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: Setting up the Determinant Expansion

To evaluate the determinant $D = \begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha \\ -\sin\beta & \cos\beta & 0 \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{vmatrix}$, we perform expansion along the first row ($R_1$).

According to the Laplace expansion formula: $D = a_{11}(C_{11}) + a_{12}(C_{12}) + a_{13}(C_{13})$, where $C_{ij}$ is the cofactor of element $a_{ij}$.

Step 2: Calculating Minor Cofactors

The expansion is given by:

$D = (\cos\alpha\cos\beta) \begin{vmatrix} \cos\beta & 0 \\ \sin\alpha\sin\beta & \cos\alpha \end{vmatrix} - (\cos\alpha\sin\beta) \begin{vmatrix} -\sin\beta & 0 \\ \sin\alpha\cos\beta & \cos\alpha \end{vmatrix} + (-\sin\alpha) \begin{vmatrix} -\sin\beta & \cos\beta \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta \end{vmatrix}$

Step 3: Solving 2x2 Sub-determinants

Evaluating each minor [By determinant definition $ad-bc$]:

• Minor 1: $(\cos\beta)(\cos\alpha) - (0)(\sin\alpha\sin\beta) = \cos\alpha\cos\beta$
• Minor 2: $(-\sin\beta)(\cos\alpha) - (0)(\sin\alpha\cos\beta) = -\cos\alpha\sin\beta$
• Minor 3: $(-\sin\beta)(\sin\alpha\sin\beta) - (\cos\beta)(\sin\alpha\cos\beta) = -\sin\alpha\sin^2\beta - \sin\alpha\cos^2\beta$

Step 4: Algebraic Simplification

Substituting these back into the expression:

$D = (\cos\alpha\cos\beta)(\cos\alpha\cos\beta) - (\cos\alpha\sin\beta)(-\cos\alpha\sin\beta) - (\sin\alpha)(-\sin\alpha\sin^2\beta - \sin\alpha\cos^2\beta)$

$D = \cos^2\alpha\cos^2\beta + \cos^2\alpha\sin^2\beta + \sin^2\alpha(\sin^2\beta + \cos^2\beta)$

Step 5: Applying Trigonometric Identities

[By Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$]:

$D = \cos^2\alpha(\cos^2\beta + \sin^2\beta) + \sin^2\alpha(1)$

$D = \cos^2\alpha(1) + \sin^2\alpha$

$D = 1$

Final Solution: 1

Step 1: Theoretical Framework for Expansion

To evaluate a $3 \times 3$ determinant, we employ the Laplace Expansion method. We expand the determinant $A = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$ along the first row ($R_1$):

$\det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$

[Per the definition of determinant expansion by minors and cofactors].

Step 2: Assigning Given Matrix Elements

Given the matrix:

$A = \begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}$

• $a_{11} = 3, a_{12} = -4, a_{13} = 5$
• $a_{21} = 1, a_{22} = 1, a_{23} = -2$
• $a_{31} = 2, a_{32} = 3, a_{33} = 1$

Step 3: Calculating Minors of Row 1

We calculate the $2 \times 2$ determinants (minors) corresponding to each element of the first row:

Element	Minor Calculation	Result
$a_{11} = 3$	$(1)(1) - (-2)(3) = 1 - (-6) = 7$	$3 \times 7 = 21$
$a_{12} = -4$	$(1)(1) - (-2)(2) = 1 - (-4) = 5$	$-(-4) \times 5 = 20$
$a_{13} = 5$	$(1)(3) - (1)(2) = 3 - 2 = 1$	$5 \times 1 = 5$

Step 4: Final Algebraic Summation

Summing the components derived from the expansion along the first row:

$\Delta = 3[(1)(1) - (-2)(3)] - (-4)[(1)(1) - (-2)(2)] + 5[(1)(3) - (1)(2)]$

$\Delta = 3[1 + 6] + 4[1 + 4] + 5[3 - 2]$

$\Delta = 3[7] + 4[5] + 5[1]$

$\Delta = 21 + 20 + 5$

$\Delta = 46$

Visual Representation of Matrix Indices

Final Solution: 46

Step 1: Matrix Representation of the System

The given system of linear equations is:

