Step 1: Computing $A^2$

Given the matrix $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, we calculate $A^2 = A \cdot A$:

$A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

$A^2 = \begin{bmatrix} (3)(3) + (1)(-1) & (3)(1) + (1)(2) \\ (-1)(3) + (2)(-1) & (-1)(1) + (2)(2) \end{bmatrix}$

$A^2 = \begin{bmatrix} 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$ [By Matrix Multiplication Rule]

Step 2: Evaluating the Expression $A^2 - 5A + 7I$

Let $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ be the identity matrix of order 2.

$5A = 5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$

$7I = 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

Now, substitute these into $A^2 - 5A + 7I$:

$\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

$= \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5-(-5)+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$ [Verified]

Step 3: Finding $A^{-1}$ using the Characteristic Equation

Starting from $A^2 - 5A + 7I = O$, we multiply by $A^{-1}$ on both sides:

$A^{-1}(A^2 - 5A + 7I) = A^{-1} \cdot O$

$(A^{-1}A)A - 5(A^{-1}A) + 7(A^{-1}I) = O$ [By Associative Law of Matrices]

$IA - 5I + 7A^{-1} = O$ [Since $A^{-1}A = I$ and $A^{-1}I = A^{-1}$]

$7A^{-1} = 5I - A$

$7A^{-1} = 5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

$7A^{-1} = \begin{bmatrix} 5-3 & 0-1 \\ 0-(-1) & 5-2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$

$A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 2/7 & -1/7 \\ 1/7 & 3/7 \end{bmatrix}$

Visual Representation of Matrix Transformation

Final Solution: $A^{-1} = \begin{bmatrix} \frac{2}{7} & -\frac{1}{7} \\ \frac{1}{7} & \frac{3}{7} \end{bmatrix}$

Step 1: Formal Definition of Adjoint

For a square matrix $A$ of order $n$, the adjoint, denoted as $\text{adj}(A)$, is the transpose of the matrix formed by the cofactors of the elements of $A$. Specifically, if $A = [a_{ij}]$, then $\text{adj}(A) = [C_{ij}]^T$, where $C_{ij}$ is the cofactor of element $a_{ij}$.

Step 2: Defining the Given Matrix

Let the given matrix be $A$:

$A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

Step 3: Calculating Co-factors ($C_{ij}$)

The cofactor $C_{ij}$ is calculated using the formula $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of the element $a_{ij}$.

• $C_{11} = (-1)^{1+1} |4| = 1 \times 4 = 4$
• $C_{12} = (-1)^{1+2} |3| = -1 \times 3 = -3$
• $C_{21} = (-1)^{2+1} |2| = -1 \times 2 = -2$
• $C_{22} = (-1)^{2+2} |1| = 1 \times 1 = 1$

Step 4: Constructing the Co-factor Matrix ($C$)

The matrix of co-factors $C$ is defined as:

$C = \begin{bmatrix} C_{11} & C_{12} \\ C_{21} & C_{22} \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}$

Step 5: Transposing the Co-factor Matrix

By definition [Transpose Operation], the rows of $C$ become the columns of $\text{adj}(A)$:

$\text{adj}(A) = C^T = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$

Step 6: Visual Representation

Final Solution: $\text{adj}(A) = \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$

Step 1: Establishing the Matrix and Existence Condition

Let the given matrix be $A$:

$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$

The inverse $A^{-1}$ exists if and only if $|A| \neq 0$ [Per the property of Non-singular Matrices].

Step 2: Calculating the Determinant $|A|$

Expanding along the first row ($R_1$):

$|A| = 1 \cdot \begin{vmatrix} \cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha \end{vmatrix} - 0 \cdot (\dots) + 0 \cdot (\dots)$ $|A| = 1 \cdot [(\cos\alpha)(-\cos\alpha) - (\sin\alpha)(\sin\alpha)]$ $|A| = -(\cos^2\alpha + \sin^2\alpha)$

Applying the Pythagorean identity $\cos^2\alpha + \sin^2\alpha = 1$ [Trigonometric Identity]:

$|A| = -1$

Since $|A| \neq 0$, the inverse $A^{-1}$ exists and is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.

Step 3: Calculating the Matrix of Cofactors $C_{ij}$

The cofactor $C_{ij}$ is defined as $(-1)^{i+j}M_{ij}$ where $M_{ij}$ is the minor of element $a_{ij}$.

• $C_{11} = +(\cos\alpha(-\cos\alpha) - \sin^2\alpha) = -1$
• $C_{12} = -(0 - 0) = 0$
• $C_{13} = +(0 - 0) = 0$
• $C_{21} = -(0 - 0) = 0$
• $C_{22} = +(1(-\cos\alpha) - 0) = -\cos\alpha$
• $C_{23} = -(1(\sin\alpha) - 0) = -\sin\alpha$
• $C_{31} = +(0 - 0) = 0$
• $C_{32} = -(1(\sin\alpha) - 0) = -\sin\alpha$
• $C_{33} = +(1(\cos\alpha) - 0) = \cos\alpha$

Step 4: Constructing the Adjoint Matrix $\text{adj}(A)$

The adjoint is the transpose of the cofactor matrix, $\text{adj}(A) = [C_{ij}]^T$:

$\text{adj}(A) = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix}$

Step 5: Final Computation of $A^{-1}$

Using $A^{-1} = \frac{1}{-1} \text{adj}(A) = -1 \cdot \text{adj}(A)$:

$A^{-1} = -1 \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos\alpha & -\sin\alpha \\ 0 & -\sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$

Final Solution: The inverse of the matrix is $\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$

Step 1: Establishing Fundamental Definitions

Let $A$ be a square matrix of order $n \times n$. A square matrix is termed nonsingular if its determinant is non-zero, i.e., $|A| \neq 0$. The adjoint of a matrix, denoted by $\text{adj } A$, is defined as the transpose of the matrix of cofactors of $A$. [Per the definition of the Adjoint of a Matrix].

