
\( \frac{\partial f}{\partial x} \cdot \frac{dx}{dx} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dx} = 0 \)
Step 2: Solve for \( \frac{dy}{dx} \)
\( \frac{\partial f}{\partial x} \cdot (1) + \frac{\partial f}{\partial y} \cdot \frac{dy}{dx} = 0 \)
\( \frac{\partial f}{\partial y} \frac{dy}{dx} = -\frac{\partial f}{\partial x} \)
\( \frac{dy}{dx} = -\frac{\partial f/\partial x}{\partial f/\partial y} \) (assuming \( \frac{\partial f}{\partial y} \neq 0 \)).
Paper 4 - Question 3 (Marks: 1)
The particular integral (PI) of \( (D^2 + 9)y = e^{3x} \) is:
Click to Reveal Answer & Solution
Correct Answer: (C) \( \frac{1}{18} e^{3x} \)
Detailed Solution:
Step 1: Write the formula for PI
The Particular Integral for \( f(D)y = e^{ax} \) is given by \( PI = \frac{1}{f(D)} e^{ax} \).
Step 2: Substitute \( D = a \)
Here, \( f(D) = D^2 + 9 \) and \( a = 3 \).
Substitute \( D = 3 \):
\( f(3) = 3^2 + 9 = 9 + 9 = 18 \).
Step 3: Write the final PI
Since \( f(3) \neq 0 \), the PI is:
\( PI = \frac{1}{18} e^{3x} \).
Paper 4 - Question 4 (Marks: 1)
The roots of the auxiliary equation (A.E) for \( (D^2 + 4)y = 0 \) are:
Click to Reveal Answer & Solution
Correct Answer: (D) \( \pm 2i \)
Detailed Solution:
Step 1: Form the Auxiliary Equation
Given the differential equation \( (D^2 + 4)y = 0 \), replace \( D \) with \( m \).
The Auxiliary Equation (A.E) is: \( m^2 + 4 = 0 \).
Step 2: Solve for \(m\)
\( m^2 = -4 \)
\( m = \pm \sqrt{-4} \)
\( m = \pm 2i \)
Paper 4 - Question 5 (Marks: 8)
Expand \( f(x,y) = e^x \cos y \) as a Taylor's series about the point \( x = 1 \) & \( y = \pi/4 \) up to second degree.
Click to Reveal Answer & Solution
Correct Answer:
\( f(x,y) \approx \frac{e}{\sqrt{2}} \left[ 1 + (x-1) - (y-\frac{\pi}{4}) + \frac{1}{2}(x-1)^2 - (x-1)(y-\frac{\pi}{4}) - \frac{1}{2}(y-\frac{\pi}{4})^2 ight] \)
Detailed Solution:
Step 1: State Taylor's Series Formula
Taylor series for \( f(x,y) \) about \( (a,b) \) up to the second degree is:
\( f(x,y) = f(a,b) + [(x-a)f_x(a,b) + (y-b)f_y(a,b)] + \frac{1}{2!}[(x-a)^2f_{xx}(a,b) + 2(x-a)(y-b)f_{xy}(a,b) + (y-b)^2f_{yy}(a,b)] + \dots \)
Step 2: Calculate the function and derivatives at \( (1, \pi/4) \)
\( f(x,y) = e^x \cos y \implies f(1, \pi/4) = e^1 \cos(\pi/4) = \frac{e}{\sqrt{2}} \)

First-order partial derivatives:
\( f_x(x,y) = e^x \cos y \implies f_x(1, \pi/4) = e^1 \cos(\pi/4) = \frac{e}{\sqrt{2}} \)
\( f_y(x,y) = -e^x \sin y \implies f_y(1, \pi/4) = -e^1 \sin(\pi/4) = -\frac{e}{\sqrt{2}} \)

Second-order partial derivatives:
\( f_{xx}(x,y) = e^x \cos y \implies f_{xx}(1, \pi/4) = \frac{e}{\sqrt{2}} \)
\( f_{xy}(x,y) = -e^x \sin y \implies f_{xy}(1, \pi/4) = -\frac{e}{\sqrt{2}} \)
\( f_{yy}(x,y) = -e^x \cos y \implies f_{yy}(1, \pi/4) = -\frac{e}{\sqrt{2}} \)
Step 3: Substitute values into the Taylor's series formula
\( f(x,y) = \frac{e}{\sqrt{2}} + \left[ (x-1)\left(\frac{e}{\sqrt{2}} ight) + (y-\frac{\pi}{4})\left(-\frac{e}{\sqrt{2}} ight) ight] + \frac{1}{2!} \left[ (x-1)^2\left(\frac{e}{\sqrt{2}} ight) + 2(x-1)(y-\frac{\pi}{4})\left(-\frac{e}{\sqrt{2}} ight) + (y-\frac{\pi}{4})^2\left(-\frac{e}{\sqrt{2}} ight) ight] \)
Step 4: Simplify the expression
Factoring out \( \frac{e}{\sqrt{2}} \) from all terms:
\( f(x,y) = \frac{e}{\sqrt{2}} \left[ 1 + (x-1) - (y-\frac{\pi}{4}) + \frac{1}{2}(x-1)^2 - (x-1)(y-\frac{\pi}{4}) - \frac{1}{2}(y-\frac{\pi}{4})^2 ight] \)
Paper 4 - Question 6 (Marks: 8)
Solve \( (D^2 - 4D - 5)y = e^{-x} + \cos 2x \).
Click to Reveal Answer & Solution
Correct Answer:
\( y = C_1 e^{5x} + C_2 e^{-x} - \frac{x}{6}e^{-x} - \frac{8}{145}\sin 2x - \frac{9}{145}\cos 2x \)
Detailed Solution:
Step 1: Find the Complementary Function (CF)
The Auxiliary Equation is \( m^2 - 4m - 5 = 0 \).
Factoring gives \( (m-5)(m+1) = 0 \implies m = 5, -1 \).
Since roots are real and distinct, \( CF = C_1 e^{5x} + C_2 e^{-x} \).
Step 2: Find Particular Integral 1 (for \( e^{-x} \))
\( PI_1 = \frac{1}{D^2 - 4D - 5} e^{-x} \).
Substitute \( D = -1 \): Denominator becomes \( (-1)^2 - 4(-1) - 5 = 1 + 4 - 5 = 0 \) (Case of Failure).
Multiply numerator by \( x \) and differentiate denominator with respect to \( D \):
\( PI_1 = x \cdot \frac{1}{2D - 4} e^{-x} \).
Substitute \( D = -1 \) again:
\( PI_1 = x \cdot \frac{1}{2(-1) - 4} e^{-x} = -\frac{x}{6} e^{-x} \).
Step 3: Find Particular Integral 2 (for \( \cos 2x \))
\( PI_2 = \frac{1}{D^2 - 4D - 5} \cos 2x \).
Replace \( D^2 \) with \( -a^2 = -(2)^2 = -4 \):
\( PI_2 = \frac{1}{-4 - 4D - 5} \cos 2x = \frac{1}{-4D - 9} \cos 2x \).
Multiply numerator and denominator by \( -4D + 9 \):
\( PI_2 = \frac{-4D + 9}{16D^2 - 81} \cos 2x \).
Replace \( D^2 \) with \( -4 \) again:
Denominator \( = 16(-4) - 81 = -64 - 81 = -145 \).
Numerator \( = -4D(\cos 2x) + 9\cos 2x = -4(-2\sin 2x) + 9\cos 2x = 8\sin 2x + 9\cos 2x \).
\( PI_2 = -\frac{1}{145}(8\sin 2x + 9\cos 2x) \).
Step 4: Write the complete solution
\( y = CF + PI_1 + PI_2 \)
\( y = C_1 e^{5x} + C_2 e^{-x} - \frac{x}{6}e^{-x} - \frac{1}{145}(8\sin 2x + 9\cos 2x) \).
Paper 4 - Question 7 (Marks: 8)
Find the extremum values (max & min) for the function \( f(x,y) = x^3 + y^2 - 6x + 12 \).
Click to Reveal Answer & Solution
Correct Answer:
The function has a minimum value of \( 12 - 4\sqrt{2} \) at the point \( (\sqrt{2}, 0) \).
The point \( (-\sqrt{2}, 0) \) is a saddle point.
Detailed Solution:
Step 1: Find the first-order partial derivatives and critical points
Given \( f(x,y) = x^3 + y^2 - 6x + 12 \).
\( f_x = \frac{\partial f}{\partial x} = 3x^2 - 6 \)
\( f_y = \frac{\partial f}{\partial y} = 2y \)
For critical points, set \( f_x = 0 \) and \( f_y = 0 \):
\( 3x^2 - 6 = 0 \implies x^2 = 2 \implies x = \pm\sqrt{2} \)
\( 2y = 0 \implies y = 0 \)
The critical points are \( (\sqrt{2}, 0) \) and \( (-\sqrt{2}, 0) \).
Step 2: Find second-order partial derivatives
\( r = f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x \)
\( s = f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \)
\( t = f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2 \)
Step 3: Analyze each critical point using \( \Delta = rt - s^2 \)
We have \( \Delta = rt - s^2 = (6x)(2) - 0 = 12x \).

