Detailed Solution:
Step 1: Characteristic Equation
The characteristic equation of a $3 \times 3$ matrix $A$ is given by: $$\lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0$$ Where:

• $S_1$ is the sum of the main diagonal elements (Trace of $A$).
  $S_1 = -2 + 1 + 0 = -1$
• $S_2$ is the sum of the minors of the main diagonal elements.
  $S_2 = \begin{vmatrix} 1 & -6 \\ -2 & 0 \end{vmatrix} + \begin{vmatrix} -2 & -3 \\ -1 & 0 \end{vmatrix} + \begin{vmatrix} -2 & 2 \\ 2 & 1 \end{vmatrix} = (0 - 12) + (0 - 3) + (-2 - 4) = -12 - 3 - 6 = -21$
• $S_3$ is the determinant of matrix $A$ ($|A|$).
  $|A| = -2(0 - 12) - 2(0 - 6) - 3(-4 + 1) = -2(-12) - 2(-6) - 3(-3) = 24 + 12 + 9 = 45$

Therefore, the characteristic equation is: $$\lambda^3 + \lambda^2 - 21\lambda - 45 = 0$$
Step 2: Finding the Eigenvalues
By trial and error (synthetic division):
Let $\lambda = 5 \implies (5)^3 + (5)^2 - 21(5) - 45 = 125 + 25 - 105 - 45 = 0$.
Hence, $\lambda = 5$ is a root.
Dividing the polynomial by $(\lambda - 5)$ yields the quadratic $\lambda^2 + 6\lambda + 9 = 0$.
Solving $(\lambda + 3)^2 = 0 \implies \lambda = -3, -3$.
Eigenvalues are $\lambda_1 = 5, \lambda_2 = -3, \lambda_3 = -3$.

Step 3: Finding the Eigenvectors
The general eigenvector equation is $(A - \lambda I)X = 0$: $$ \begin{pmatrix} -2-\lambda & 2 & -3 \\ 2 & 1-\lambda & -6 \\ -1 & -2 & -\lambda \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$ Case 1: For $\lambda = 5$
Substituting $\lambda = 5$ gives: $$ \begin{pmatrix} -7 & 2 & -3 \\ 2 & -4 & -6 \\ -1 & -2 & -5 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$ Using Cross-Multiplication on the first two rows: $-7x_1 + 2x_2 - 3x_3 = 0$
$2x_1 - 4x_2 - 6x_3 = 0$
$$ \frac{x_1}{-12 - 12} = \frac{x_2}{-6 - 42} = \frac{x_3}{28 - 4} $$ $$ \frac{x_1}{-24} = \frac{x_2}{-48} = \frac{x_3}{24} \implies \frac{x_1}{-1} = \frac{x_2}{-2} = \frac{x_3}{1} $$ $X_1 = \begin{pmatrix} -1 \\ -2 \\ 1 \end{pmatrix}$

Case 2: For $\lambda = -3$ (Repeated Root)
Substituting $\lambda = -3$ gives: $$ \begin{pmatrix} 1 & 2 & -3 \\ 2 & 4 & -6 \\ -1 & -2 & 3 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$ All three equations are linearly dependent and reduce to a single equation: $$x_1 + 2x_2 - 3x_3 = 0$$ Since we have 1 equation and 3 unknowns, there are two free variables. Let $x_2 = 0, x_3 = 1 \implies x_1 = 3$.
Wait, let's substitute correctly: $x_1 = -2(0) + 3(1) = 3$. So vector is $(3, 0, 1)^T$. (Or $(-3, 0, 1)$? Let's check: $x_1 + 2(0) - 3(1) = 0 \implies x_1 = 3$). Let's set $x_3 = 1, x_2 = 0 \implies x_1 = 3$. So $\begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix}$. Let's set $x_3 = 0, x_2 = -1 \implies x_1 = 2$. So $\begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix}$.
Let $x_3 = 0, x_2 = -1 \implies x_1 = 2 \implies X_2 = \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix}$
Let $x_2 = 0, x_3 = 1 \implies x_1 = 3 \implies X_3 = \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix}$