Step 2: Utilizing the Fundamental Identity

The relationship between a square matrix $A$ and its adjoint is governed by the following identity: $A(\text{adj } A) = (\text{adj } A)A = |A|I_n$ where $I_n$ is the identity matrix of order $n$. [By the Property of Adjoint Matrices].

Step 3: Applying Determinant Properties

Taking the determinant on both sides of the identity: $|A(\text{adj } A)| = | |A|I_n |$ Applying the multiplicative property of determinants ($|XY| = |X||Y|$): $|A| \cdot |\text{adj } A| = | |A|I_n |$

Step 4: Scalar Multiplication in Determinants

For a scalar $k$ and a matrix $I_n$ of order $n$, the determinant property states $|kI_n| = k^n|I_n|$. Since $|I_n| = 1$: $| |A|I_n | = |A|^n \cdot |I_n| = |A|^n$ Substituting this into the equation from Step 3: $|A| \cdot |\text{adj } A| = |A|^n$

Step 5: Solving for $|\text{adj } A|$

Since the matrix $A$ is nonsingular, $|A| \neq 0$. We can divide both sides by $|A|$: $|\text{adj } A| = \frac{|A|^n}{|A|} = |A|^{n-1}$

Step 6: Substituting Specific Parameters

The problem specifies that $A$ is a square matrix of order $3 \times 3$, implying $n = 3$. $|\text{adj } A| = |A|^{3-1}$ $|\text{adj } A| = |A|^2$

Final Solution: |adj A| = |A|^2

Given Variables & Initial Setup

Let $A$ be an invertible square matrix of order $2 \times 2$. By definition, for any square matrix $A$, the product of the matrix and its inverse is the identity matrix $I$:

$A \cdot A^{-1} = I$ [Definition of Matrix Inversion]

Step 1: Applying the Determinant Property

Take the determinant on both sides of the equation:

$\det(A \cdot A^{-1}) = \det(I)$

By the Multiplicative Property of Determinants, which states that $\det(AB) = \det(A) \cdot \det(B)$, we have:

$\det(A) \cdot \det(A^{-1}) = \det(I)$ [Multiplicative Property of Determinants]

Step 2: Evaluating the Determinant of Identity Matrix

For a square matrix of order $2$, the identity matrix $I$ is defined as:

$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

The determinant of $I$ is calculated as:

$\det(I) = (1 \times 1) - (0 \times 0) = 1$

Step 3: Algebraic Isolation of $\det(A^{-1})$

Substitute $\det(I) = 1$ into the equation derived in Step 1:

$\det(A) \cdot \det(A^{-1}) = 1$

Since $A$ is given as an invertible matrix, $\det(A) \neq 0$. Therefore, we can divide both sides by $\det(A)$:

$\det(A^{-1}) = \frac{1}{\det(A)}$

Geometric/Visual Interpretation (Linear Scaling)

Summary of Determinant Properties

Property Name	Mathematical Formulation
Product Property	$\det(AB) = \det(A)\det(B)$
Identity Property	$\det(I_n) = 1$
Inverse Determinant	$\det(A^{-1}) = (\det(A))^{-1}$

Final Solution: The value of $\det(A^{-1})$ is equal to $\frac{1}{\det(A)}$ or $(\det(A))^{-1}$.

Step 1: Establishing the Theoretical Framework

To find the inverse of a square matrix $A$, denoted as $A^{-1}$, we must apply the formula:

$A^{-1} = \frac{1}{|A|} \text{adj}(A)$

This operation is defined only if the determinant $|A| \neq 0$, in which case the matrix is termed non-singular and invertible [Per the Invertibility Criterion for Square Matrices].

Step 2: Calculating the Determinant

Given the matrix $A = \begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}$, we calculate the determinant $|A|$ using the expansion formula for a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, where $|A| = ad - bc$.

$|A| = (-1)(2) - (5)(-3)$ $|A| = -2 - (-15)$ $|A| = -2 + 15$ $|A| = 13$

Since $|A| = 13 \neq 0$, the inverse exists [By the Non-Singularity Theorem].

Step 3: Determining the Adjoint Matrix

For a $2 \times 2$ matrix, the adjugate (adjoint) is found by interchanging the diagonal elements and changing the signs of the off-diagonal elements [Per the definition of the Adjoint of a $2 \times 2$ matrix].