Case 1: At point \( (\sqrt{2}, 0) \)
\( \Delta = 12(\sqrt{2}) > 0 \).
Since \( \Delta > 0 \) and \( r = 6\sqrt{2} > 0 \), the function has a local minimum at this point.
Minimum value \( = f(\sqrt{2}, 0) = (\sqrt{2})^3 + 0^2 - 6(\sqrt{2}) + 12 = 2\sqrt{2} - 6\sqrt{2} + 12 = 12 - 4\sqrt{2} \).

Case 2: At point \( (-\sqrt{2}, 0) \)
\( \Delta = 12(-\sqrt{2}) = -12\sqrt{2} < 0 \).
Since \( \Delta < 0 \), this point is a saddle point.
Paper 4 - Question 8 (Marks: 15)
Find the shortest and largest distance of the point \( (1,2,-1) \) from the sphere \( x^2 + y^2 + z^2 = 24 \), using Lagrange method of multiplier.
Click to Reveal Answer & Solution
Correct Answer:
Shortest distance = \( \sqrt{6} \) (at point \( (2, 4, -2) \)).
Largest distance = \( 3\sqrt{6} \) (at point \( (-2, -4, 2) \)).
Detailed Solution:
Step 1: Define the function and constraint
Let the distance squared from point \( (x,y,z) \) on the sphere to \( (1,2,-1) \) be our objective function to optimize to avoid square roots.
Objective function: \( f(x,y,z) = (x-1)^2 + (y-2)^2 + (z+1)^2 \).
Constraint (the sphere): \( g(x,y,z) = x^2 + y^2 + z^2 - 24 = 0 \).
Step 2: Construct the Lagrangian function
\( F(x,y,z,\lambda) = f(x,y,z) + \lambda g(x,y,z) \)
\( F = (x-1)^2 + (y-2)^2 + (z+1)^2 + \lambda(x^2 + y^2 + z^2 - 24) \)
Step 3: Find the partial derivatives and set to zero
\( \frac{\partial F}{\partial x} = 2(x-1) + 2\lambda x = 0 \implies x(1+\lambda) = 1 \implies x = \frac{1}{1+\lambda} \)
\( \frac{\partial F}{\partial y} = 2(y-2) + 2\lambda y = 0 \implies y(1+\lambda) = 2 \implies y = \frac{2}{1+\lambda} \)
\( \frac{\partial F}{\partial z} = 2(z+1) + 2\lambda z = 0 \implies z(1+\lambda) = -1 \implies z = \frac{-1}{1+\lambda} \)
Step 4: Substitute into the constraint to find \( \lambda \)
Substitute \( x, y, z \) into \( x^2 + y^2 + z^2 = 24 \):
\( \left(\frac{1}{1+\lambda} ight)^2 + \left(\frac{2}{1+\lambda} ight)^2 + \left(\frac{-1}{1+\lambda} ight)^2 = 24 \)
\( \frac{1 + 4 + 1}{(1+\lambda)^2} = 24 \)
\( \frac{6}{(1+\lambda)^2} = 24 \implies (1+\lambda)^2 = \frac{6}{24} = \frac{1}{4} \)
\( 1+\lambda = \pm \frac{1}{2} \)
Step 5: Find points and calculate distances
Case 1: \( 1+\lambda = \frac{1}{2} \)
\( x = \frac{1}{1/2} = 2 \), \( y = \frac{2}{1/2} = 4 \), \( z = \frac{-1}{1/2} = -2 \).
Point is \( (2, 4, -2) \).
Distance squared: \( f(2,4,-2) = (2-1)^2 + (4-2)^2 + (-2+1)^2 = 1 + 4 + 1 = 6 \).
Distance = \( \sqrt{6} \) (This is the Shortest Distance).