Detailed Solution:
Step 1: Use the Trace property of Matrices
We know that the sum of the eigenvalues of a matrix is equal to its trace (the sum of the principal diagonal elements).
Let the eigenvalues of matrix $A$ be $\lambda_1, \lambda_2,$ and $\lambda_3$. We are given $\lambda_1 = 2$ and $\lambda_2 = 3$.
$\text{Trace}(A) = 3 + (-3) + 7 = 7$
Therefore, $\lambda_1 + \lambda_2 + \lambda_3 = 7$
$2 + 3 + \lambda_3 = 7$
$5 + \lambda_3 = 7 \implies \lambda_3 = 2$
The third eigenvalue of $A$ is 2.
Step 2: Find eigenvalues of $A^{-1}$
By the properties of eigenvalues, if $\lambda_1, \lambda_2, \lambda_3$ are the eigenvalues of a non-singular matrix $A$, then the eigenvalues of its inverse $A^{-1}$ are $\frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \frac{1}{\lambda_3}$.
Since the eigenvalues of $A$ are $2, 3, 2$, the eigenvalues of $A^{-1}$ are: $$ \frac{1}{2}, \frac{1}{3}, \text{ and } \frac{1}{2} $$

Detailed Solution:
Step 1: State Taylor's Series Formula for two variables
The Taylor's series expansion of $f(x,y)$ about the point $(a, b)$ up to the second degree is: $$ f(x,y) = f(a,b) + \left[ (x-a)f_x(a,b) + (y-b)f_y(a,b) \right] + \frac{1}{2!} \left[ (x-a)^2 f_{xx}(a,b) + 2(x-a)(y-b)f_{xy}(a,b) + (y-b)^2 f_{yy}(a,b) \right] $$ Here, the point is $(a, b) = (1, \frac{\pi}{4})$.
Step 2: Evaluate the Function and its Partial Derivatives at $(1, \frac{\pi}{4})$

Derivative	Function Expression	Value at $(1, \frac{\pi}{4})$
$f(x,y)$	$e^x \sin y$	$e^1 \sin(\frac{\pi}{4}) = \frac{e}{\sqrt{2}}$
$f_x(x,y)$	$e^x \sin y$	$e^1 \sin(\frac{\pi}{4}) = \frac{e}{\sqrt{2}}$
$f_y(x,y)$	$e^x \cos y$	$e^1 \cos(\frac{\pi}{4}) = \frac{e}{\sqrt{2}}$
$f_{xx}(x,y)$	$e^x \sin y$	$\frac{e}{\sqrt{2}}$
$f_{xy}(x,y)$	$e^x \cos y$	$\frac{e}{\sqrt{2}}$
$f_{yy}(x,y)$	$-e^x \sin y$	$-\frac{e}{\sqrt{2}}$

Step 3: Substitute into the Formula
$$ f(x,y) = \frac{e}{\sqrt{2}} + \left[ (x-1)\left(\frac{e}{\sqrt{2}}\right) + (y - \frac{\pi}{4})\left(\frac{e}{\sqrt{2}}\right) \right] + \frac{1}{2} \left[ (x-1)^2 \left(\frac{e}{\sqrt{2}}\right) + 2(x-1)(y-\frac{\pi}{4}) \left(\frac{e}{\sqrt{2}}\right) - (y-\frac{\pi}{4})^2 \left(\frac{e}{\sqrt{2}}\right) \right] $$ Simplifying gives the final answer: $$ f(x,y) = \frac{e}{\sqrt{2}} + \frac{e}{\sqrt{2}}(x-1) + \frac{e}{\sqrt{2}}(y - \frac{\pi}{4}) + \frac{e}{2\sqrt{2}}(x-1)^2 + \frac{e}{\sqrt{2}}(x-1)(y-\frac{\pi}{4}) - \frac{e}{2\sqrt{2}}(y-\frac{\pi}{4})^2 $$