Given $A = \begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}$:

• Swap $a_{11}$ and $a_{22}$: $(-1 \leftrightarrow 2)$
• Negate $a_{12}$ and $a_{21}$: $(5 \rightarrow -5, -3 \rightarrow 3)$

$\text{adj}(A) = \begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}$

Step 4: Constructing the Inverse Matrix

Substitute the results from Step 2 and Step 3 into the inverse formula:

$A^{-1} = \frac{1}{13} \begin{bmatrix} 2 & -5 \\ 3 & -1 \end{bmatrix}$ $A^{-1} = \begin{bmatrix} \frac{2}{13} & -\frac{5}{13} \\ \frac{3}{13} & -\frac{1}{13} \end{bmatrix}$

Step 5: Visual Representation of Transformation

Final Solution: The inverse of the matrix is $\begin{bmatrix} \frac{2}{13} & -\frac{5}{13} \\ \frac{3}{13} & -\frac{1}{13} \end{bmatrix}$.

Step 1: Establishing the Theoretical Framework

To find the inverse of a square matrix $A$, denoted as $A^{-1}$, we must satisfy the condition defined by the formula:

$A^{-1} = \frac{1}{|A|} \text{adj}(A)$

[Per the definition of the inverse of a square matrix, where $|A| \neq 0$].

Step 2: Calculating the Determinant $|A|$

Given the matrix $A = \begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}$, the determinant is calculated as follows:

$|A| = (2 \times 3) - (-2 \times 4)$

$|A| = 6 - (-8)$

$|A| = 6 + 8 = 14$

[By the property of $2 \times 2$ determinant evaluation: $\det \begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc$].

Since $|A| = 14 \neq 0$, the matrix $A$ is non-singular, and therefore, $A^{-1}$ exists.

Step 3: Calculating the Adjoint Matrix $\text{adj}(A)$

For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is found by interchanging the diagonal elements ($a$ and $d$) and changing the signs of the off-diagonal elements ($b$ and $c$).

$\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

Applying this to matrix $A$:

$\text{adj}(A) = \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$

Step 4: Final Computation of the Inverse

Substitute the values back into the inverse formula:

$A^{-1} = \frac{1}{14} \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} \frac{3}{14} & \frac{2}{14} \\ \frac{-4}{14} & \frac{2}{14} \end{bmatrix}$

Simplifying the fractions:

$A^{-1} = \begin{bmatrix} \frac{3}{14} & \frac{1}{7} \\ -\frac{2}{7} & \frac{1}{7} \end{bmatrix}$

Visual Representation of Matrix Transformation

Final Solution: $A^{-1} = \begin{bmatrix} \frac{3}{14} & \frac{1}{7} \\ -\frac{2}{7} & \frac{1}{7} \end{bmatrix}$

Step 1: Calculating $A^2$

To find $A^2$, we compute $A \times A$:

$A^2 = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$

Performing matrix multiplication:

• $R_1 \to C_1: (2)(2) + (-1)(-1) + (1)(1) = 4 + 1 + 1 = 6$
• $R_1 \to C_2: (2)(-1) + (-1)(2) + (1)(-1) = -2 - 2 - 1 = -5$
• $R_1 \to C_3: (2)(1) + (-1)(-1) + (1)(2) = 2 + 1 + 2 = 5$
• ...Repeating for all entries, we obtain:

$A^2 = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix}$

Step 2: Calculating $A^3$

We compute $A^3 = A^2 \times A$:

$A^3 = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$

Performing matrix multiplication:

$A^3 = \begin{bmatrix} 12+5+5 & -6-10-5 & 6+5+10 \\ -10-6-5 & 5+12+5 & -5-6-10 \\ 10+5+6 & -5-10-6 & 5+5+12 \end{bmatrix} = \begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix}$

Step 3: Verification of $A^3 - 6A^2 + 9A - 4I = O$

Substitute the matrices into the characteristic equation:

$\begin{bmatrix} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{bmatrix} - 6 \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} + 9 \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} - 4 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Simplifying the scalars:

$\begin{bmatrix} 22-36+18-4 & -21+30-9+0 & 21-30+9+0 \\ -21+30-9+0 & 22-36+18-4 & -21+30-9+0 \\ 21-30+9+0 & -21+30-9+0 & 22-36+18-4 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O$

[Verified as per the Cayley-Hamilton Theorem]

Step 4: Finding $A^{-1}$

Using the identity $A^3 - 6A^2 + 9A - 4I = O$, multiply by $A^{-1}$ throughout [Per Property of Inverses]:

$A^2 - 6A + 9I - 4A^{-1} = O$

Rearranging for $A^{-1}$:

$4A^{-1} = A^2 - 6A + 9I$ $4A^{-1} = \begin{bmatrix} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{bmatrix} - 6 \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} + \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix}$ $4A^{-1} = \begin{bmatrix} 6-12+9 & -5+6+0 & 5-6+0 \\ -5+6+0 & 6-12+9 & -5+6+0 \\ 5-6+0 & -5+6+0 & 6-12+9 \end{bmatrix} = \begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}$ $A^{-1} = \frac{1}{4} \begin{bmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{bmatrix}$

Final Solution: $A^{-1} = \begin{bmatrix} 3/4 & 1/4 & -1/4 \\ 1/4 & 3/4 & 1/4 \\ -1/4 & 1/4 & 3/4 \end{bmatrix}$

Step 1: Establishing the Matrix and Determinant

Let the given matrix be denoted as $A$:

$A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$

First, we calculate the determinant $|A|$ by expanding along the second column, as it contains two zeros [Per Laplace Expansion Theorem]:

$|A| = (-1)^{1+2}(-1) \begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix} + (-1)^{2+2}(0) \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} + (-1)^{3+2}(0) \begin{vmatrix} 1 & 2 \\ 3 & -2 \end{vmatrix}$

$|A| = (1) [ (3)(3) - (-2)(1) ] + 0 + 0$

$|A| = 9 + 2 = 11$

Step 2: Computing the Adjoint of A

The adjoint matrix, $adj(A)$, is the transpose of the cofactor matrix $C$. Let $C_{ij}$ be the cofactor of element $a_{ij}$.

