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.

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

Step 1: Matrix Representation of the System

The given system of linear equations is:

$4x - 3y = 3$

$3x - 5y = 7$

We represent this system in the matrix form $AX = B$, where:

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

Step 2: Determinant Calculation and Non-Singularity Check

To determine if the system has a unique solution, we calculate the determinant of matrix $A$ [Per the property of inverse matrices, a unique solution exists if $|A| \neq 0$].

$|A| = \det\begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix} = (4)(-5) - (-3)(3)$

$|A| = -20 - (-9) = -20 + 9 = -11$

Since $|A| \neq 0$, matrix $A$ is non-singular and invertible. The solution is given by $X = A^{-1}B$.

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

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

$adj(A) = \begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix}$

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

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

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

Step 5: Computing the Variable Vector $X$

$X = A^{-1}B = \begin{bmatrix} \frac{5}{11} & -\frac{3}{11} \\ \frac{3}{11} & -\frac{4}{11} \end{bmatrix} \begin{bmatrix} 3 \\ 7 \end{bmatrix}$

Performing matrix multiplication:

$x = (\frac{5}{11})(3) + (-\frac{3}{11})(7) = \frac{15}{11} - \frac{21}{11} = -\frac{6}{11}$

$y = (\frac{3}{11})(3) + (-\frac{4}{11})(7) = \frac{9}{11} - \frac{28}{11} = -\frac{19}{11}$

Final Solution: $x = -\frac{6}{11}, y = -\frac{19}{11}$

Step 1: Defining Variables and System of Equations

Let $x$, $y$, and $z$ represent the cost per kg (in ₹) of onion, wheat, and rice, respectively. Based on the provided information, we formulate the following system of linear equations:

• $4x + 3y + 2z = 60$
• $2x + 4y + 6z = 90$
• $6x + 2y + 3z = 70$

Step 2: Matrix Representation

The system is expressed as $AX = B$, where:

$A = \begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, $B = \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix}$

Step 3: Determinant Calculation

We calculate $|A|$ to ensure the matrix is invertible [Per the condition $AX=B \implies X=A^{-1}B$]:

$|A| = 4(4 \times 3 - 6 \times 2) - 3(2 \times 3 - 6 \times 6) + 2(2 \times 2 - 4 \times 6)$

$|A| = 4(12 - 12) - 3(6 - 36) + 2(4 - 24) = 4(0) - 3(-30) + 2(-20) = 0 + 90 - 40 = 50$

Since $|A| = 50 \neq 0$, $A^{-1}$ exists.

Step 4: Adjoint Matrix Calculation ($Adj A$)

We find the cofactor matrix $C_{ij}$ and transpose it to obtain $Adj A$.

Element	Cofactor Calculation	Value
$C_{11}$	$+(12-12)$	0
$C_{12}$	$-(6-36)$	30
$C_{13}$	$+(4-24)$	-20
$C_{21}$	$-(9-4)$	-5
$C_{22}$	$+(12-12)$	0
$C_{23}$	$-(8-18)$	10
$C_{31}$	$+(18-8)$	10
$C_{32}$	$-(24-4)$	-20
$C_{33}$	$+(16-6)$	10

$Adj A = \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}$

Step 5: Solving for $X$

$X = A^{-1}B = \frac{1}{|A|} Adj A \cdot B$

$X = \frac{1}{50} \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix} \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix}$

$X = \frac{1}{50} \begin{bmatrix} (0)(60) + (-5)(90) + (10)(70) \\ (30)(60) + (0)(90) + (-20)(70) \\ (-20)(60) + (10)(90) + (10)(70) \end{bmatrix} = \frac{1}{50} \begin{bmatrix} 250 \\ 400 \\ 400 \end{bmatrix}$

$X = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix}$

Final Solution: The cost of onion is ₹ 5 per kg, wheat is ₹ 8 per kg, and rice is ₹ 8 per kg.