Case 2: \( 1+\lambda = -\frac{1}{2} \)
\( x = \frac{1}{-1/2} = -2 \), \( y = \frac{2}{-1/2} = -4 \), \( z = \frac{-1}{-1/2} = 2 \).
Point is \( (-2, -4, 2) \).
Distance squared: \( f(-2,-4,2) = (-2-1)^2 + (-4-2)^2 + (2+1)^2 = 9 + 36 + 9 = 54 \).
Distance = \( \sqrt{54} = 3\sqrt{6} \) (This is the Largest Distance).
Paper 4 - Question 9 (i) (Marks: 10)
Solve \( (D^2 + 4D + 8)y = x - 3 \).
Click to Reveal Answer & Solution
Correct Answer:
\( y = e^{-2x}(C_1 \cos 2x + C_2 \sin 2x) + \frac{2x - 7}{16} \)
Detailed Solution:
Step 1: Find the Complementary Function (CF)
Auxiliary Equation: \( m^2 + 4m + 8 = 0 \).
Use quadratic formula: \( m = \frac{-4 \pm \sqrt{16 - 32}}{2} = \frac{-4 \pm \sqrt{-16}}{2} = \frac{-4 \pm 4i}{2} = -2 \pm 2i \).
Since roots are complex (\( \alpha \pm i\beta \)), the CF is:
\( CF = e^{-2x}(C_1 \cos 2x + C_2 \sin 2x) \).
Step 2: Find the Particular Integral (PI)
\( PI = \frac{1}{D^2 + 4D + 8} (x - 3) \).
Since the RHS is a polynomial, we expand the operator in ascending powers of D using Binomial expansion.
Factor out the constant 8 from the denominator:
\( PI = \frac{1}{8\left[1 + \left(\frac{D^2 + 4D}{8} ight) ight]} (x - 3) \)
\( PI = \frac{1}{8} \left[ 1 + \left(\frac{D^2 + 4D}{8} ight) ight]^{-1} (x - 3) \)
Step 3: Apply Binomial Expansion
Using \( (1 + z)^{-1} \approx 1 - z + z^2 - \dots \), we only need terms up to \( D^1 \) since the highest power of \( x \) is 1 (higher derivatives will be zero).
\( PI = \frac{1}{8} \left[ 1 - \frac{D^2 + 4D}{8} + \dots ight] (x - 3) \)
\( PI = \frac{1}{8} \left[ (x - 3) - \frac{4D(x - 3)}{8} ight] \)
Note that \( D(x - 3) = 1 \).
\( PI = \frac{1}{8} \left[ x - 3 - \frac{4(1)}{8} ight] = \frac{1}{8} \left[ x - 3 - \frac{1}{2} ight] \)
\( PI = \frac{1}{8} \left( x - \frac{7}{2} ight) = \frac{2x - 7}{16} \).
Step 4: Write the General Solution
\( y = CF + PI \)
\( y = e^{-2x}(C_1 \cos 2x + C_2 \sin 2x) + \frac{2x - 7}{16} \).
Paper 4 - Question 9 (ii) (Marks: 5)
Using method of variation of parameter solve \( (D^2 + a^2)y = \tan ax \).
Click to Reveal Answer & Solution
Correct Answer:
\( y = C_1 \cos ax + C_2 \sin ax - \frac{\cos ax}{a^2} \ln|\sec ax + \tan ax| \)
Detailed Solution:
Step 1: Find the Complementary Function (CF)
A.E: \( m^2 + a^2 = 0 \implies m = \pm ai \).
\( CF = C_1 \cos ax + C_2 \sin ax \).
Let \( y_1 = \cos ax \) and \( y_2 = \sin ax \).
RHS function: \( X = \tan ax \).
Step 2: Find the Wronskian (\( W \))
\( W = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} \cos ax & \sin ax \ -a\sin ax & a\cos ax \end{vmatrix} \)
\( W = (\cos ax)(a\cos ax) - (\sin ax)(-a\sin ax) = a\cos^2 ax + a\sin^2 ax = a(\cos^2 ax + \sin^2 ax) = a \).
Step 3: Apply the Variation of Parameters formula for PI
\( PI = -y_1 \int \frac{y_2 X}{W} dx + y_2 \int \frac{y_1 X}{W} dx \)

First Integral:
\( \int \frac{y_2 X}{W} dx = \int \frac{\sin ax \tan ax}{a} dx = \frac{1}{a} \int \frac{\sin^2 ax}{\cos ax} dx \)
Using \( \sin^2 ax = 1 - \cos^2 ax \):
\( \frac{1}{a} \int \frac{1 - \cos^2 ax}{\cos ax} dx = \frac{1}{a} \int (\sec ax - \cos ax) dx \)
\( = \frac{1}{a} \left[ \frac{1}{a}\ln|\sec ax + \tan ax| - \frac{1}{a}\sin ax ight] = \frac{1}{a^2} \ln|\sec ax + \tan ax| - \frac{1}{a^2} \sin ax \).

Second Integral:
\( \int \frac{y_1 X}{W} dx = \int \frac{\cos ax \tan ax}{a} dx = \frac{1}{a} \int \sin ax dx = \frac{1}{a} \left( -\frac{1}{a}\cos ax ight) = -\frac{1}{a^2} \cos ax \).
Step 4: Substitute integrals back into PI
\( PI = -\cos ax \left( \frac{1}{a^2} \ln|\sec ax + \tan ax| - \frac{1}{a^2} \sin ax ight) + \sin ax \left( -\frac{1}{a^2} \cos ax ight) \)
\( PI = -\frac{\cos ax}{a^2} \ln|\sec ax + \tan ax| + \frac{\sin ax \cos ax}{a^2} - \frac{\sin ax \cos ax}{a^2} \)
\( PI = -\frac{\cos ax}{a^2} \ln|\sec ax + \tan ax| \).
Step 5: Write the complete solution
\( y = CF + PI \)
\( y = C_1 \cos ax + C_2 \sin ax - \frac{\cos ax}{a^2} \ln|\sec ax + \tan ax| \).
Paper 4 - Question 10 (i) (Marks: 10)
Solve \( x^2 y'' - 3xy' - 5y = \cos(\log x) \).
Click to Reveal Answer & Solution
Correct Answer:
\( y = C_1 x^5 + \frac{C_2}{x} - \frac{1}{26} [ 2\sin(\log x) + 3\cos(\log x) ] \)
Detailed Solution:
Step 1: Convert Cauchy-Euler equation to Constant Coefficients
This is a Cauchy-Euler Differential Equation.
Let \( x = e^z \implies z = \log x \). Let \( \theta = \frac{d}{dz} \).
Then \( x\frac{dy}{dx} = \theta y \) and \( x^2\frac{d^2y}{dx^2} = \theta(\theta - 1)y \).
Substituting these into the given equation:
\( [\theta(\theta - 1) - 3\theta - 5]y = \cos z \)
\( (\theta^2 - \theta - 3\theta - 5)y = \cos z \)
\( (\theta^2 - 4\theta - 5)y = \cos z \)
Step 2: Find the Complementary Function (CF)
A.E: \( m^2 - 4m - 5 = 0 \)
\( (m-5)(m+1) = 0 \implies m = 5, -1 \).
\( CF (in z) = C_1 e^{5z} + C_2 e^{-z} \).
Substituting back \( x = e^z \):
\( CF (in x) = C_1 x^5 + C_2 x^{-1} = C_1 x^5 + \frac{C_2}{x} \).
Step 3: Find the Particular Integral (PI)
\( PI = \frac{1}{\theta^2 - 4\theta - 5} \cos z \).
For \( \cos az \), substitute \( \theta^2 = -a^2 = -1^2 = -1 \):
\( PI = \frac{1}{-1 - 4\theta - 5} \cos z = \frac{1}{-4\theta - 6} \cos z \).
Factor out \( -1/2 \):
\( PI = -\frac{1}{2} \cdot \frac{1}{2\theta + 3} \cos z \).
Multiply numerator and denominator by \( 2\theta - 3 \):
\( PI = -\frac{1}{2} \frac{2\theta - 3}{4\theta^2 - 9} \cos z \).
Substitute \( \theta^2 = -1 \) again:
\( PI = -\frac{1}{2} \frac{2\theta(\cos z) - 3\cos z}{4(-1) - 9} = -\frac{1}{2} \frac{-2\sin z - 3\cos z}{-13} = \frac{-2\sin z - 3\cos z}{26} = -\frac{1}{26}(2\sin z + 3\cos z) \).
Substituting back \( z = \log x \):
\( PI = -\frac{1}{26} [2\sin(\log x) + 3\cos(\log x)] \).
Step 4: Write the complete solution
\( y = CF + PI \)
\( y = C_1 x^5 + \frac{C_2}{x} - \frac{1}{26} [ 2\sin(\log x) + 3\cos(\log x) ] \).
Paper 4 - Question 10 (ii) (Marks: 5)
If \( u = (x - y)(y - z)(z - x) \), then show that \( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = 0 \).
Click to Reveal Answer & Solution
Correct Answer: Proved that \( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = 0 \).
Detailed Solution:
Step 1: Expand the function \(u\)
Given \( u = (x - y)(y - z)(z - x) \).
Let's multiply the first two terms:
\( (x - y)(y - z) = xy - xz - y^2 + yz \).
Now multiply by the third term \( (z - x) \):
\( u = (xy - xz - y^2 + yz)(z - x) \)
\( u = xyz - x^2y - xz^2 + x^2z - y^2z + xy^2 + yz^2 - xyz \)
The \(xyz\) terms cancel out:
\( u = -x^2y - xz^2 + x^2z - y^2z + xy^2 + yz^2 \)
Step 2: Find the partial derivatives
Differentiating partially with respect to \(x\) (treating \(y, z\) as constants):
\( \frac{\partial u}{\partial x} = -2xy - z^2 + 2xz + y^2 \)
Differentiating partially with respect to \(y\) (treating \(x, z\) as constants):
\( \frac{\partial u}{\partial y} = -x^2 - 2yz + 2xy + z^2 \)
Differentiating partially with respect to \(z\) (treating \(x, y\) as constants):
\( \frac{\partial u}{\partial z} = -2xz + x^2 - y^2 + 2yz \)
Step 3: Sum the partial derivatives
\( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = (-2xy - z^2 + 2xz + y^2) + (-x^2 - 2yz + 2xy + z^2) + (-2xz + x^2 - y^2 + 2yz) \)
Group like terms together:
\( (-2xy + 2xy) + (-z^2 + z^2) + (2xz - 2xz) + (y^2 - y^2) + (-x^2 + x^2) + (-2yz + 2yz) \)
\( = 0 + 0 + 0 + 0 + 0 + 0 = 0 \)
Hence Proved.