$x - y + z = 4$

$2x + y - 3z = 0$

$x + y + z = 2$

This can be expressed in the matrix form $AX = B$, where:

$A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$

Step 2: Determinant of Matrix A

Calculate $|A|$ to determine if the inverse exists [Per the condition for non-singular matrices]:

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

$|A| = 1(4) + 1(5) + 1(1) = 4 + 5 + 1 = 10$

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

Step 3: Calculating the Adjoint of A

Find the cofactor matrix $C = [C_{ij}]$.

• $C_{11} = +(1+3) = 4, C_{12} = -(2+3) = -5, C_{13} = +(2-1) = 1$
• $C_{21} = -(-1-1) = 2, C_{22} = +(1-1) = 0, C_{23} = -(1+1) = -2$
• $C_{31} = +(3-1) = 2, C_{32} = -(-3-2) = 5, C_{33} = +(1+2) = 3$

$\text{adj}(A) = C^T = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$

Step 4: Finding the Inverse and Solving for X

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

$A^{-1} = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$

Now, $X = A^{-1}B$:

$X = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$

$X = \frac{1}{10} \begin{bmatrix} (4 \times 4) + (2 \times 0) + (2 \times 2) \\ (-5 \times 4) + (0 \times 0) + (5 \times 2) \\ (1 \times 4) + (-2 \times 0) + (3 \times 2) \end{bmatrix}$

$X = \frac{1}{10} \begin{bmatrix} 16 + 0 + 4 \\ -20 + 0 + 10 \\ 4 + 0 + 6 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$

Visual Verification (Coordinate Mapping)

Final Solution: $x = 2, y = -1, z = 1$

Step 1: Matrix Representation of the System

The system of linear equations is given by:

• $2x + 3y + 3z = 5$
• $x - 2y + z = -4$
• $3x - y - 2z = 3$

This system can be written in the form $AX = B$, where:

$A = \begin{pmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, $B = \begin{pmatrix} 5 \\ -4 \\ 3 \end{pmatrix}$

Step 2: Determining Invertibility via Determinant Analysis

We calculate the determinant $|A|$ to ensure the inverse $A^{-1}$ exists [Condition for unique solution: $|A| \neq 0$].

$|A| = 2((-2)(-2) - (1)(-1)) - 3((1)(-2) - (1)(3)) + 3((1)(-1) - (-2)(3))$ $|A| = 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6)$ $|A| = 2(5) - 3(-5) + 3(5) = 10 + 15 + 15 = 40$

Since $|A| = 40 \neq 0$, the matrix $A$ is non-singular, and the system has a unique solution $X = A^{-1}B$.

Step 3: Computing the Adjoint of Matrix A

The inverse is calculated as $A^{-1} = \frac{1}{|A|} \text{adj}(A)$. We find the cofactor matrix $C_{ij}$:

$C_{11} = +((-2)(-2) - (-1)(1)) = 5$	$C_{12} = -((1)(-2) - (3)(1)) = 5$	$C_{13} = +((1)(-1) - (3)(-2)) = 5$
$C_{21} = -((3)(-2) - (-1)(3)) = 3$	$C_{22} = +((2)(-2) - (3)(3)) = -13$	$C_{23} = -((2)(-1) - (3)(3)) = 11$
$C_{31} = +((3)(1) - (-2)(3)) = 9$	$C_{32} = -((2)(1) - (1)(3)) = 1$	$C_{33} = +((2)(-2) - (1)(3)) = -7$

Transposing the cofactor matrix gives $\text{adj}(A)$:

$\text{adj}(A) = \begin{pmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{pmatrix}$

Step 4: Calculating the Variable Vector X

Applying $X = \frac{1}{|A|} \text{adj}(A)B$:

$X = \frac{1}{40} \begin{pmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{pmatrix} \begin{pmatrix} 5 \\ -4 \\ 3 \end{pmatrix}$ $X = \frac{1}{40} \begin{pmatrix} (5)(5) + (3)(-4) + (9)(3) \\ (5)(5) + (-13)(-4) + (1)(3) \\ (5)(5) + (11)(-4) + (-7)(3) \end{pmatrix}$ $X = \frac{1}{40} \begin{pmatrix} 25 - 12 + 27 \\ 25 + 52 + 3 \\ 25 - 44 - 21 \end{pmatrix} = \frac{1}{40} \begin{pmatrix} 40 \\ 80 \\ -40 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$

Visual Representation of Matrix Transformation

Final Solution: x = 1, y = 2, z = -1

Step 1: Matrix Representation of the System

The system of linear equations can be represented in the matrix form $AX = B$, where:

$A = \begin{pmatrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{pmatrix}, X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, B = \begin{pmatrix} 5 \\ 2 \\ -1 \end{pmatrix}$

Step 2: Calculating the Determinant of Matrix A

A system of linear equations is consistent if $|A| \neq 0$ (unique solution) or if $|A| = 0$ and $(adj A)B = O$ (infinitely many or no solutions). We first calculate $|A|$:

$|A| = 5 \begin{vmatrix} 3 & 5 \\ -2 & 6 \end{vmatrix} - (-1) \begin{vmatrix} 2 & 5 \\ 5 & 6 \end{vmatrix} + 4 \begin{vmatrix} 2 & 3 \\ 5 & -2 \end{vmatrix}$

$|A| = 5[(3)(6) - (5)(-2)] + 1[(2)(6) - (5)(5)] + 4[(2)(-2) - (3)(5)]$

$|A| = 5[18 + 10] + 1[12 - 25] + 4[-4 - 15]$

$|A| = 5(28) + 1(-13) + 4(-19)$

$|A| = 140 - 13 - 76$

$|A| = 140 - 89 = 51$

[Since $|A| = 51 \neq 0$, the matrix $A$ is non-singular and invertible.]

Step 3: Determining Consistency

According to the theorem of linear systems [Per Cramer's Rule/Matrix Inversion Method]:

• If $|A| \neq 0$, the system is consistent and possesses a unique solution defined by $X = A^{-1}B$.

Since the determinant of the coefficient matrix $A$ is $51$, which is non-zero, the system of equations has a unique solution.

Visual Representation of the System Determinant

Final Solution: The system of equations is consistent because the determinant of the coefficient matrix $A$ is $51$, which is not equal to zero, indicating that the system has a unique solution.

Given Vertices & Coordinate Setup

The vertices of the triangle are provided as follows:

• $A(x_1, y_1) = (-2, -3)$
• $B(x_2, y_2) = (3, 2)$
• $C(x_3, y_3) = (-1, -8)$

Step 1: Establishing the Determinant Formula

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the determinant representation [Per the definition of area in coordinate geometry]:

$\text{Area} = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$

Step 2: Substituting Values into the Matrix

Substituting the given coordinates into the determinant structure:

$\text{Area} = \frac{1}{2} \begin{vmatrix} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{vmatrix}$

Step 3: Calculating the Determinant Value

Expanding the determinant along the first row ($R_1$):

$\Delta = -2 \begin{vmatrix} 2 & 1 \\ -8 & 1 \end{vmatrix} - (-3) \begin{vmatrix} 3 & 1 \\ -1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 3 & 2 \\ -1 & -8 \end{vmatrix}$

Calculating the $2 \times 2$ determinants:

• $\begin{vmatrix} 2 & 1 \\ -8 & 1 \end{vmatrix} = (2)(1) - (-8)(1) = 2 + 8 = 10$
• $\begin{vmatrix} 3 & 1 \\ -1 & 1 \end{vmatrix} = (3)(1) - (-1)(1) = 3 + 1 = 4$
• $\begin{vmatrix} 3 & 2 \\ -1 & -8 \end{vmatrix} = (3)(-8) - (-1)(2) = -24 + 2 = -22$

Substituting back into the expansion:

$\text{Area} = \frac{1}{2} [ -2(10) + 3(4) + 1(-22) ]$

$\text{Area} = \frac{1}{2} [ -20 + 12 - 22 ]$

$\text{Area} = \frac{1}{2} [ -30 ]$

Step 4: Applying Absolute Magnitude

Since the area must be a positive scalar quantity [By definition of spatial area], we take the absolute value of the determinant result:

$\text{Area} = | -15 | = 15$

Visualization of Vertices

Final Solution: 15 \text{ square units}

Theoretical Foundation: Laplace Expansion of a Determinant

The value of a determinant of order 3, denoted as $\Delta$, is defined as the sum of the products of the elements of any row (or column) and their corresponding cofactors. Let $A_{ij}$ represent the cofactor of the element $a_{ij}$. The cofactor $A_{ij}$ is defined as $(-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of the element $a_{ij}$.

Step 1: Formal Definition of the Determinant

By the Laplace Expansion theorem, the value of the determinant $\Delta$ can be computed by expanding along any row $i$ or any column $j$. Specifically, for any row $i$ ($i=1, 2, 3$):

$\Delta = \sum_{j=1}^{3} a_{ij} A_{ij}$

Step 2: Explicit Expansion for Each Row

Based on the structure of the determinant $\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$, the value is derived as follows:

Expansion Reference	Mathematical Expression
Expanding along Row 1	$\Delta = a_{11}A_{11} + a_{12}A_{12} + a_{13}A_{13}$
Expanding along Row 2	$\Delta = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23}$
Expanding along Row 3	$\Delta = a_{31}A_{31} + a_{32}A_{32} + a_{33}A_{33}$

Analytical Note: Properties of Cofactors

It is important to distinguish the expansion formula from the property of cofactors where the sum of products of elements of one row with the cofactors of another row is zero [Per the properties of determinants in Linear Algebra]. Specifically, if $i \neq k$, then $a_{i1}A_{k1} + a_{i2}A_{k2} + a_{i3}A_{k3} = 0$.

Visual Representation of Matrix Indices

Final Solution: The value of $\Delta$ is given by the sum of the product of elements of any row (or column) and their corresponding cofactors: $\Delta = a_{i1}A_{i1} + a_{i2}A_{i2} + a_{i3}A_{i3}$ for any row $i = 1, 2, \text{ or } 3$.

Step 1: Computing the Determinant of Matrix $A$

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

The determinant $|A|$ is calculated by expanding along the first row [Laplace Expansion Theorem]:

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

$|A| = 1(15 - 1) + 2(-10 - 1) + 1(-2 - 3)$

$|A| = 1(14) + 2(-11) + 1(-5) = 14 - 22 - 5 = -13$

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

Step 2: Computing the Adjoint of Matrix $A$

The adjoint, $adj(A)$, is the transpose of the cofactor matrix $C = [C_{ij}]$.

• $C_{11} = +(15-1) = 14$; $C_{12} = -(-10-1) = 11$; $C_{13} = +(-2-3) = -5$
• $C_{21} = -(-10-1) = 11$; $C_{22} = +(5-1) = 4$; $C_{23} = -(1+2) = -3$
• $C_{31} = +(-2-3) = -5$; $C_{32} = -(1+2) = -3$; $C_{33} = +(3-4) = -1$

$adj(A) = \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix}^T = \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix}$

Step 3: Calculating $[adj A]^{-1}$

Let $B = adj(A)$. To find $B^{-1} = \frac{1}{|B|} adj(B)$, we first find $|B|$:

$|B| = 14(-4 - 9) - 11(-11 - 15) - 5(-33 + 20) = 14(-13) - 11(-26) - 5(-13)$

$|B| = -182 + 286 + 65 = 169$

Note that $|adj(A)| = |A|^{n-1} = (-13)^{3-1} = (-13)^2 = 169$ [Property: $|adj(A)| = |A|^{n-1}$].