Step 1: Representing the System as a Matrix Equation

The given system of linear equations is:

• $2x + y + z = 1$
• $x - 2y - z = 1.5$
• $0x + 3y - 5z = 9$

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

$A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{pmatrix}, X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, B = \begin{pmatrix} 1 \\ 1.5 \\ 9 \end{pmatrix}$

Step 2: Determining Invertibility via Determinant Calculation

Calculate $|A|$ by expanding along the first row [Per Laplace Expansion]:

$|A| = 2((-2)(-5) - (-1)(3)) - 1((1)(-5) - (-1)(0)) + 1((1)(3) - (-2)(0))$ $|A| = 2(10 + 3) - 1(-5 - 0) + 1(3 - 0)$ $|A| = 2(13) + 5 + 3 = 26 + 8 = 34$

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

Step 3: Calculating the Adjoint of Matrix A

Find the matrix of cofactors $C_{ij}$:

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

The Adjoint matrix is the transpose of the cofactor matrix [Per Definition of Adjoint]:

$adj(A) = \begin{pmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{pmatrix}$

Step 4: Computing the Inverse and Final Solution

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

$X = \frac{1}{34} \begin{pmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{pmatrix} \begin{pmatrix} 1 \\ 1.5 \\ 9 \end{pmatrix}$

Perform the matrix multiplication:

$x = \frac{1}{34} (13(1) + 8(1.5) + 1(9)) = \frac{1}{34} (13 + 12 + 9) = \frac{34}{34} = 1$ $y = \frac{1}{34} (5(1) - 10(1.5) + 3(9)) = \frac{1}{34} (5 - 15 + 27) = \frac{17}{34} = 0.5$ $z = \frac{1}{34} (3(1) - 6(1.5) - 5(9)) = \frac{1}{34} (3 - 9 - 45) = \frac{-51}{34} = -1.5$

Final Solution: $x = 1, y = \frac{1}{2}, z = -\frac{3}{2}$

Step 1: Representation in Matrix Form

The given system of linear equations is:

$5x + 2y = 3$

$3x + 2y = 5$

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

$A = \begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, $B = \begin{bmatrix} 3 \\ 5 \end{bmatrix}$

Step 2: Verification of Matrix Invertibility

To determine if $X = A^{-1}B$ exists, we must calculate the determinant of $A$ [Per the property of non-singular matrices].

$|A| = (5 \times 2) - (2 \times 3) = 10 - 6 = 4$

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

Step 3: Determination of Adjoint Matrix

The adjoint of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

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

Step 4: Calculation of Inverse Matrix

The inverse is defined as $A^{-1} = \frac{1}{|A|} adj(A)$ [By the definition of matrix inversion].

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

Step 5: Solving for X

Applying the formula $X = A^{-1}B$:

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

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

$X = \frac{1}{4} \begin{bmatrix} 6 - 10 \\ -9 + 25 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} -4 \\ 16 \end{bmatrix}$

$X = \begin{bmatrix} -1 \\ 4 \end{bmatrix}$

Geometric Visualization

Final Solution: $x = -1, y = 4$

Step 1: Calculating the Determinant $|A|$

For the matrix $A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$, we expand along the first row:

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

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

$|A| = 2(0) + 3(-2) + 5(1) = 0 - 6 + 5 = -1$

[Since $|A| \neq 0$, $A^{-1}$ exists.]

Step 2: Finding the Adjoint Matrix $adj(A)$

Calculate the cofactor matrix $C_{ij}$:

• $C_{11} = +(-4 - (-4)) = 0$; $C_{12} = -(-6 - (-4)) = 2$; $C_{13} = +(3 - 2) = 1$
• $C_{21} = -(6 - 5) = -1$; $C_{22} = +(-4 - 5) = -9$; $C_{23} = -(2 - (-3)) = -5$
• $C_{31} = +(12 - 10) = 2$; $C_{32} = -(-8 - 15) = 23$; $C_{33} = +(4 - (-9)) = 13$

$adj(A) = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix} = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}$

Step 3: Calculating Inverse $A^{-1}$

$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-1} \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}$

Step 4: Solving the System of Equations

The system is $AX = B$, where $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$.