• $C_{11} = + \begin{vmatrix} 0 & -2 \\ 0 & 3 \end{vmatrix} = 0$
• $C_{12} = - \begin{vmatrix} 3 & -2 \\ 1 & 3 \end{vmatrix} = -(9 + 2) = -11$
• $C_{13} = + \begin{vmatrix} 3 & 0 \\ 1 & 0 \end{vmatrix} = 0$
• $C_{21} = - \begin{vmatrix} -1 & 2 \\ 0 & 3 \end{vmatrix} = -(-3) = 3$
• $C_{22} = + \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} = (3 - 2) = 1$
• $C_{23} = - \begin{vmatrix} 1 & -1 \\ 1 & 0 \end{vmatrix} = -(0 - (-1)) = -1$
• $C_{31} = + \begin{vmatrix} -1 & 2 \\ 0 & -2 \end{vmatrix} = (2 - 0) = 2$
• $C_{32} = - \begin{vmatrix} 1 & 2 \\ 3 & -2 \end{vmatrix} = -(-2 - 6) = 8$
• $C_{33} = + \begin{vmatrix} 1 & -1 \\ 3 & 0 \end{vmatrix} = (0 - (-3)) = 3$

Therefore, $adj(A) = C^T = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}$

Step 3: Verification of $A(adj A)$

$A(adj A) = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix}$

$= \begin{bmatrix} (0+11+0) & (3-1-2) & (2-8+6) \\ (0+0+0) & (9+0+2) & (6+0-6) \\ (0+0+0) & (3+0-3) & (2+0+9) \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} = 11I$

Step 4: Verification of $(adj A)A$

$(adj A)A = \begin{bmatrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$

$= \begin{bmatrix} (0+9+2) & (0+0+0) & (0-6+6) \\ (-11+3+8) & (11+0+0) & (-22-2+24) \\ (0-3+3) & (0+0+0) & (0+2+9) \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} = 11I$

Since $|A| = 11$ and $|A|I = 11 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$, we observe that $A(adj A) = (adj A)A = |A|I$ holds true [Property of Adjoint Matrices].

Final Solution: $A(adj A) = (adj A)A = \begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix} = 11I$. Hence verified.

Step 1: Identifying the Matrix and Calculating the Determinant

Given the matrix $A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$.

The determinant $|A|$ is calculated as: $|A| = (2)(-6) - (3)(-4) = -12 - (-12) = -12 + 12 = 0$ [Per the definition of the determinant of a $2 \times 2$ matrix, $|A| = ad - bc$].

Step 2: Calculating the Adjoint of Matrix A

For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint is given by $adj(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

Substituting the values from $A$: $adj(A) = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix}$

Step 3: Verifying the Identity $A(adj A) = |A|I$

$A(adj A) = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix}$

Performing matrix multiplication: $= \begin{bmatrix} (2)(-6) + (3)(4) & (2)(-3) + (3)(2) \\ (-4)(-6) + (-6)(4) & (-4)(-3) + (-6)(2) \end{bmatrix}$ $= \begin{bmatrix} -12 + 12 & -6 + 6 \\ 24 - 24 & 12 - 12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ Since $|A|I = 0 \cdot \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$, it is verified that $A(adj A) = |A|I$.

Step 4: Verifying the Identity $(adj A)A = |A|I$

$(adj A)A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$

Performing matrix multiplication: $= \begin{bmatrix} (-6)(2) + (-3)(-4) & (-6)(3) + (-3)(-6) \\ (4)(2) + (2)(-4) & (4)(3) + (2)(-6) \end{bmatrix}$ $= \begin{bmatrix} -12 + 12 & -18 + 18 \\ 8 - 8 & 12 - 12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ Since $|A|I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$, it is verified that $(adj A)A = |A|I$.

Final Solution: $A(adj A) = (adj A)A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = |A|I$. The identity is verified.

Step 1: Calculating $A^2$

To compute $A^2$, we perform matrix multiplication $A \times A$:

$A^2 = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$
Using row-by-column multiplication:

• $R_1: (1+1+2, 1+2-1, 1-3+3) = (4, 2, 1)$
• $R_2: (1+2-6, 1+4+3, 1-6-9) = (-3, 8, -14)$
• $R_3: (2-1+6, 2-2-3, 2+3+9) = (7, -3, 14)$

$A^2 = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix}$

Step 2: Calculating $A^3$

$A^3 = A^2 \times A = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$

• $R_1: (4+2+2, 4+4-1, 4-6+3) = (8, 7, 1)$
• $R_2: (-3+8-28, -3+16+14, -3-24-42) = (-23, 27, -69)$
• $R_3: (7-3+28, 7-6-14, 7+9+42) = (32, -13, 58)$

$A^3 = \begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix}$

Step 3: Verification of the Characteristic Equation

Evaluate $A^3 - 6A^2 + 5A + 11I$:

$A^3$	$-6A^2$	$+5A$	$+11I$
$\begin{bmatrix} 8 & 7 & 1 \\ -23 & 27 & -69 \\ 32 & -13 & 58 \end{bmatrix}$	$\begin{bmatrix} -24 & -12 & -6 \\ 18 & -48 & 84 \\ -42 & 18 & -84 \end{bmatrix}$	$\begin{bmatrix} 5 & 5 & 5 \\ 5 & 10 & -15 \\ 10 & -5 & 15 \end{bmatrix}$	$\begin{bmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{bmatrix}$

Summing the components:
Element $(1,1) = 8 - 24 + 5 + 11 = 0$
Element $(1,2) = 7 - 12 + 5 + 0 = 0$
Element $(2,1) = -23 + 18 + 5 + 0 = 0$
Following this arithmetic for all elements, the resultant is $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O$. [Per Cayley-Hamilton Theorem]

Step 4: Deriving $A^{-1}$

Given $A^3 - 6A^2 + 5A + 11I = O$, multiply by $A^{-1}$ throughout:

$A^3A^{-1} - 6A^2A^{-1} + 5AA^{-1} + 11IA^{-1} = O \cdot A^{-1}$ $A^2 - 6A + 5I + 11A^{-1} = O$ $11A^{-1} = -A^2 + 6A - 5I$ $A^{-1} = -\frac{1}{11} (A^2 - 6A + 5I)$

Compute bracketed term:

$A^2 - 6A + 5I = \begin{bmatrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{bmatrix} - \begin{bmatrix} 6 & 6 & 6 \\ 6 & 12 & -18 \\ 12 & -6 & 18 \end{bmatrix} + \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} = \begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix}$

Therefore, $A^{-1} = -\frac{1}{11} \begin{bmatrix} 3 & -4 & -5 \\ -9 & 1 & 4 \\ -5 & 3 & 1 \end{bmatrix} = \begin{bmatrix} -3/11 & 4/11 & 5/11 \\ 9/11 & -1/11 & -4/11 \\ 5/11 & -3/11 & -1/11 \end{bmatrix}$

Final Solution: $A^{-1} = \frac{1}{11} \begin{bmatrix} -3 & 4 & 5 \\ 9 & -1 & -4 \\ 5 & -3 & -1 \end{bmatrix}$

Step 1: Identifying the Matrix and Checking for Invertibility

Let the given matrix be $A$.

$A = \begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}$

A matrix is invertible if and only if its determinant $|A| \neq 0$ [Per the property of non-singular matrices].

Expanding along the first row ($R_1$):

$|A| = 2 \begin{vmatrix} -1 & 0 \\ 2 & 1 \end{vmatrix} - 1 \begin{vmatrix} 4 & 0 \\ -7 & 1 \end{vmatrix} + 3 \begin{vmatrix} 4 & -1 \\ -7 & 2 \end{vmatrix}$

$|A| = 2(-1 - 0) - 1(4 - 0) + 3(8 - 7)$

$|A| = 2(-1) - 1(4) + 3(1) = -2 - 4 + 3 = -3$

Since $|A| = -3 \neq 0$, the inverse $A^{-1}$ exists.

Step 2: Calculating the Matrix of Cofactors ($C_{ij}$)

The cofactor $C_{ij}$ is given by $(-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor.

Cofactor	Calculation	Result
$C_{11}$	$+(-1-0)$	$-1$
$C_{12}$	$-(4-0)$	$-4$
$C_{13}$	$+(8-7)$	$1$
$C_{21}$	$-(1-6)$	$5$
$C_{22}$	$+(2+21)$	$23$
$C_{23}$	$-(4+7)$	$-11$
$C_{31}$	$+(0+3)$	$3$
$C_{32}$	$-(0-12)$	$12$
$C_{33}$	$+(-2-4)$	$-6$

Step 3: Adjoint and Inverse Formulation

The adjoint matrix $\text{adj}(A)$ is the transpose of the cofactor matrix [Per the definition of the adjoint of a matrix].

Cofactor Matrix $= \begin{bmatrix} -1 & -4 & 1 \\ 5 & 23 & -11 \\ 3 & 12 & -6 \end{bmatrix}$

$\text{adj}(A) = \begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix}$

Step 4: Computing the Inverse

The formula for the inverse is $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.

$A^{-1} = \frac{1}{-3} \begin{bmatrix} -1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & -11 & -6 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} 1/3 & -5/3 & -1 \\ 4/3 & -23/3 & -4 \\ -1/3 & 11/3 & 2 \end{bmatrix}$

Final Solution: $A^{-1} = \begin{bmatrix} 1/3 & -5/3 & -1 \\ 4/3 & -23/3 & -4 \\ -1/3 & 11/3 & 2 \end{bmatrix}$

Step 1: Establishing the Matrix and Checking for Invertibility

Given the matrix $A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$.

An inverse $A^{-1}$ exists if and only if $\det(A) \neq 0$ [Per the definition of non-singular matrices].

Expanding along the first column:

$\det(A) = 1 \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} - 0 \begin{vmatrix} -1 & 2 \\ -2 & 4 \end{vmatrix} + 3 \begin{vmatrix} -1 & 2 \\ 2 & -3 \end{vmatrix}$
$\det(A) = 1(8 - 6) - 0 + 3(3 - 4)$
$\det(A) = 2 + 3(-1) = 2 - 3 = -1$

Since $\det(A) = -1 \neq 0$, the matrix $A$ is non-singular and the inverse $A^{-1}$ exists.