Paper 5: FT-III (13-11-2024)

Course Code: 21MAB101T Course Title: Calculus and Linear Algebra Date: 13.11.2024 Max Marks: 50 Duration: 100 Minutes

PART A (4 × 1 = 4 Marks)

Question 1 (Marks: 1)
If \( f(x,y) = x^3 + y^3 - 3xy \), then \( \frac{\partial f}{\partial x} \) is
  • (A) \( 3y^2 - 3x \)
  • (B) \( 3y^2 - 3y \)
  • (C) \( 3x^2 - 3x \)
  • (D) \( 3x^2 - 3y \) (Correct)
Click to Reveal Answer & Solution
Correct Answer: (D) \( 3x^2 - 3y \)
Detailed Solution:
Step 1: Understand the partial derivative.
To find the partial derivative of \( f(x,y) \) with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), we must differentiate the function with respect to \( x \) while treating \( y \) as a constant.

Step 2: Differentiate each term.
Given \( f(x,y) = x^3 + y^3 - 3xy \):
- Derivative of \( x^3 \) with respect to \( x \) is \( 3x^2 \).
- Derivative of \( y^3 \) with respect to \( x \) is \( 0 \) (since \( y \) is treated as a constant).
- Derivative of \( -3xy \) with respect to \( x \) is \( -3y \cdot \frac{d}{dx}(x) = -3y(1) = -3y \).

Step 3: Combine the results.
\( \frac{\partial f}{\partial x} = 3x^2 + 0 - 3y = 3x^2 - 3y \).
Therefore, the correct option is (D).
Question 2 (Marks: 1)
If \( x e^{-y} = 1 \), then \( \frac{dy}{dx} \) using partial derivatives is
  • (A) \( \frac{1}{x} \) (Correct)
  • (B) \( x \)
  • (C) \( -\frac{1}{x} \)
  • (D) \( -x \)
Click to Reveal Answer & Solution
Correct Answer: (A) \( \frac{1}{x} \)
Detailed Solution:
Step 1: Define the implicit function \( f(x,y) \).
Let \( f(x,y) = x e^{-y} - 1 = 0 \).

Step 2: State the formula for implicit differentiation using partial derivatives.
For an implicit function \( f(x,y) = 0 \), the derivative is given by:
\( \frac{dy}{dx} = -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}} \)

Step 3: Calculate the partial derivatives.
- Differentiate \( f \) with respect to \( x \) (treating \( y \) as constant):
\( \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (x e^{-y} - 1) = e^{-y} \)
- Differentiate \( f \) with respect to \( y \) (treating \( x \) as constant):
\( \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (x e^{-y} - 1) = x(-e^{-y}) = -x e^{-y} \)

Step 4: Substitute into the formula.
\( \frac{dy}{dx} = -\frac{e^{-y}}{-x e^{-y}} \)
Cancel the negative signs and \( e^{-y} \) from numerator and denominator:
\( \frac{dy}{dx} = \frac{1}{x} \).
Therefore, the correct option is (A).
Question 3 (Marks: 1)
The roots of the auxiliary equation of \( (D^2 + 6D + 9)y = 0 \) are
  • (A) \( m = 3, 3 \)
  • (B) \( m_1 = -3, m_2 = 3 \)
  • (C) \( m = -3, -3 \) (Correct)
  • (D) \( m_1 = 3, m_2 = 6 \)
Click to Reveal Answer & Solution
Correct Answer: (C) \( m = -3, -3 \)
Detailed Solution:
Step 1: Write the Auxiliary Equation (AE).
For a linear differential equation \( f(D)y = 0 \), the auxiliary equation is obtained by replacing \( D \) with \( m \) and equating to zero.
The AE is: \( m^2 + 6m + 9 = 0 \).

Step 2: Solve the quadratic equation.
The equation can be factored as a perfect square:
\( (m + 3)^2 = 0 \)
\( m + 3 = 0 \implies m = -3 \)

Step 3: State the roots.
Since it is a repeated root of a quadratic equation, the roots are \( m = -3, -3 \).
Therefore, the correct option is (C).
Question 4 (Marks: 1)
Find the Particular Integral of \( (D^2 + 2D + 1)y = 3 \).
  • (A) 1
  • (B) 2
  • (C) 3 (Correct)
  • (D) 4
Click to Reveal Answer & Solution
Correct Answer: (C) 3
Detailed Solution:
Step 1: Write the expression for the Particular Integral (P.I.).
\( P.I. = \frac{1}{f(D)} X(x) \)
\( P.I. = \frac{1}{D^2 + 2D + 1} (3) \)

Step 2: Use the standard rule for constant terms.
A constant \( k \) can be written as \( k \cdot e^{0x} \). Here, \( 3 = 3e^{0x} \).
The rule states that for \( \frac{1}{f(D)} e^{ax} \), we substitute \( D = a \), provided \( f(a) \neq 0 \).

Step 3: Evaluate by substituting \( D = 0 \).
\( P.I. = 3 \cdot \frac{1}{0^2 + 2(0) + 1} e^{0x} \)
\( P.I. = 3 \cdot \frac{1}{1} \cdot 1 = 3 \).

Therefore, the Particular Integral is 3. The correct option is (C).