Computing the adjugate of $B$ (cofactors of $adj(A)$):

$[adj(adj(A))] = \begin{bmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{bmatrix} \implies [adj A]^{-1} = \frac{1}{169} \begin{bmatrix} -13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65 \end{bmatrix} = \begin{bmatrix} -1/13 & 2/13 & -1/13 \\ 2/13 & -3/13 & -1/13 \\ -1/13 & -1/13 & -5/13 \end{bmatrix}$

Step 4: Calculating $adj(A^{-1})$

Using the property $A^{-1} = \frac{1}{|A|} adj(A) = -\frac{1}{13} \begin{bmatrix} 14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1 \end{bmatrix}$:

$A^{-1} = \begin{bmatrix} -14/13 & -11/13 & 5/13 \\ -11/13 & -4/13 & 3/13 \\ 5/13 & 3/13 & 1/13 \end{bmatrix}$

The inverse of a matrix $X$ is $\frac{1}{|X|}adj(X)$. For a $3 \times 3$ matrix, $adj(A^{-1}) = |A^{-1}| (A^{-1})^{-1}$.

Since $(A^{-1})^{-1} = A$ and $|A^{-1}| = \frac{1}{|A|} = -\frac{1}{13}$:

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

Comparison Table

Element	$[adj A]^{-1}$	$adj(A^{-1})$
Result	$\begin{bmatrix} -1/13 & 2/13 & -1/13 \\ 2/13 & -3/13 & -1/13 \\ -1/13 & -1/13 & -5/13 \end{bmatrix}$	$\begin{bmatrix} -1/13 & 2/13 & -1/13 \\ 2/13 & -3/13 & -1/13 \\ -1/13 & -1/13 & -5/13 \end{bmatrix}$

Final Solution: Since the matrices are identical, the identity $[adj A]^{-1} = adj (A^{-1})$ is verified.

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

Given Determinant and Structural Identification

The given determinant is $A = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = \begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$.

The elements are identified as follows:

• $a_{11} = 2$
• $a_{12} = -4$
• $a_{21} = 0$
• $a_{22} = 3$

Step 1: Calculating Minors ($M_{ij}$)

The minor $M_{ij}$ of an element $a_{ij}$ is the determinant of the $1 \times 1$ matrix obtained by deleting the $i^{th}$ row and $j^{th}$ column [Per definition of minors in determinants].

• $M_{11} = |a_{22}| = |3| = 3$
• $M_{12} = |a_{21}| = |0| = 0$
• $M_{21} = |a_{12}| = |-4| = -4$
• $M_{22} = |a_{11}| = |2| = 2$

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

The cofactor $C_{ij}$ of an element $a_{ij}$ is defined as $C_{ij} = (-1)^{i+j} M_{ij}$ [By definition of cofactor expansion].

• $C_{11} = (-1)^{1+1} M_{11} = (-1)^2 (3) = 1 \times 3 = 3$
• $C_{12} = (-1)^{1+2} M_{12} = (-1)^3 (0) = -1 \times 0 = 0$
• $C_{21} = (-1)^{2+1} M_{21} = (-1)^3 (-4) = -1 \times (-4) = 4$
• $C_{22} = (-1)^{2+2} M_{22} = (-1)^4 (2) = 1 \times 2 = 2$

Visual Representation of Matrix Indices

Data Synthesis

Element	Minor ($M_{ij}$)	Cofactor ($C_{ij}$)
$a_{11} = 2$	$3$	$3$
$a_{12} = -4$	$0$	$0$
$a_{21} = 0$	$-4$	$4$
$a_{22} = 3$	$2$	$2$

Final Solution: The minors are $M_{11}=3, M_{12}=0, M_{21}=-4, M_{22}=2$ and the corresponding cofactors are $C_{11}=3, C_{12}=0, C_{21}=4, C_{22}=2$.

Step 1: Establishing the Coordinate Geometry Formula

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the determinant formula:

$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

[By the Determinant Definition of Area of a Triangle]

Step 2: Substitution of Given Variables

Given vertices are $(x_1, y_1) = (2, -6)$, $(x_2, y_2) = (5, 4)$, and $(x_3, y_3) = (k, 4)$. The area is given as $35$ sq units.

