Question

If matrix A = [■(1&2&0@-1&1&1@2&-1&3)] , find A-1 and use it to solve the following system of linear equations: x+2y =4, -x+y+z=1, 2x-y+3z=7

Answer 1

x = 12/7, y = 8/7, and z = 11/7

Answer 2

To find the inverse of matrix A and use it to solve the system of linear equations, we follow these steps:

Part 1: Find A⁻¹

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

Step 1: Calculate the determinant of A (det(A)).

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

det(A) = $1 \times (1 \times 3 - 1 \times (-1)) - 2 \times ((-1) \times 3 - 1 \times 2) + 0$

det(A) = $1 \times (3 + 1) - 2 \times (-3 - 2)$

det(A) = $1 \times 4 - 2 \times (-5)$

det(A) = $4 + 10 = 14$

Since det(A) ≠ 0, A⁻¹ exists.

Step 2: Calculate the cofactor matrix of A (C).

$C_{11} = M_{11} = (1 \times 3 - 1 \times (-1)) = 4$

$C_{12} = -M_{12} = -((-1) \times 3 - 1 \times 2) = -(-5) = 5$

$C_{13} = M_{13} = ((-1) \times (-1) - 1 \times 2) = -1$

$C_{21} = -M_{21} = -(2 \times 3 - 0 \times (-1)) = -6$

$C_{22} = M_{22} = (1 \times 3 - 0 \times 2) = 3$

$C_{23} = -M_{23} = -(1 \times (-1) - 2 \times 2) = -(-5) = 5$

$C_{31} = M_{31} = (2 \times 1 - 0 \times 1) = 2$

$C_{32} = -M_{32} = -(1 \times 1 - 0 \times (-1)) = -1$

$C_{33} = M_{33} = (1 \times 1 - 2 \times (-1)) = 3$

The cofactor matrix C is:

$C = \begin{bmatrix} 4 & 5 & -1 \\ -6 & 3 & 5 \\ 2 & -1 & 3 \end{bmatrix}$

Step 3: Calculate the adjoint of A (adj(A)).

The adjoint of A is the transpose of the cofactor matrix:

adj(A) = $C^T = \begin{bmatrix} 4 & -6 & 2 \\ 5 & 3 & -1 \\ -1 & 5 & 3 \end{bmatrix}$

Step 4: Calculate A⁻¹.

A⁻¹ = (1/det(A)) * adj(A)

A⁻¹ = $\frac{1}{14} \begin{bmatrix} 4 & -6 & 2 \\ 5 & 3 & -1 \\ -1 & 5 & 3 \end{bmatrix}$

A⁻¹ = $\begin{bmatrix} 4/14 & -6/14 & 2/14 \\ 5/14 & 3/14 & -1/14 \\ -1/14 & 5/14 & 3/14 \end{bmatrix} = \begin{bmatrix} 2/7 & -3/7 & 1/7 \\ 5/14 & 3/14 & -1/14 \\ -1/14 & 5/14 & 3/14 \end{bmatrix}$

Part 2: Use A⁻¹ to solve the system of linear equations

The given system of equations is:

x + 2y = 4

-x + y + z = 1

2x - y + 3z = 7

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

A = $\begin{bmatrix} 1 & 2 & 0 \\ -1 & 1 & 1 \\ 2 & -1 & 3 \end{bmatrix}$ , X = $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ , B = $\begin{bmatrix} 4 \\ 1 \\ 7 \end{bmatrix}$

To solve for X, we use the relation X = A⁻¹B.

X = $\frac{1}{14} \begin{bmatrix} 4 & -6 & 2 \\ 5 & 3 & -1 \\ -1 & 5 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 1 \\ 7 \end{bmatrix}$

X = $\frac{1}{14} \begin{bmatrix} (4)(4) + (-6)(1) + (2)(7) \\ (5)(4) + (3)(1) + (-1)(7) \\ (-1)(4) + (5)(1) + (3)(7) \end{bmatrix}$

X = $\frac{1}{14} \begin{bmatrix} 16 - 6 + 14 \\ 20 + 3 - 7 \\ -4 + 5 + 21 \end{bmatrix}$

X = $\frac{1}{14} \begin{bmatrix} 24 \\ 16 \\ 22 \end{bmatrix}$

Therefore,

x = 24/14 = 12/7

y = 16/14 = 8/7

z = 22/14 = 11/7

The solution to the system of equations is x = 12/7, y = 8/7, and z = 11/7.

Explanation of the solution:

1. Calculate Determinant: Find det(A). If det(A) = 0, the inverse does not exist, and the system might have no solution or infinitely many solutions. Here, det(A) = 14.

2. Find Cofactor Matrix: Calculate the cofactor for each element $A_{ij}$ as $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor.

3. Find Adjoint Matrix: The adjoint matrix, adj(A), is the transpose of the cofactor matrix.

4. Calculate Inverse: A⁻¹ = (1/det(A)) * adj(A).

5. Matrix Form of Equations: Represent the system of linear equations as AX = B.

6. Solve for Variables: Multiply both sides by A⁻¹ to get X = A⁻¹B. Perform the matrix multiplication to find the values of x, y, and z.

Answer:

The inverse of matrix A is:

$A^{-1} = \frac{1}{14} \begin{bmatrix} 4 & -6 & 2 \\ 5 & 3 & -1 \\ -1 & 5 & 3 \end{bmatrix}$

The solution to the system of linear equations is:

x = 12/7

y = 8/7

z = 11/7