PART B (2 × 8 = 16 Marks)

Question 5 (Marks: 8)
Expand \( \tan^{-1}\left(\frac{y}{x}\right) \) using Taylor's series near \( (1,1) \) up to second degree terms.
Click to Reveal Answer & Solution
Correct Answer: \( \frac{\pi}{4} - \frac{1}{2}(x-1) + \frac{1}{2}(y-1) + \frac{1}{4}(x-1)^2 - \frac{1}{4}(y-1)^2 \)
Detailed Solution:
Step 1: State Taylor's series formula for two variables.
The Taylor's series expansion of \( f(x,y) \) around \( (a,b) \) up to the second degree is:
\( f(x,y) \approx f(a,b) + [(x-a)f_x(a,b) + (y-b)f_y(a,b)] + \frac{1}{2!} \left[ (x-a)^2 f_{xx}(a,b) + 2(x-a)(y-b)f_{xy}(a,b) + (y-b)^2 f_{yy}(a,b) \right] \)
Here, \( f(x,y) = \tan^{-1}\left(\frac{y}{x}\right) \) and the point is \( (a,b) = (1,1) \).

Step 2: Evaluate the function at \( (1,1) \).
\( f(1,1) = \tan^{-1}\left(\frac{1}{1}\right) = \tan^{-1}(1) = \frac{\pi}{4} \).

Step 3: Calculate the first-order partial derivatives at \( (1,1) \).
- \( f_x = \frac{\partial f}{\partial x} = \frac{1}{1 + (y/x)^2} \cdot \frac{\partial}{\partial x}\left(\frac{y}{x}\right) = \frac{x^2}{x^2 + y^2} \cdot \left(\frac{-y}{x^2}\right) = \frac{-y}{x^2 + y^2} \).
At \( (1,1) \): \( f_x(1,1) = \frac{-1}{1^2 + 1^2} = -\frac{1}{2} \).

- \( f_y = \frac{\partial f}{\partial y} = \frac{1}{1 + (y/x)^2} \cdot \frac{\partial}{\partial y}\left(\frac{y}{x}\right) = \frac{x^2}{x^2 + y^2} \cdot \left(\frac{1}{x}\right) = \frac{x}{x^2 + y^2} \).
At \( (1,1) \): \( f_y(1,1) = \frac{1}{1^2 + 1^2} = \frac{1}{2} \).

Step 4: Calculate the second-order partial derivatives at \( (1,1) \).
- \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}\left( \frac{-y}{x^2 + y^2} \right) = -y \cdot \left( \frac{-1}{(x^2+y^2)^2} \right) \cdot 2x = \frac{2xy}{(x^2 + y^2)^2} \).
At \( (1,1) \): \( f_{xx}(1,1) = \frac{2(1)(1)}{(1^2 + 1^2)^2} = \frac{2}{4} = \frac{1}{2} \).

- \( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial y}\left( \frac{-y}{x^2 + y^2} \right) \)
Using Quotient Rule: \( \frac{\partial}{\partial y}\left( \frac{-y}{x^2 + y^2} \right) = \frac{-(x^2 + y^2) - (-y)(2y)}{(x^2 + y^2)^2} = \frac{-x^2 - y^2 + 2y^2}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2} \).
At \( (1,1) \): \( f_{xy}(1,1) = \frac{1^2 - 1^2}{(1^2 + 1^2)^2} = \frac{0}{4} = 0 \).

- \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}\left( \frac{x}{x^2 + y^2} \right) = x \cdot \left( \frac{-1}{(x^2+y^2)^2} \right) \cdot 2y = \frac{-2xy}{(x^2 + y^2)^2} \).
At \( (1,1) \): \( f_{yy}(1,1) = \frac{-2(1)(1)}{(1^2 + 1^2)^2} = \frac{-2}{4} = -\frac{1}{2} \).

Step 5: Substitute all values into the Taylor's series expansion.
\( f(x,y) = \frac{\pi}{4} + \left[ (x-1)\left(-\frac{1}{2}\right) + (y-1)\left(\frac{1}{2}\right) \right] + \frac{1}{2!} \left[ (x-1)^2 \left(\frac{1}{2}\right) + 2(x-1)(y-1)(0) + (y-1)^2 \left(-\frac{1}{2}\right) \right] \)

Simplifying:
\( f(x,y) = \frac{\pi}{4} - \frac{1}{2}(x-1) + \frac{1}{2}(y-1) + \frac{1}{2} \left[ \frac{1}{2}(x-1)^2 - \frac{1}{2}(y-1)^2 \right] \)
\( f(x,y) = \frac{\pi}{4} - \frac{1}{2}(x-1) + \frac{1}{2}(y-1) + \frac{1}{4}(x-1)^2 - \frac{1}{4}(y-1)^2 \).
This is the required Taylor's series expansion up to the second degree.
Question 6 (Marks: 8)
Determine the extreme values of the function \( x^3 y^2 (12 - x - y) \) where \( x > 0 \), \( y > 0 \).
Click to Reveal Answer & Solution
Correct Answer: The function has a maximum value of 6912 at the point \( (6,4) \).
Detailed Solution:
Step 1: Define the function and expand it.
Let \( f(x,y) = x^3 y^2 (12 - x - y) = 12x^3y^2 - x^4y^2 - x^3y^3 \).

Step 2: Find the first-order partial derivatives.
- \( p = \frac{\partial f}{\partial x} = 36x^2y^2 - 4x^3y^2 - 3x^2y^3 = x^2y^2(36 - 4x - 3y) \).
- \( q = \frac{\partial f}{\partial y} = 24x^3y - 2x^4y - 3x^3y^2 = x^3y(24 - 2x - 3y) \).

Step 3: Equate \( p = 0 \) and \( q = 0 \) to find stationary points.
For extreme values, we must have:
\( x^2y^2(36 - 4x - 3y) = 0 \)
\( x^3y(24 - 2x - 3y) = 0 \)
Since we are given that \( x > 0 \) and \( y > 0 \), we can divide by \( x^2y^2 \) and \( x^3y \):
\( 36 - 4x - 3y = 0 \implies 4x + 3y = 36 \) --- (Equation 1)
\( 24 - 2x - 3y = 0 \implies 2x + 3y = 24 \) --- (Equation 2)

Step 4: Solve the simultaneous equations.
Subtracting Equation 2 from Equation 1:
\( (4x + 3y) - (2x + 3y) = 36 - 24 \)
\( 2x = 12 \implies x = 6 \).
Substitute \( x = 6 \) into Equation 2:
\( 2(6) + 3y = 24 \implies 12 + 3y = 24 \implies 3y = 12 \implies y = 4 \).
Thus, the only stationary point in the domain is \( (6, 4) \).

Step 5: Calculate the second-order partial derivatives.
- \( r = f_{xx} = \frac{\partial}{\partial x} (36x^2y^2 - 4x^3y^2 - 3x^2y^3) = 72xy^2 - 12x^2y^2 - 6xy^3 = 6xy^2(12 - 2x - y) \).
- \( s = f_{xy} = \frac{\partial}{\partial y} (36x^2y^2 - 4x^3y^2 - 3x^2y^3) = 72x^2y - 8x^3y - 9x^2y^2 = x^2y(72 - 8x - 9y) \).
- \( t = f_{yy} = \frac{\partial}{\partial y} (24x^3y - 2x^4y - 3x^3y^2) = 24x^3 - 2x^4 - 6x^3y = 2x^3(12 - x - 3y) \).