Detailed Solution:
Step 1: Construct the Symmetric Matrix of the Quadratic Form
The general form is $x^T A x$. The diagonal elements of $A$ are the coefficients of the squared terms. The off-diagonal elements $a_{ij}$ are half of the coefficients of the $x_i x_j$ terms. $$ A = \begin{pmatrix} \text{coeff of } x_1^2 & \frac{1}{2}(\text{coeff of } x_1 x_2) & \frac{1}{2}(\text{coeff of } x_1 x_3) \\ \frac{1}{2}(\text{coeff of } x_1 x_2) & \text{coeff of } x_2^2 & \frac{1}{2}(\text{coeff of } x_2 x_3) \\ \frac{1}{2}(\text{coeff of } x_1 x_3) & \frac{1}{2}(\text{coeff of } x_2 x_3) & \text{coeff of } x_3^2 \end{pmatrix} $$ $$ A = \begin{pmatrix} 3 & -1 & -3 \\ -1 & 3 & -3 \\ -3 & -3 & 5 \end{pmatrix} $$
Step 2: Evaluate the Principal Minors
To determine the nature without finding eigenvalues, we calculate the leading principal minors $D_1, D_2, D_3$.

• 1. First Principal Minor ($D_1$):
  $D_1 = |3| = 3$
  Since $3 > 0$, $D_1$ is Positive.


• 2. Second Principal Minor ($D_2$):
  $D_2 = \begin{vmatrix} 3 & -1 \\ -1 & 3 \end{vmatrix} = (3)(3) - (-1)(-1) = 9 - 1 = 8$
  Since $8 > 0$, $D_2$ is Positive.


• 3. Third Principal Minor ($D_3 = |A|$):
  $D_3 = \begin{vmatrix} 3 & -1 & -3 \\ -1 & 3 & -3 \\ -3 & -3 & 5 \end{vmatrix}$
  $D_3 = 3(15 - 9) - (-1)(-5 - 9) + (-3)(3 - (-9))$
  $D_3 = 3(6) + 1(-14) - 3(12)$
  $D_3 = 18 - 14 - 36 = -32$
  Since $-32 < 0$, $D_3$ is Negative.

Step 3: Conclusion on Nature
The conditions for nature using principal minors are:

• Positive Definite: $D_1 > 0, D_2 > 0, D_3 > 0$
• Negative Definite: $D_1 < 0, D_2 > 0, D_3 < 0$
• Indefinite: When the signs do not follow the pattern for positive or negative definite.

Here, $D_1 > 0, D_2 > 0,$ but $D_3 < 0$. Therefore, the quadratic form is Indefinite.

Detailed Solution:
Step 1: Construct the Matrix of the Quadratic Form
$$ A = \begin{pmatrix} 2 & 1 & -1 \\ 1 & 1 & -2 \\ -1 & -2 & 1 \end{pmatrix} $$
Step 2: Find the Characteristic Equation and Eigenvalues
The characteristic equation is $\lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0$.
$S_1 = \text{Trace}(A) = 2 + 1 + 1 = 4$
$S_2 = \begin{vmatrix} 1 & -2 \\ -2 & 1 \end{vmatrix} + \begin{vmatrix} 2 & -1 \\ -1 & 1 \end{vmatrix} + \begin{vmatrix} 2 & 1 \\ 1 & 1 \end{vmatrix} = (1 - 4) + (2 - 1) + (2 - 1) = -3 + 1 + 1 = -1$
$S_3 = |A| = 2(1 - 4) - 1(1 - 2) - 1(-2 + 1) = 2(-3) - (-1) - (-1) = -6 + 1 + 1 = -4$
Characteristic equation: $\lambda^3 - 4\lambda^2 - \lambda + 4 = 0$
Using synthetic division, $\lambda = 1$ is a root: $(1) - 4(1) - 1 + 4 = 0$.
Dividing by $(\lambda - 1)$ gives $\lambda^2 - 3\lambda - 4 = 0 \implies (\lambda - 4)(\lambda + 1) = 0$.
Eigenvalues are $\lambda = 1, -1, 4$.
Step 3: Find the Eigenvectors