Substituting into the determinant form:

$35 = \pm \frac{1}{2} \begin{vmatrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{vmatrix}$

[Note: Since area is an absolute value, we must account for both positive and negative results when removing the modulus bars.]

Step 3: Evaluating the Determinant

Expand along the first row:

$\Delta = 2(4(1) - 4(1)) - (-6)(5(1) - k(1)) + 1(5(4) - k(4))$

$\Delta = 2(0) + 6(5 - k) + (20 - 4k)$

$\Delta = 30 - 6k + 20 - 4k$

$\Delta = 50 - 10k$

Step 4: Solving for $k$

From the formula $\text{Area} = \frac{1}{2} |\Delta|$:

$35 = \frac{1}{2} |50 - 10k|$

$70 = |50 - 10k|$

This yields two linear equations:

• Case 1: $50 - 10k = 70$
  $-10k = 20 \implies k = -2$
• Case 2: $50 - 10k = -70$
  $-10k = -120 \implies k = 12$

Step 5: Visual Representation

Final Solution: The possible values for $k$ are $12$ or $-2$.

Step 1: Analyzing the Matrix Structure

The given matrix $A$ is a diagonal matrix of the form:

$A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$

where $x, y, z \neq 0$. [By Definition of a Diagonal Matrix, all non-diagonal elements are zero].

Step 2: Calculating the Determinant

To determine if the inverse exists, we calculate the determinant $|A|$ [Per the Invertibility Criterion, $A^{-1}$ exists if $|A| \neq 0$].

$|A| = \begin{vmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{vmatrix}$

Expanding along the first row:

$|A| = x(yz - 0) - 0(0 - 0) + 0(0 - 0) = xyz$

Since $x, y, z \neq 0$, it follows that $xyz \neq 0$, thus $A$ is non-singular and invertible.

Step 3: Finding the Adjoint Matrix

The adjoint of a matrix, $\text{adj}(A)$, is the transpose of the cofactor matrix $C_{ij}$, where $C_{ij} = (-1)^{i+j}M_{ij}$ [By Definition of Adjoint].

The cofactors are:

• $C_{11} = +(yz - 0) = yz$
• $C_{12} = -(0 - 0) = 0$
• $C_{13} = +(0 - 0) = 0$
• $C_{21} = -(0 - 0) = 0$
• $C_{22} = +(xz - 0) = xz$
• $C_{23} = -(0 - 0) = 0$
• $C_{31} = +(0 - 0) = 0$
• $C_{32} = -(0 - 0) = 0$
• $C_{33} = +(xy - 0) = xy$

Therefore, the cofactor matrix is:

$C = \begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix}$

Since $C$ is symmetric, $\text{adj}(A) = C^T = \begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix}$.

Step 4: Computing the Inverse Matrix

The inverse is calculated using the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$ [By Inverse Matrix Theorem].

$A^{-1} = \frac{1}{xyz} \begin{bmatrix} yz & 0 & 0 \\ 0 & xz & 0 \\ 0 & 0 & xy \end{bmatrix}$

Distributing the scalar $\frac{1}{xyz}$ into the matrix:

$A^{-1} = \begin{bmatrix} \frac{yz}{xyz} & 0 & 0 \\ 0 & \frac{xz}{xyz} & 0 \\ 0 & 0 & \frac{xy}{xyz} \end{bmatrix} = \begin{bmatrix} \frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z} \end{bmatrix}$

Final Solution: $A^{-1} = \begin{bmatrix} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & z^{-1} \end{bmatrix}$

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

Step 1: Evaluation of the Determinant on the Left-Hand Side (LHS)

For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is defined as $|A| = ad - bc$ [By the definition of the determinant of a $2 \times 2$ matrix].