Step 6: Evaluate \( r, s, t \) at the stationary point \( (6, 4) \).
- \( r = 6(6)(4^2)(12 - 2(6) - 4) = 36(16)(12 - 12 - 4) = 576(-4) = -2304 \).
- \( s = (6^2)(4)(72 - 8(6) - 9(4)) = 36(4)(72 - 48 - 36) = 144(72 - 84) = 144(-12) = -1728 \).
- \( t = 2(6^3)(12 - 6 - 3(4)) = 2(216)(12 - 6 - 12) = 432(-6) = -2592 \).

Step 7: Apply the second derivative test.
Calculate \( rt - s^2 \):
\( rt - s^2 = (-2304)(-2592) - (-1728)^2 = 5971968 - 2985984 = 2985984 \).
Since \( rt - s^2 > 0 \) and \( r < 0 \) (since \(-2304 < 0\)), the function has a maximum at \( (6, 4) \).

Step 8: Find the extreme maximum value.
Substitute \( x=6, y=4 \) back into the original function:
\( f_{max} = (6^3)(4^2)(12 - 6 - 4) = (216)(16)(2) = 3456(2) = 6912 \).
Question 7 (Marks: 8)
Solve \( (D^2 + 6D + 8)y = e^{-2x} + \sin x \).
Click to Reveal Answer & Solution
Correct Answer: \( y = c_1 e^{-2x} + c_2 e^{-4x} + \frac{x}{2} e^{-2x} + \frac{7\sin x - 6\cos x}{85} \)
Detailed Solution:
The given differential equation is \( (D^2 + 6D + 8)y = e^{-2x} + \sin x \).
The complete solution is \( y = C.F. + P.I. \).

Step 1: Find the Complementary Function (C.F.).
The Auxiliary Equation (AE) is \( m^2 + 6m + 8 = 0 \).
Factorizing the quadratic equation:
\( m^2 + 4m + 2m + 8 = 0 \implies m(m+4) + 2(m+4) = 0 \implies (m+2)(m+4) = 0 \).
The roots are real and distinct: \( m = -2, -4 \).
Therefore, \( C.F. = c_1 e^{-2x} + c_2 e^{-4x} \), where \( c_1 \) and \( c_2 \) are arbitrary constants.

Step 2: Find the Particular Integral for \( e^{-2x} \) (Let's call it \( P.I._1 \)).
\( P.I._1 = \frac{1}{D^2 + 6D + 8} e^{-2x} \)
According to the rule for \( e^{ax} \), we substitute \( D = a = -2 \).
Denominator = \( (-2)^2 + 6(-2) + 8 = 4 - 12 + 8 = 0 \). This is a case of failure.
Multiply the numerator by \( x \) and differentiate the denominator with respect to \( D \):
\( P.I._1 = x \cdot \frac{1}{2D + 6} e^{-2x} \)
Substitute \( D = -2 \) again:
\( P.I._1 = x \cdot \frac{1}{2(-2) + 6} e^{-2x} = x \cdot \frac{1}{-4 + 6} e^{-2x} = \frac{x}{2} e^{-2x} \).

Step 3: Find the Particular Integral for \( \sin x \) (Let's call it \( P.I._2 \)).
\( P.I._2 = \frac{1}{D^2 + 6D + 8} \sin(1x) \)
For \( \sin(ax) \), the rule is to substitute \( D^2 = -a^2 \). Here, \( a=1 \), so \( D^2 = -1^2 = -1 \).
\( P.I._2 = \frac{1}{-1 + 6D + 8} \sin x = \frac{1}{6D + 7} \sin x \).
To eliminate \( D \) from the denominator, multiply and divide by its conjugate \( 6D - 7 \):
\( P.I._2 = \frac{6D - 7}{(6D + 7)(6D - 7)} \sin x = \frac{6D - 7}{36D^2 - 49} \sin x \).
Again, substitute \( D^2 = -1 \) in the denominator:
\( P.I._2 = \frac{6D - 7}{36(-1) - 49} \sin x = \frac{6D - 7}{-36 - 49} \sin x = \frac{6D - 7}{-85} \sin x \).
Now, operate the numerator on \( \sin x \) (remembering \( D \equiv \frac{d}{dx} \)):
\( P.I._2 = -\frac{1}{85} (6D(\sin x) - 7\sin x) = -\frac{1}{85} (6\cos x - 7\sin x) = \frac{7\sin x - 6\cos x}{85} \).

Step 4: Write the complete solution.
\( y = C.F. + P.I._1 + P.I._2 \)
\( y = c_1 e^{-2x} + c_2 e^{-4x} + \frac{x}{2} e^{-2x} + \frac{7\sin x - 6\cos x}{85} \).

PART C (2 × 15 = 30 Marks)

Question 8 (Marks: 15)
Find the maximum and minimum distance of the point \( (3, 4, 12) \) from the sphere \( x^2 + y^2 + z^2 = 1 \) using Lagrange's method of multipliers.
Click to Reveal Answer & Solution
Correct Answer: Maximum distance = 14, Minimum distance = 12
Detailed Solution:
Step 1: Set up the objective function and constraint.
Let \( P(x, y, z) \) be any point on the sphere \( x^2 + y^2 + z^2 = 1 \).
The distance \( D \) between \( P(x, y, z) \) and the point \( A(3, 4, 12) \) is:
\( D = \sqrt{(x-3)^2 + (y-4)^2 + (z-12)^2} \).
To simplify differentiation, we will extremize the square of the distance. Let the objective function be:
\( f(x, y, z) = D^2 = (x-3)^2 + (y-4)^2 + (z-12)^2 \).
The constraint equation representing the sphere is:
\( \phi(x, y, z) = x^2 + y^2 + z^2 - 1 = 0 \).

Step 2: Construct the Lagrange multiplier equation.
Define the Lagrangian function \( L(x, y, z, \lambda) \):
\( L(x, y, z, \lambda) = f(x, y, z) + \lambda \phi(x, y, z) \)
\( L(x, y, z, \lambda) = (x-3)^2 + (y-4)^2 + (z-12)^2 + \lambda(x^2 + y^2 + z^2 - 1) \).

Step 3: Find the partial derivatives and equate to zero for stationary points.
\( \frac{\partial L}{\partial x} = 2(x-3) + 2\lambda x = 0 \implies x-3 + \lambda x = 0 \implies x(1+\lambda) = 3 \implies x = \frac{3}{1+\lambda} \).

\( \frac{\partial L}{\partial y} = 2(y-4) + 2\lambda y = 0 \implies y-4 + \lambda y = 0 \implies y(1+\lambda) = 4 \implies y = \frac{4}{1+\lambda} \).

\( \frac{\partial L}{\partial z} = 2(z-12) + 2\lambda z = 0 \implies z-12 + \lambda z = 0 \implies z(1+\lambda) = 12 \implies z = \frac{12}{1+\lambda} \).

Step 4: Substitute \( x, y, z \) into the constraint equation to find \( \lambda \).
\( x^2 + y^2 + z^2 = 1 \)
\( \left(\frac{3}{1+\lambda}\right)^2 + \left(\frac{4}{1+\lambda}\right)^2 + \left(\frac{12}{1+\lambda}\right)^2 = 1 \)
\( \frac{9}{(1+\lambda)^2} + \frac{16}{(1+\lambda)^2} + \frac{144}{(1+\lambda)^2} = 1 \)
\( \frac{169}{(1+\lambda)^2} = 1 \implies (1+\lambda)^2 = 169 \)
Taking the square root on both sides:
\( 1+\lambda = \pm 13 \).