Step 2: Calculating the Matrix of Cofactors

The cofactor $C_{ij}$ is given by $(-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of element $a_{ij}$.

• $C_{11} = + \begin{vmatrix} 2 & -3 \\ -2 & 4 \end{vmatrix} = 8 - 6 = 2$
• $C_{12} = - \begin{vmatrix} 0 & -3 \\ 3 & 4 \end{vmatrix} = -(0 - (-9)) = -9$
• $C_{13} = + \begin{vmatrix} 0 & 2 \\ 3 & -2 \end{vmatrix} = 0 - 6 = -6$
• $C_{21} = - \begin{vmatrix} -1 & 2 \\ -2 & 4 \end{vmatrix} = -(-4 - (-4)) = 0$
• $C_{22} = + \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = 4 - 6 = -2$
• $C_{23} = - \begin{vmatrix} 1 & -1 \\ 3 & -2 \end{vmatrix} = -(-2 - (-3)) = -1$
• $C_{31} = + \begin{vmatrix} -1 & 2 \\ 2 & -3 \end{vmatrix} = 3 - 4 = -1$
• $C_{32} = - \begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix} = -(-3 - 0) = 3$
• $C_{33} = + \begin{vmatrix} 1 & -1 \\ 0 & 2 \end{vmatrix} = 2 - 0 = 2$

Step 3: Finding the Adjoint Matrix $adj(A)$

The adjoint matrix is the transpose of the cofactor matrix [Per the properties of the Adjoint of a matrix]:

$adj(A) = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix} = \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix}$

Step 4: Calculating $A^{-1}$

Using the formula $A^{-1} = \frac{1}{\det(A)} adj(A)$:

$A^{-1} = \frac{1}{-1} \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix}$
$A^{-1} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$

Final Solution: $A^{-1} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$

Step 1: Evaluation of the Determinant

Given the matrix $A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$, we must first confirm the existence of $A^{-1}$ by calculating the determinant $|A|$. A matrix is invertible if and only if $|A| \neq 0$ [By the Invertibility Criterion].

Expanding along the first row ($R_1$):

$|A| = 1 \begin{vmatrix} 3 & 0 \\ 2 & -1 \end{vmatrix} - 0 \begin{vmatrix} 3 & 0 \\ 5 & -1 \end{vmatrix} + 0 \begin{vmatrix} 3 & 3 \\ 5 & 2 \end{vmatrix}$

$|A| = 1[(3)(-1) - (0)(2)] - 0 + 0$

$|A| = 1[-3 - 0] = -3$

Since $|A| = -3 \neq 0$, the inverse matrix $A^{-1}$ exists.

Step 2: Calculation of the Adjoint Matrix

The inverse is defined as $A^{-1} = \frac{1}{|A|} \text{adj}(A)$. We compute the cofactor matrix $C_{ij} = (-1)^{i+j} M_{ij}$.

• $C_{11} = + \begin{vmatrix} 3 & 0 \\ 2 & -1 \end{vmatrix} = -3$
• $C_{12} = - \begin{vmatrix} 3 & 0 \\ 5 & -1 \end{vmatrix} = -(-3) = 3$
• $C_{13} = + \begin{vmatrix} 3 & 3 \\ 5 & 2 \end{vmatrix} = (6 - 15) = -9$
• $C_{21} = - \begin{vmatrix} 0 & 0 \\ 2 & -1 \end{vmatrix} = 0$
• $C_{22} = + \begin{vmatrix} 1 & 0 \\ 5 & -1 \end{vmatrix} = -1$
• $C_{23} = - \begin{vmatrix} 1 & 0 \\ 5 & 2 \end{vmatrix} = -(2 - 0) = -2$
• $C_{31} = + \begin{vmatrix} 0 & 0 \\ 3 & 0 \end{vmatrix} = 0$
• $C_{32} = - \begin{vmatrix} 1 & 0 \\ 3 & 0 \end{vmatrix} = 0$
• $C_{33} = + \begin{vmatrix} 1 & 0 \\ 3 & 3 \end{vmatrix} = 3$

The cofactor matrix is $C = \begin{bmatrix} -3 & 3 & -9 \\ 0 & -1 & -2 \\ 0 & 0 & 3 \end{bmatrix}$. The adjoint is the transpose of the cofactor matrix, $\text{adj}(A) = C^T$.

$\text{adj}(A) = \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}$

Step 3: Determination of the Inverse

$A^{-1} = \frac{1}{-3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} \frac{-3}{-3} & \frac{0}{-3} & \frac{0}{-3} \\ \frac{3}{-3} & \frac{-1}{-3} & \frac{0}{-3} \\ \frac{-9}{-3} & \frac{-2}{-3} & \frac{3}{-3} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1/3 & 0 \\ 3 & 2/3 & -1 \end{bmatrix}$

Visual Representation of Matrix Transformation

Final Solution: $A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & \frac{1}{3} & 0 \\ 3 & \frac{2}{3} & -1 \end{bmatrix}$

Step 1: Calculating $A^2$

Given the matrix $A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$. To compute $A^2$, we perform matrix multiplication $A \cdot A$:

$A^2 = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$

Performing row-column multiplication [Definition of Matrix Multiplication]:

• Element (1,1): $(3 \times 3) + (2 \times 1) = 9 + 2 = 11$
• Element (1,2): $(3 \times 2) + (2 \times 1) = 6 + 2 = 8$
• Element (2,1): $(1 \times 3) + (1 \times 1) = 3 + 1 = 4$
• Element (2,2): $(1 \times 2) + (1 \times 1) = 2 + 1 = 3$

Thus, $A^2 = \begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix}$.