The solution is given by $X = A^{-1}B$ [Per Matrix Algebra theorem for non-singular systems]:

$X = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$

Step 5: Matrix Multiplication

$x = (0)(11) + (1)(-5) + (-2)(-3) = 0 - 5 + 6 = 1$

$y = (-2)(11) + (9)(-5) + (-23)(-3) = -22 - 45 + 69 = 2$

$z = (-1)(11) + (5)(-5) + (-13)(-3) = -11 - 25 + 39 = 3$

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

Step 1: Representing the System in Matrix Form

A system of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ can be expressed as $AX = B$, where:

$A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $B = \begin{bmatrix} 5 \\ 8 \end{bmatrix}$

Step 2: Calculating the Determinant of Matrix $A$

To determine consistency, we must evaluate $\det(A)$ or $|A|$. [Per the property of consistency: If $|A| \neq 0$, the system is consistent with a unique solution. If $|A| = 0$, the system may be inconsistent or have infinitely many solutions.]

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

$|A| = 6 - 6 = 0$

Step 3: Evaluating Consistency via the Adjoint Matrix Method

Since $|A| = 0$, we must compute the matrix $adj(A) \times B$. [Condition for consistency when $|A| = 0$: If $(adj A)B = O$, the system is consistent (infinitely many solutions); if $(adj A)B \neq O$, the system is inconsistent.]

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

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

Now, calculate $(adj A)B$:

$(adj A)B = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ 8 \end{bmatrix} = \begin{bmatrix} (6 \times 5) + (-3 \times 8) \\ (-2 \times 5) + (1 \times 8) \end{bmatrix}$

$(adj A)B = \begin{bmatrix} 30 - 24 \\ -10 + 8 \end{bmatrix} = \begin{bmatrix} 6 \\ -2 \end{bmatrix}$

Step 4: Conclusion of Consistency

Since $(adj A)B \neq \begin{bmatrix} 0 \\ 0 \end{bmatrix}$, the system of equations has no solution. Geometrically, the two lines represented by the equations are parallel and non-coincident.

Final Solution: The system of equations is inconsistent.

Step 1: Representing the System in Matrix Form

A system of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ can be expressed in the matrix equation form $AX = B$, where:

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

[Per the matrix representation of linear systems $AX = B$].

Step 2: Calculating the Determinant of Matrix A

To determine consistency, we must evaluate the determinant $|A|$:

$|A| = \det \begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix}$

$|A| = (2)(1) - (-1)(1)$ [By the definition of a $2 \times 2$ determinant: $ad - bc$]

$|A| = 2 + 1 = 3$

Step 3: Analyzing Consistency Criteria

A system $AX = B$ is consistent if $|A| \neq 0$ or if $|A| = 0$ and $(\text{adj } A)B = O$.

• Since $|A| = 3$, and $3 \neq 0$, the matrix $A$ is non-singular.
• A non-singular matrix implies the existence of a unique solution given by $X = A^{-1}B$.

[Per the theorem: If $|A| \neq 0$, the system of equations is consistent and has a unique solution].

Step 4: Geometric Representation (Visual Verification)

Step 5: Theoretical Conclusion

Because the determinant of the coefficient matrix is non-zero ($|A| = 3 \neq 0$), the system is algebraically guaranteed to possess a unique solution. Consequently, the system of equations is classified as consistent.

Final Solution: The system of equations is consistent as $|A| = 3 \neq 0$.

Step 1: Matrix Representation of the System

The system of linear equations is given by:

• $x + y + z = 1$
• $2x + 3y + 2z = 2$
• $ax + ay + 2az = 4$

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

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

Step 2: Calculating the Determinant of Matrix A

To determine consistency, we must evaluate $|A|$ [Per Cramer’s Rule and the property of non-singular matrices].