Step 5: Determine the points on the sphere.
Case 1: If \( 1+\lambda = 13 \)
\( x = \frac{3}{13} \), \( y = \frac{4}{13} \), \( z = \frac{12}{13} \).
So, point \( P_1 = \left(\frac{3}{13}, \frac{4}{13}, \frac{12}{13}\right) \).

Case 2: If \( 1+\lambda = -13 \)
\( x = \frac{3}{-13} = -\frac{3}{13} \), \( y = \frac{4}{-13} = -\frac{4}{13} \), \( z = \frac{12}{-13} = -\frac{12}{13} \).
So, point \( P_2 = \left(-\frac{3}{13}, -\frac{4}{13}, -\frac{12}{13}\right) \).

Step 6: Calculate the distance from A(3, 4, 12) to \( P_1 \) and \( P_2 \).
Distance to \( P_1 \):
\( d_1 = \sqrt{\left(\frac{3}{13} - 3\right)^2 + \left(\frac{4}{13} - 4\right)^2 + \left(\frac{12}{13} - 12\right)^2} \)
\( d_1 = \sqrt{\left(\frac{3 - 39}{13}\right)^2 + \left(\frac{4 - 52}{13}\right)^2 + \left(\frac{12 - 156}{13}\right)^2} \)
\( d_1 = \sqrt{\left(\frac{-36}{13}\right)^2 + \left(\frac{-48}{13}\right)^2 + \left(\frac{-144}{13}\right)^2} \)
\( d_1 = \frac{1}{13} \sqrt{1296 + 2304 + 20736} = \frac{1}{13} \sqrt{24336} = \frac{156}{13} = 12 \).

Distance to \( P_2 \):
\( d_2 = \sqrt{\left(-\frac{3}{13} - 3\right)^2 + \left(-\frac{4}{13} - 4\right)^2 + \left(-\frac{12}{13} - 12\right)^2} \)
\( d_2 = \sqrt{\left(\frac{-3 - 39}{13}\right)^2 + \left(\frac{-4 - 52}{13}\right)^2 + \left(\frac{-12 - 156}{13}\right)^2} \)
\( d_2 = \sqrt{\left(\frac{-42}{13}\right)^2 + \left(\frac{-56}{13}\right)^2 + \left(\frac{-168}{13}\right)^2} \)
\( d_2 = \frac{1}{13} \sqrt{1764 + 3136 + 28224} = \frac{1}{13} \sqrt{33124} = \frac{182}{13} = 14 \).

Step 7: Conclude the results.
Comparing \( d_1 \) and \( d_2 \):
The maximum distance is 14 and the minimum distance is 12.
Question 9 (i) (Marks: 10)
Solve \( \frac{d^2y}{dx^2} + 9y = \sec 3x \) using the method of variation of parameters.
Click to Reveal Answer & Solution
Correct Answer: \( y = c_1 \cos 3x + c_2 \sin 3x - \frac{1}{9} (\cos 3x) \log|\sec 3x| + \frac{x}{3} \sin 3x \)
Detailed Solution:
The given differential equation can be written as \( (D^2 + 9)y = \sec 3x \).

Step 1: Find the Complementary Function (C.F.).
The Auxiliary Equation is \( m^2 + 9 = 0 \implies m^2 = -9 \implies m = \pm 3i \).
The roots are complex conjugates \( \alpha \pm i\beta \) with \( \alpha = 0, \beta = 3 \).
\( C.F. = c_1 \cos 3x + c_2 \sin 3x \).

Step 2: Identify \( y_1, y_2 \), and \( X \).
Let \( y_1 = \cos 3x \) and \( y_2 = \sin 3x \).
The right-hand side function is \( X = \sec 3x \).

Step 3: Calculate the Wronskian (\( W \)).
\( W = y_1 y_2' - y_1' y_2 \)
\( y_1' = -3\sin 3x \) and \( y_2' = 3\cos 3x \).
\( W = (\cos 3x)(3\cos 3x) - (-3\sin 3x)(\sin 3x) \)
\( W = 3\cos^2 3x + 3\sin^2 3x = 3(\cos^2 3x + \sin^2 3x) = 3(1) = 3 \).
Since \( W \neq 0 \), \( y_1 \) and \( y_2 \) are linearly independent.

Step 4: Apply the variation of parameters formula for the Particular Integral (P.I.).
\( P.I. = -y_1 \int \frac{y_2 X}{W} dx + y_2 \int \frac{y_1 X}{W} dx \)

Part 4a: Evaluate the first integral.
\( \int \frac{y_2 X}{W} dx = \int \frac{\sin 3x \sec 3x}{3} dx \)
Since \( \sec 3x = \frac{1}{\cos 3x} \):
\( = \frac{1}{3} \int \frac{\sin 3x}{\cos 3x} dx = \frac{1}{3} \int \tan 3x dx \)
We know \( \int \tan(ax) dx = \frac{1}{a}\log|\sec(ax)| \):
\( = \frac{1}{3} \left( \frac{1}{3} \log|\sec 3x| \right) = \frac{1}{9} \log|\sec 3x| \).

Part 4b: Evaluate the second integral.
\( \int \frac{y_1 X}{W} dx = \int \frac{\cos 3x \sec 3x}{3} dx \)
Since \( \cos 3x \sec 3x = 1 \):
\( = \int \frac{1}{3} dx = \frac{1}{3} x = \frac{x}{3} \).

Step 5: Assemble the P.I.
Substituting the evaluated integrals back into the formula:
\( P.I. = -\cos 3x \left( \frac{1}{9} \log|\sec 3x| \right) + \sin 3x \left( \frac{x}{3} \right) \)
\( P.I. = -\frac{1}{9} \cos 3x \log|\sec 3x| + \frac{x}{3} \sin 3x \).

Step 6: Write the complete solution.
\( y = C.F. + P.I. \)
\( y = c_1 \cos 3x + c_2 \sin 3x - \frac{1}{9} \cos 3x \log|\sec 3x| + \frac{x}{3} \sin 3x \).
Question 9 (ii) (Marks: 5)
Solve \( (D^2 + 3D + 2)y = x + 1 \).
Click to Reveal Answer & Solution
Correct Answer: \( y = c_1 e^{-x} + c_2 e^{-2x} + \frac{2x - 1}{4} \)
Detailed Solution:
Step 1: Find the Complementary Function (C.F.).
The Auxiliary Equation is \( m^2 + 3m + 2 = 0 \).
Factorizing: \( (m+1)(m+2) = 0 \implies m = -1, -2 \).
\( C.F. = c_1 e^{-x} + c_2 e^{-2x} \).