Step 2: Constructing the Matrix Equation

We are given the equation $A^2 + aA + bI = O$, where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $O$ is the zero matrix $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$. Substituting the known values:

$\begin{bmatrix} 11 & 8 \\ 4 & 3 \end{bmatrix} + a\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} + b\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

Step 3: Forming the System of Linear Equations

By performing scalar multiplication and matrix addition [Properties of Matrix Algebra], we combine the matrices into a single expression:

$\begin{bmatrix} 11 + 3a + b & 8 + 2a + 0 \\ 4 + a + 0 & 3 + a + b \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

Equating corresponding elements, we derive the following equations:

1. $11 + 3a + b = 0$
2. $8 + 2a = 0$
3. $4 + a = 0$
4. $3 + a + b = 0$

Step 4: Solving for $a$ and $b$

From equation (3): $4 + a = 0 \implies a = -4$.

Verify this with equation (2): $8 + 2(-4) = 8 - 8 = 0$. (Consistency confirmed).

Substitute $a = -4$ into equation (4):

$3 + (-4) + b = 0$

$-1 + b = 0 \implies b = 1$.

Verify with equation (1): $11 + 3(-4) + 1 = 11 - 12 + 1 = 0$. (Consistency confirmed).

Visual Representation of Matrix Transformation

Final Solution: $a = -4, b = 1$

Step 1: Establishing the Matrix and Existence Criterion

Let the given matrix be denoted by $A$.

$A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}$

For the inverse $A^{-1}$ to exist, the matrix $A$ must be non-singular, i.e., $|A| \neq 0$ [Per the condition for invertibility of a matrix].

Calculating the determinant $|A|$ by expanding along the first column $C_1$:

$|A| = 1 \begin{vmatrix} 2 & 4 \\ 0 & 5 \end{vmatrix} - 0 \begin{vmatrix} 2 & 3 \\ 0 & 5 \end{vmatrix} + 0 \begin{vmatrix} 2 & 3 \\ 2 & 4 \end{vmatrix}$ $|A| = 1(2 \times 5 - 4 \times 0) = 10$

Since $|A| = 10 \neq 0$, the inverse $A^{-1}$ exists.

Step 2: Calculating the Matrix of Cofactors

The cofactor $C_{ij}$ is defined as $(-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor of element $a_{ij}$.

• $C_{11} = + \begin{vmatrix} 2 & 4 \\ 0 & 5 \end{vmatrix} = 10$
• $C_{12} = - \begin{vmatrix} 0 & 4 \\ 0 & 5 \end{vmatrix} = 0$
• $C_{13} = + \begin{vmatrix} 0 & 2 \\ 0 & 0 \end{vmatrix} = 0$
• $C_{21} = - \begin{vmatrix} 2 & 3 \\ 0 & 5 \end{vmatrix} = -10$
• $C_{22} = + \begin{vmatrix} 1 & 3 \\ 0 & 5 \end{vmatrix} = 5$
• $C_{23} = - \begin{vmatrix} 1 & 2 \\ 0 & 0 \end{vmatrix} = 0$
• $C_{31} = + \begin{vmatrix} 2 & 3 \\ 2 & 4 \end{vmatrix} = (8 - 6) = 2$
• $C_{32} = - \begin{vmatrix} 1 & 3 \\ 0 & 4 \end{vmatrix} = -4$
• $C_{33} = + \begin{vmatrix} 1 & 2 \\ 0 & 2 \end{vmatrix} = 2$

Step 3: Constructing the Adjoint Matrix $adj(A)$

The adjoint matrix is the transpose of the cofactor matrix, $adj(A) = [C_{ij}]^T$ [Per the definition of the Adjoint of a Matrix].

$adj(A) = \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}$

Step 4: Computing the Inverse $A^{-1}$

The inverse is given by the formula $A^{-1} = \frac{1}{|A|} adj(A)$.

$A^{-1} = \frac{1}{10} \begin{bmatrix} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{bmatrix}$

Distributing the scalar $\frac{1}{10}$ into each element:

$A^{-1} = \begin{bmatrix} 1 & -1 & 0.2 \\ 0 & 0.5 & -0.4 \\ 0 & 0 & 0.2 \end{bmatrix}$

Summary Visualization of Matrix $A$

Final Solution: $A^{-1} = \begin{bmatrix} 1 & -1 & 0.2 \\ 0 & 0.5 & -0.4 \\ 0 & 0 & 0.2 \end{bmatrix}$

Step 1: Compute the Product Matrix $AB$

Given $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$.