• For $\lambda = 1$: $(A - 1I)X = 0 \implies \begin{pmatrix} 1 & 1 & -1 \\ 1 & 0 & -2 \\ -1 & -2 & 0 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0$
  $x_1 + x_2 - x_3 = 0$
  $x_1 - 2x_3 = 0$
  From equation 2, $x_1 = 2x_3$. Substitute into equation 1: $2x_3 + x_2 - x_3 = 0 \implies x_2 = -x_3$.
  Let $x_3 = 1 \implies x_1 = 2, x_2 = -1$.
  $X_1 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$


• For $\lambda = -1$: $(A + 1I)X = 0 \implies \begin{pmatrix} 3 & 1 & -1 \\ 1 & 2 & -2 \\ -1 & -2 & 2 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0$
  $3x_1 + x_2 - x_3 = 0$
  $x_1 + 2x_2 - 2x_3 = 0$
  Cross-multiplication: $\frac{x_1}{-2 - (-2)} = \frac{x_2}{-1 - (-6)} = \frac{x_3}{6 - 1} \implies \frac{x_1}{0} = \frac{x_2}{5} = \frac{x_3}{5}$.
  $X_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$


• For $\lambda = 4$: $(A - 4I)X = 0 \implies \begin{pmatrix} -2 & 1 & -1 \\ 1 & -3 & -2 \\ -1 & -2 & -3 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0$
  $-2x_1 + x_2 - x_3 = 0$
  $x_1 - 3x_2 - 2x_3 = 0$
  Cross-multiplication: $\frac{x_1}{-2 - 3} = \frac{x_2}{-1 - 4} = \frac{x_3}{6 - 1} \implies \frac{x_1}{-5} = \frac{x_2}{-5} = \frac{x_3}{5}$.
  $X_3 = \begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix}$

Step 4: Normalize Eigenvectors and Form Modal Matrix $N$
The vectors are already mutually orthogonal because the eigenvalues are distinct symmetric real matrix properties. We normalize each by dividing by its magnitude: $|X_1| = \sqrt{4+1+1} = \sqrt{6}$, $|X_2| = \sqrt{0+1+1} = \sqrt{2}$, $|X_3| = \sqrt{1+1+1} = \sqrt{3}$. $$ N = \begin{pmatrix} \frac{2}{\sqrt{6}} & 0 & \frac{-1}{\sqrt{3}} \\ \frac{-1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \end{pmatrix} $$ The transformation $X = NY$ converts $A$ into a diagonal matrix $D = N^T A N$ whose elements are the eigenvalues: $$ D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 4 \end{pmatrix} $$
Step 5: Write the Canonical Form and Analyze
Canonical Form = $Y^T D Y = 1y_1^2 - 1y_2^2 + 4y_3^2$.

• Rank (r): Number of non-zero terms in Canonical Form = 3.
• Index (p): Number of positive terms in Canonical Form = 2.
• Signature (s): Number of positive terms - Number of negative terms = $p - (r-p) = 2p - r = 4 - 3 = 1$.
• Nature: Since eigenvalues are a mix of positive (1, 4) and negative (-1), the nature is Indefinite.