Step 2: Find the Particular Integral (P.I.).
The RHS is an algebraic function (a polynomial), so we expand the differential operator using the Binomial Theorem.
\( P.I. = \frac{1}{D^2 + 3D + 2} (x + 1) \)
Take out the lowest degree term (the constant term) from the denominator to form \( (1 + \phi(D)) \):
\( P.I. = \frac{1}{2\left(1 + \frac{3D + D^2}{2}\right)} (x + 1) \)
\( P.I. = \frac{1}{2} \left[1 + \left(\frac{3D + D^2}{2}\right)\right]^{-1} (x + 1) \)

Expand the series up to the degree of the polynomial (which is 1 here, so we only need up to \( D^1 \)). The binomial expansion is \( (1+z)^{-1} = 1 - z + z^2 - \dots \)
\( P.I. = \frac{1}{2} \left[ 1 - \left(\frac{3D + D^2}{2}\right) + \text{higher order terms} \right] (x + 1) \)
Neglecting \( D^2 \) and higher powers because the second derivative of \( (x+1) \) is 0:
\( P.I. = \frac{1}{2} \left[ 1 - \frac{3}{2}D \right] (x + 1) \)
Operate on \( (x+1) \):
\( P.I. = \frac{1}{2} \left[ 1(x+1) - \frac{3}{2} D(x+1) \right] \)
\( P.I. = \frac{1}{2} \left[ (x+1) - \frac{3}{2}(1) \right] = \frac{1}{2} \left[ x + 1 - \frac{3}{2} \right] \)
\( P.I. = \frac{1}{2} \left[ x - \frac{1}{2} \right] = \frac{2x - 1}{4} \).

Step 3: Write the complete solution.
\( y = C.F. + P.I. \)
\( y = c_1 e^{-x} + c_2 e^{-2x} + \frac{2x - 1}{4} \).
Question 10 (i) (Marks: 10)
Solve \( (x^2 D^2 + 4xD + 2)y = x \log x \).
Click to Reveal Answer & Solution
Correct Answer: \( y = c_1 x^{-1} + c_2 x^{-2} + \frac{x(6\log x - 5)}{36} \)
Detailed Solution:
This is a Cauchy-Euler linear differential equation with variable coefficients.

Step 1: Transform to a linear differential equation with constant coefficients.
Let \( x = e^z \implies z = \log x \).
Let the new differential operator be \( \theta = \frac{d}{dz} \).
The standard transformations are:
\( xD = \theta \)
\( x^2 D^2 = \theta(\theta - 1) = \theta^2 - \theta \)
Substitute these into the given equation:
\( (\theta^2 - \theta + 4\theta + 2)y = e^z \cdot z \)
\( (\theta^2 + 3\theta + 2)y = z e^z \).
This is now a linear ODE with constant coefficients.

Step 2: Find the Complementary Function (C.F.).
The Auxiliary Equation is \( m^2 + 3m + 2 = 0 \).
Factorizing: \( (m+1)(m+2) = 0 \implies m = -1, -2 \).
So, \( C.F. (\text{in terms of } z) = c_1 e^{-z} + c_2 e^{-2z} \).
Substituting \( e^z = x \):
\( C.F. = c_1 (e^z)^{-1} + c_2 (e^z)^{-2} = c_1 x^{-1} + c_2 x^{-2} = \frac{c_1}{x} + \frac{c_2}{x^2} \).

Step 3: Find the Particular Integral (P.I.).
\( P.I. = \frac{1}{\theta^2 + 3\theta + 2} (z e^z) \)
Use the exponential shift rule: \( \frac{1}{f(\theta)} (e^{az} V) = e^{az} \frac{1}{f(\theta + a)} V \). Here, \( a=1 \) and \( V=z \).
\( P.I. = e^z \frac{1}{(\theta+1)^2 + 3(\theta+1) + 2} (z) \)
\( P.I. = e^z \frac{1}{(\theta^2 + 2\theta + 1) + 3\theta + 3 + 2} (z) \)
\( P.I. = e^z \frac{1}{\theta^2 + 5\theta + 6} (z) \)
Now, expand the operator using the Binomial Theorem by taking the constant term 6 common:
\( P.I. = e^z \frac{1}{6\left(1 + \frac{5\theta + \theta^2}{6}\right)} (z) \)
\( P.I. = \frac{e^z}{6} \left[ 1 + \left(\frac{5\theta + \theta^2}{6}\right) \right]^{-1} (z) \)
Expand up to the first power of \( \theta \) because the degree of \( z \) is 1:
\( P.I. = \frac{e^z}{6} \left[ 1 - \frac{5\theta}{6} + \text{higher order terms} \right] (z) \)
Operate on \( z \):
\( P.I. = \frac{e^z}{6} \left[ z - \frac{5}{6}\theta(z) \right] = \frac{e^z}{6} \left[ z - \frac{5}{6}(1) \right] \)
\( P.I. = \frac{e^z}{6} \left( \frac{6z - 5}{6} \right) = \frac{e^z (6z - 5)}{36} \).

Step 4: Convert P.I. back to the original independent variable \( x \).
Substitute \( z = \log x \) and \( e^z = x \):
\( P.I. = \frac{x (6\log x - 5)}{36} \).

Step 5: Write the complete solution.
\( y = C.F. + P.I. \)
\( y = \frac{c_1}{x} + \frac{c_2}{x^2} + \frac{x(6\log x - 5)}{36} \).
Question 10 (ii) (Marks: 5)
If \( x = r \cos 2\theta \), \( y = r \sin 2\theta \), then find \( \frac{\partial(x,y)}{\partial(r,\theta)} \).
Click to Reveal Answer & Solution
Correct Answer: \( 2r \)
Detailed Solution:
Step 1: Define the Jacobian determinant formula.
The Jacobian of \( x, y \) with respect to \( r, \theta \) is denoted by \( J \) and given by the determinant:
\( J = \frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} \)

Step 2: Find all four first-order partial derivatives.
Given functions:
\( x = r \cos 2\theta \)
\( y = r \sin 2\theta \)

- Differentiate \( x \) with respect to \( r \) (treating \( \theta \) as constant):
\( \frac{\partial x}{\partial r} = \cos 2\theta \)
- Differentiate \( x \) with respect to \( \theta \) (treating \( r \) as constant):
\( \frac{\partial x}{\partial \theta} = r(-\sin 2\theta) \cdot 2 = -2r \sin 2\theta \)
- Differentiate \( y \) with respect to \( r \) (treating \( \theta \) as constant):
\( \frac{\partial y}{\partial r} = \sin 2\theta \)
- Differentiate \( y \) with respect to \( \theta \) (treating \( r \) as constant):
\( \frac{\partial y}{\partial \theta} = r(\cos 2\theta) \cdot 2 = 2r \cos 2\theta \)

Step 3: Construct and evaluate the determinant.
Substitute the derivatives into the Jacobian determinant:
\( J = \begin{vmatrix} \cos 2\theta & -2r \sin 2\theta \\ \sin 2\theta & 2r \cos 2\theta \end{vmatrix} \)

Expand the \( 2 \times 2 \) determinant (\( ad - bc \)):
\( J = (\cos 2\theta)(2r \cos 2\theta) - (-2r \sin 2\theta)(\sin 2\theta) \)
\( J = 2r \cos^2 2\theta + 2r \sin^2 2\theta \)

Step 4: Simplify using trigonometric identities.
Take out the common factor \( 2r \):
\( J = 2r (\cos^2 2\theta + \sin^2 2\theta) \)
Using the fundamental Pythagorean trigonometric identity \( \cos^2 A + \sin^2 A = 1 \) (here \( A = 2\theta \)):
\( J = 2r (1) = 2r \).

Therefore, the value of the Jacobian is \( 2r \).