The product $AB$ is calculated as:

$AB = \begin{bmatrix} (3 \times 6) + (7 \times 7) & (3 \times 8) + (7 \times 9) \\ (2 \times 6) + (5 \times 7) & (2 \times 8) + (5 \times 9) \end{bmatrix}$

$AB = \begin{bmatrix} 18 + 49 & 24 + 63 \\ 12 + 35 & 16 + 45 \end{bmatrix} = \begin{bmatrix} 67 & 87 \\ 47 & 61 \end{bmatrix}$

Step 2: Calculate $(AB)^{-1}$

For a matrix $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $M^{-1} = \frac{1}{\det(M)} \text{adj}(M)$, where $\det(M) = ad - bc$ and $\text{adj}(M) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ [By the Property of Matrix Inverses].

$\det(AB) = (67 \times 61) - (87 \times 47) = 4087 - 4089 = -2$.

$\text{adj}(AB) = \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix}$.

$(AB)^{-1} = -\frac{1}{2} \begin{bmatrix} 61 & -87 \\ -47 & 67 \end{bmatrix} = \begin{bmatrix} -30.5 & 43.5 \\ 23.5 & -33.5 \end{bmatrix}$.

Step 3: Calculate $A^{-1}$ and $B^{-1}$

For $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$, $\det(A) = (3 \times 5) - (7 \times 2) = 15 - 14 = 1$.

$A^{-1} = \frac{1}{1} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$.

For $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$, $\det(B) = (6 \times 9) - (8 \times 7) = 54 - 56 = -2$.

$B^{-1} = -\frac{1}{2} \begin{bmatrix} 9 & -8 \\ -7 & 6 \end{bmatrix} = \begin{bmatrix} -4.5 & 4 \\ 3.5 & -3 \end{bmatrix}$.

Step 4: Calculate $B^{-1} A^{-1}$ and Verify Equality

$B^{-1} A^{-1} = \begin{bmatrix} -4.5 & 4 \\ 3.5 & -3 \end{bmatrix} \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$

$B^{-1} A^{-1} = \begin{bmatrix} (-4.5 \times 5) + (4 \times -2) & (-4.5 \times -7) + (4 \times 3) \\ (3.5 \times 5) + (-3 \times -2) & (3.5 \times -7) + (-3 \times 3) \end{bmatrix}$

$B^{-1} A^{-1} = \begin{bmatrix} -22.5 - 8 & 31.5 + 12 \\ 17.5 + 6 & -24.5 - 9 \end{bmatrix} = \begin{bmatrix} -30.5 & 43.5 \\ 23.5 & -33.5 \end{bmatrix}$

Final Solution: Since $(AB)^{-1} = \begin{bmatrix} -30.5 & 43.5 \\ 23.5 & -33.5 \end{bmatrix}$ and $B^{-1} A^{-1} = \begin{bmatrix} -30.5 & 43.5 \\ 23.5 & -33.5 \end{bmatrix}$, it is verified that $(AB)^{-1} = B^{-1} A^{-1}$.

Given Matrix and Objective

Let $A$ be the given $3 \times 3$ matrix:

$A = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}$

The objective is to find the adjoint of matrix $A$, denoted as $\text{adj}(A)$, which is defined as the transpose of the cofactor matrix $C$, where $C_{ij} = (-1)^{i+j} M_{ij}$. [Per the definition of the Adjoint of a Matrix].

Step 1: Calculating the Cofactors ($C_{ij}$)

We calculate the cofactors for each element $a_{ij}$ of matrix $A$.

• $C_{11} = + \begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix} = (3)(1) - (0)(5) = 3$
• $C_{12} = - \begin{vmatrix} 2 & 5 \\ -2 & 1 \end{vmatrix} = -[(2)(1) - (-2)(5)] = -[2 + 10] = -12$
• $C_{13} = + \begin{vmatrix} 2 & 3 \\ -2 & 0 \end{vmatrix} = (2)(0) - (-2)(3) = 6$
• $C_{21} = - \begin{vmatrix} -1 & 2 \\ 0 & 1 \end{vmatrix} = -[(-1)(1) - (0)(2)] = -[-1] = 1$
• $C_{22} = + \begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix} = (1)(1) - (-2)(2) = 1 + 4 = 5$
• $C_{23} = - \begin{vmatrix} 1 & -1 \\ -2 & 0 \end{vmatrix} = -[(1)(0) - (-2)(-1)] = -[0 - 2] = 2$
• $C_{31} = + \begin{vmatrix} -1 & 2 \\ 3 & 5 \end{vmatrix} = (-1)(5) - (3)(2) = -5 - 6 = -11$
• $C_{32} = - \begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix} = -[(1)(5) - (2)(2)] = -[5 - 4] = -1$
• $C_{33} = + \begin{vmatrix} 1 & -1 \\ 2 & 3 \end{vmatrix} = (1)(3) - (2)(-1) = 3 + 2 = 5$

Step 2: Constructing the Cofactor Matrix and Transposing

The cofactor matrix $C$ is formed by placing the calculated cofactors in their respective positions:

$C = \begin{bmatrix} 3 & -12 & 6 \\ 1 & 5 & 2 \\ -11 & -1 & 5 \end{bmatrix}$

By definition, $\text{adj}(A) = C^T$ (the transpose of the cofactor matrix). We swap the rows and columns [Per matrix transposition rules]:

$\text{adj}(A) = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}$

Visual Representation of Matrix Transformation

Final Solution: $\text{adj}(A) = \begin{bmatrix} 3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5 \end{bmatrix}$