Given the LHS: $\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}$

Calculation: $(2 \times 1) - (4 \times 5)$

$= 2 - 20$

$= -18$

Step 2: Evaluation of the Determinant on the Right-Hand Side (RHS)

Given the RHS: $\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$

Calculation: $(2x \times x) - (4 \times 6)$

$= 2x^2 - 24$

Step 3: Equating LHS and RHS to Form the Algebraic Equation

According to the problem statement, the determinants are equal:

$-18 = 2x^2 - 24$

Add $24$ to both sides [Addition Property of Equality]:

$-18 + 24 = 2x^2$

$6 = 2x^2$

Step 4: Solving for $x$

Divide both sides by $2$:

$\frac{6}{2} = x^2$

$3 = x^2$

Apply the square root property: $x = \pm\sqrt{3}$

Final Solution: $x = \pm\sqrt{3}$

Step 1: Matrix Representation of the System

Given the system of linear equations:

• $x - y + 2z = 7$
• $3x + 4y - 5z = -5$
• $2x - y + 3z = 12$

The system can be represented as $AX = B$, where:

$A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}$

Step 2: Verification of Non-Singularity

We calculate the determinant $|A|$ to determine invertibility [Per the Invertibility Criterion for Square Matrices]:

$|A| = 1(4 \cdot 3 - (-5)(-1)) - (-1)(3 \cdot 3 - (-5)(2)) + 2(3(-1) - 4 \cdot 2)$ $|A| = 1(12 - 5) + 1(9 + 10) + 2(-3 - 8)$ $|A| = 1(7) + 1(19) + 2(-11) = 7 + 19 - 22 = 4$

Since $|A| = 4 \neq 0$, $A^{-1}$ exists, and the system has a unique solution.

Step 3: Determination of the Adjoint Matrix

To find $A^{-1}$, we compute the cofactor matrix $C_{ij}$.

$C_{11} = +(12-5) = 7, \quad C_{12} = -(9+10) = -19, \quad C_{13} = +(-3-8) = -11$ $C_{21} = -(-3+2) = 1, \quad C_{22} = +(3-4) = -1, \quad C_{23} = -(-1+2) = -1$ $C_{31} = +(5-8) = -3, \quad C_{32} = -(-5-6) = 11, \quad C_{33} = +(4+3) = 7$

The Adjoint matrix $Adj(A)$ is the transpose of the cofactor matrix:

$Adj(A) = \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}$

Step 4: Calculation of Inverse and Solution Matrix

Per $A^{-1} = \frac{1}{|A|} Adj(A)$:

$A^{-1} = \frac{1}{4} \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}$

The solution is $X = A^{-1}B$:

$X = \frac{1}{4} \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix} \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}$ $x = \frac{1}{4} [7(7) + 1(-5) + (-3)(12)] = \frac{1}{4} [49 - 5 - 36] = \frac{8}{4} = 2$ $y = \frac{1}{4} [-19(7) + (-1)(-5) + 11(12)] = \frac{1}{4} [-133 + 5 + 132] = \frac{4}{4} = 1$ $z = \frac{1}{4} [-11(7) + (-1)(-5) + 7(12)] = \frac{1}{4} [-77 + 5 + 84] = \frac{12}{4} = 3$

Visual Representation of System Geometry

Final Solution: x = 2, y = 1, z = 3

Given Determinant and Structural Analysis

We are given the determinant:

$\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}$

The objective is to evaluate $\Delta$ by expanding along the third column ($C_3$). The elements of $C_3$ are $a_{13} = yz$, $a_{23} = zx$, and $a_{33} = xy$.

Step 1: Calculating Cofactors of the Third Column

The cofactor $C_{ij}$ of an element $a_{ij}$ is defined as $(-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor obtained by deleting the $i$-th row and $j$-th column. [Per definition of Cofactor in Determinant Theory].

For $a_{13} = yz$:

$C_{13} = (-1)^{1+3} \begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix} = (1)(z - y) = z - y$

For $a_{23} = zx$:

$C_{23} = (-1)^{2+3} \begin{vmatrix} 1 & x \\ 1 & z \end{vmatrix} = -(1)(z - x) = x - z$

For $a_{33} = xy$:

$C_{33} = (-1)^{3+3} \begin{vmatrix} 1 & x \\ 1 & y \end{vmatrix} = (1)(y - x) = y - x$

Step 2: Expansion along the Third Column

The determinant is evaluated as $\Delta = a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33}$. [Per Laplace Expansion Theorem].