$|A| = 1(3 \cdot 2a - 2 \cdot a) - 1(2 \cdot 2a - 2 \cdot a) + 1(2 \cdot a - 3 \cdot a)$

$|A| = 1(6a - 2a) - 1(4a - 2a) + 1(2a - 3a)$

$|A| = 4a - 2a - a = a$

Step 3: Evaluating Consistency Conditions

A system is consistent if $|A| \neq 0$ (unique solution) or if $|A| = 0$ and $(adj A)B = O$ (infinitely many solutions).

Condition	Result
If $a \neq 0$	$|A| \neq 0$, the system has a unique solution and is consistent.
If $a = 0$	$|A| = 0$, we must check $(adj A)B$.

Step 4: Analysis for $a = 0$

If $a = 0$, the matrix $A$ becomes:

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

The system of equations becomes:

• $x + y + z = 1$
• $2x + 3y + 2z = 2$
• $0x + 0y + 0z = 4 \implies 0 = 4$

Since $0 = 4$ is a mathematical contradiction [By the principle of logical consistency], there are no values of $x, y, z$ that satisfy the third equation.

Step 5: Final Conclusion

The system of equations is consistent for all $a \in \mathbb{R} \setminus \{0\}$. For $a = 0$, the system is inconsistent as it leads to a contradictory statement.

Final Solution: The system is consistent if $a \neq 0$ and inconsistent if $a = 0$.

Step 1: Matrix Representation of the System

The given system of linear equations is:

$2x - y = -2$

$3x + 4y = 3$

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

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

Step 2: Calculating the Determinant of Matrix A

To determine if the system has a unique solution, calculate $|A|$ [Per the Invertible Matrix Theorem]:

$|A| = \det\begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} = (2 \times 4) - (-1 \times 3)$

$|A| = 8 + 3 = 11$

Since $|A| \neq 0$, the matrix $A$ is non-singular and therefore invertible. The unique solution exists and is given by $X = A^{-1}B$.

Step 3: Determining the Adjoint of Matrix A

For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the adjoint is $\text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ [By definition of Adjoint Matrix].

$\text{adj}(A) = \begin{pmatrix} 4 & 1 \\ -3 & 2 \end{pmatrix}$

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

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

$A^{-1} = \frac{1}{11} \begin{pmatrix} 4 & 1 \\ -3 & 2 \end{pmatrix}$

Step 5: Solving for X

Using $X = A^{-1}B$:

$X = \frac{1}{11} \begin{pmatrix} 4 & 1 \\ -3 & 2 \end{pmatrix} \begin{pmatrix} -2 \\ 3 \end{pmatrix}$

$X = \frac{1}{11} \begin{pmatrix} (4 \times -2) + (1 \times 3) \\ (-3 \times -2) + (2 \times 3) \end{pmatrix}$

$X = \frac{1}{11} \begin{pmatrix} -8 + 3 \\ 6 + 6 \end{pmatrix} = \frac{1}{11} \begin{pmatrix} -5 \\ 12 \end{pmatrix}$

$X = \begin{pmatrix} -5/11 \\ 12/11 \end{pmatrix}$

Final Solution: $x = -\frac{5}{11}, y = \frac{12}{11}$

Step 1: Representing the System as a Matrix Equation

The given system of linear equations is:

$5x + 2y = 4$

$7x + 3y = 5$

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

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

[Per the matrix representation of linear systems $AX = B$]

Step 2: Determining Invertibility via Determinant Analysis

To solve for $X$, we must calculate $|A|$ to ensure the matrix is non-singular. [Per the property that $X = A^{-1}B$ exists if $|A| \neq 0$]

$|A| = (5 \times 3) - (2 \times 7)$

$|A| = 15 - 14 = 1$

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

Step 3: Calculating the Adjoint and Inverse of Matrix A

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

$\text{adj}(A) = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$

The inverse is defined as $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.