Detailed Solution:
Step 1: Find the Characteristic Equation
The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation.
Characteristic equation is $\lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0$.
$S_1 = \text{Trace}(A) = 2 + 1 + 2 = 5$
$S_2 = \begin{vmatrix} 1 & 2 \\ -1 & 2 \end{vmatrix} + \begin{vmatrix} 2 & -1 \\ -1 & 2 \end{vmatrix} + \begin{vmatrix} 2 & 1 \\ 2 & 1 \end{vmatrix} = (2 - (-2)) + (4 - 1) + (2 - 2) = 4 + 3 + 0 = 7$
$S_3 = |A| = 2(2 - (-2)) - 1(4 - (-2)) - 1(-2 - (-1)) = 2(4) - 1(6) - 1(-1) = 8 - 6 + 1 = 3$
Characteristic equation: $\lambda^3 - 5\lambda^2 + 7\lambda - 3 = 0$.
By Cayley-Hamilton theorem, we must prove: $$ A^3 - 5A^2 + 7A - 3I = 0 $$
Step 2: Calculate $A^2$ and $A^3$
$$ A^2 = A \cdot A = \begin{pmatrix} 2 & 1 & -1 \\ 2 & 1 & 2 \\ -1 & -1 & 2 \end{pmatrix} \begin{pmatrix} 2 & 1 & -1 \\ 2 & 1 & 2 \\ -1 & -1 & 2 \end{pmatrix} $$ $$ A^2 = \begin{pmatrix} (4+2+1) & (2+1+1) & (-2+2-2) \\ (4+2-2) & (2+1-2) & (-2+2+4) \\ (-2-2-2) & (-1-1-2) & (1-2+4) \end{pmatrix} = \begin{pmatrix} 7 & 4 & -2 \\ 4 & 1 & 4 \\ -6 & -4 & 3 \end{pmatrix} $$ $$ A^3 = A^2 \cdot A = \begin{pmatrix} 7 & 4 & -2 \\ 4 & 1 & 4 \\ -6 & -4 & 3 \end{pmatrix} \begin{pmatrix} 2 & 1 & -1 \\ 2 & 1 & 2 \\ -1 & -1 & 2 \end{pmatrix} $$ $$ A^3 = \begin{pmatrix} (14+8+2) & (7+4+2) & (-7+8-4) \\ (8+2-4) & (4+1-4) & (-4+2+8) \\ (-12-8-3) & (-6-4-3) & (6-8+6) \end{pmatrix} = \begin{pmatrix} 24 & 13 & -3 \\ 6 & 1 & 6 \\ -23 & -13 & 4 \end{pmatrix} $$
Step 3: Verification
Substitute the matrices into $A^3 - 5A^2 + 7A - 3I$: $$ \begin{pmatrix} 24 & 13 & -3 \\ 6 & 1 & 6 \\ -23 & -13 & 4 \end{pmatrix} - 5\begin{pmatrix} 7 & 4 & -2 \\ 4 & 1 & 4 \\ -6 & -4 & 3 \end{pmatrix} + 7\begin{pmatrix} 2 & 1 & -1 \\ 2 & 1 & 2 \\ -1 & -1 & 2 \end{pmatrix} - 3\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ Evaluate element (1,1): $24 - 35 + 14 - 3 = 0$.
Evaluate element (1,2): $13 - 20 + 7 - 0 = 0$.
Evaluate element (3,3): $4 - 15 + 14 - 3 = 0$.
Computing for all elements gives the Zero Matrix $\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$. Hence verified.
Step 4: Finding Inverse $A^{-1}$
Multiply the equation $A^3 - 5A^2 + 7A - 3I = 0$ by $A^{-1}$: $$ A^2 - 5A + 7I - 3A^{-1} = 0 \implies 3A^{-1} = A^2 - 5A + 7I $$ $$ 3A^{-1} = \begin{pmatrix} 7 & 4 & -2 \\ 4 & 1 & 4 \\ -6 & -4 & 3 \end{pmatrix} - \begin{pmatrix} 10 & 5 & -5 \\ 10 & 5 & 10 \\ -5 & -5 & 10 \end{pmatrix} + \begin{pmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7 \end{pmatrix} $$ $$ 3A^{-1} = \begin{pmatrix} (7 - 10 + 7) & (4 - 5 + 0) & (-2 + 5 + 0) \\ (4 - 10 + 0) & (1 - 5 + 7) & (4 - 10 + 0) \\ (-6 + 5 + 0) & (-4 + 5 + 0) & (3 - 10 + 7) \end{pmatrix} = \begin{pmatrix} 4 & -1 & 3 \\ -6 & 3 & -6 \\ -1 & 1 & 0 \end{pmatrix} $$ $$ A^{-1} = \frac{1}{3} \begin{pmatrix} 4 & -1 & 3 \\ -6 & 3 & -6 \\ -1 & 1 & 0 \end{pmatrix} $$