$\Delta = (yz)(z - y) + (zx)(x - z) + (xy)(y - x)$

Step 3: Algebraic Simplification

Expanding the terms:

$\Delta = yz^2 - y^2z + zx^2 - z^2x + xy^2 - x^2y$

Group terms to facilitate factorization:

$\Delta = -(x^2y - xy^2) + (zx^2 - z^2x) + (yz^2 - y^2z)$

$\Delta = -xy(x - y) + zx(x - z) + yz(z - y)$

Continuing the expansion to reach the standard cyclic form:

$\Delta = yz^2 - y^2z + zx^2 - z^2x + xy^2 - x^2y$

Rearranging for factoring: $(x-y)(y-z)(z-x)$

$\Delta = x^2y - x^2z - xy^2 + y^2z + xz^2 - yz^2$

Factoring step-by-step results in: $\Delta = -(x - y)(y - z)(z - x)$

Visual Representation of Matrix Components

Final Solution: $\Delta = -(x - y)(y - z)(z - x) = (x - y)(y - z)(x - z)$ or $(x - y)(y - z)(z - x)$ expressed as $-(x-y)(y-z)(z-x)$

Step 1: Evaluation of the Determinant on the Left-Hand Side (LHS)

To evaluate a $2 \times 2$ determinant of the form $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$, we utilize the standard formula: $ad - bc$ [By definition of the determinant of a $2 \times 2$ matrix].

For the LHS determinant $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}$:

LHS $= (2 \times 5) - (3 \times 4)$

LHS $= 10 - 12$

LHS $= -2$

Step 2: Evaluation of the Determinant on the Right-Hand Side (RHS)

Applying the same determinant property to the RHS $\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$:

RHS $= (x \times 5) - (3 \times 2x)$

RHS $= 5x - 6x$

RHS $= -x$

Step 3: Equating LHS and RHS to solve for $x$

Given the equation: $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$

Substituting the results from Step 1 and Step 2 into the equation:

$-2 = -x$

Step 4: Algebraic Simplification

Multiply both sides of the equation by $-1$ [By the property of equality]:

$x = 2$

Final Solution: $x = 2$

Step 1: Definition of Given Matrix and Determinant Calculation

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

To calculate the determinant of $A$, denoted as $|A|$, we apply the formula for a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ where $|A| = ad - bc$ [Per the definition of the determinant of a square matrix].

$|A| = (1 \times 2) - (2 \times 4)$

$|A| = 2 - 8$

$|A| = -6$

Step 2: Calculating 4|A|

Multiply the result obtained in Step 1 by the scalar $4$:

$4|A| = 4 \times (-6)$

$4|A| = -24$

Step 3: Evaluating the Matrix 2A

To find $2A$, we multiply every element of matrix $A$ by the scalar $2$ [Per the definition of scalar multiplication of a matrix]:

$2A = 2 \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 2 \times 1 & 2 \times 2 \\ 2 \times 4 & 2 \times 2 \end{bmatrix}$

$2A = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$

Step 4: Calculating the Determinant |2A|

Using the determinant formula for the resultant matrix $2A$:

$|2A| = \begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix}$

$|2A| = (2 \times 4) - (4 \times 8)$

$|2A| = 8 - 32$

$|2A| = -24$

Step 5: Verification and Conclusion

Comparing the results from Step 2 and Step 4:

Expression	Result
$|2A|$	$-24$
$4|A|$	$-24$

Since $|2A| = -24$ and $4|A| = -24$, it is demonstrated that $|2A| = 4|A|$. This is a specific application of the property $|kA| = k^n|A|$, where $n$ is the order of the square matrix [Per the properties of determinants for a matrix of order $n=2$].

Final Solution: $|2A| = 4|A| = -24$