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

Step 4: Solving for Variables X

Using the formula $X = A^{-1}B$:

$X = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} \begin{bmatrix} 4 \\ 5 \end{bmatrix}$

Performing matrix multiplication:

$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} (3 \times 4) + (-2 \times 5) \\ (-7 \times 4) + (5 \times 5) \end{bmatrix}$

$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 12 - 10 \\ -28 + 25 \end{bmatrix}$

$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}$

Visual Representation of Linear Intersections

Final Solution: $x = 2$, $y = -3$

Step 1: Representing the System in Matrix Form

The given system of linear equations is:

$x + 2y = 2$

$2x + 3y = 3$

This system can be represented as the matrix equation $AX = B$, where:

$A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $B = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$

[Per the matrix representation of linear systems $AX=B$]

Step 2: Calculating the Determinant of Matrix A

To determine the consistency of the system, we must evaluate the determinant $|A|$. A system is consistent if $|A| \neq 0$ (unique solution) or if $|A| = 0$ and $(adj A)B = O$ (infinitely many solutions).

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

$|A| = (1 \times 3) - (2 \times 2)$

$|A| = 3 - 4 = -1$

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

Step 3: Analytical Inference

Since $|A| = -1$, it follows that $|A| \neq 0$.

[By the Non-Singular Matrix Theorem: If the determinant of the coefficient matrix is non-zero, the matrix $A$ is invertible ($A^{-1}$ exists), and the system possesses a unique solution defined by $X = A^{-1}B$.]

Step 4: Visualizing the System Geometry

Step 5: Final Conclusion

Because $|A| \neq 0$, the matrix $A$ is non-singular. Consequently, the system of equations is consistent and has a unique solution.

Final Solution: The system of equations is consistent.

Step 1: Matrix Representation of the System

The system of linear equations is given by:

• $3x - y - 2z = 2$
• $0x + 2y - z = -1$
• $3x - 5y + 0z = 3$

We represent the system in the form $AX = B$, where:

$A = \begin{bmatrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$

Step 2: Calculating the Determinant of Matrix A

To determine consistency, we first evaluate $|A|$ [Per the property of linear systems: if $|A| \neq 0$, the system is consistent with a unique solution].

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

$|A| = 3(0 - 5) + 1(0 - (-3)) - 2(0 - 6)$
$|A| = 3(-5) + 1(3) - 2(-6)$
$|A| = -15 + 3 + 12$
$|A| = 0$

Step 3: Analyzing Consistency when $|A| = 0$

Since $|A| = 0$, the system is either inconsistent (no solution) or consistent with infinitely many solutions. We must calculate $Adj(A) \cdot B$ [Per the criteria for singularity in linear systems].

First, find the cofactor matrix $C_{ij}$:

• $C_{11} = +(0-5) = -5$
• $C_{12} = -(0 - (-3)) = -3$
• $C_{13} = +(0-6) = -6$
• $C_{21} = -(0 - 10) = 10$
• $C_{22} = +(0 - (-6)) = 6$
• $C_{23} = -(-15 - (-3)) = 12$
• $C_{31} = +(1 - (-4)) = 5$
• $C_{32} = -(-3 - 0) = 3$
• $C_{33} = +(6 - 0) = 6$

$Adj(A) = \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix}$

Step 4: Evaluating $(Adj A) \cdot B$

$(Adj A) \cdot B = \begin{bmatrix} -5 & 10 & 5 \\ -3 & 6 & 3 \\ -6 & 12 & 6 \end{bmatrix} \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$
$= \begin{bmatrix} (-5)(2) + (10)(-1) + (5)(3) \\ (-3)(2) + (6)(-1) + (3)(3) \\ (-6)(2) + (12)(-1) + (6)(3) \end{bmatrix} = \begin{bmatrix} -10 - 10 + 15 \\ -6 - 6 + 9 \\ -12 - 12 + 18 \end{bmatrix} = \begin{bmatrix} -5 \\ -3 \\ -6 \end{bmatrix}$

Since $(Adj A) \cdot B \neq \mathbf{0}$, the system has no solution [By the consistency criteria for $|A|=0$].

Visual Representation of Determinant Space

Final Solution: The system of equations is inconsistent.