Given the matrix $A = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{pmatrix}$, we need to: 1. Calculate the determinant of $A$ and determine if $A$ is invertible. If it is, find the inverse matrix $A^{-1}$.

1. Problem Description

Given the matrix $A = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{pmatrix}$, we need to:

1. Calculate the determinant of $A$ and determine if $A$ is invertible. If it is, find the inverse matrix $A^{-1}$.

2. Solve the system of equations:

$x + z = -3$

$2x + y = 1$

$-x + 2y + 3z = 2$

using the inverse matrix $A^{-1}$.

2. Solution Steps

1. Calculate the determinant of A:

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

$det(A) = 1(3) - 0(6) + 1(4 + 1)$

$det(A) = 3 + 5 = 8$

Since $det(A) = 8 \neq 0$, the matrix $A$ is invertible.

2. Calculate the inverse matrix $A^{-1}$:

First, find the matrix of cofactors $C$:

$C_{11} = (1*3 - 0*2) = 3$

$C_{12} = -(2*3 - 0*(-1)) = -6$

$C_{13} = (2*2 - 1*(-1)) = 5$

$C_{21} = -(0*3 - 1*2) = 2$

$C_{22} = (1*3 - 1*(-1)) = 4$

$C_{23} = -(1*2 - 0*(-1)) = -2$

$C_{31} = (0*0 - 1*1) = -1$

$C_{32} = -(1*0 - 1*2) = 2$

$C_{33} = (1*1 - 0*2) = 1$

So the matrix of cofactors is:

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

Next, find the adjugate (transpose of the cofactor matrix):

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

Then, $A^{-1} = \frac{1}{det(A)} * adj(A)$:

$A^{-1} = \frac{1}{8} \begin{pmatrix} 3 & 2 & -1 \\ -6 & 4 & 2 \\ 5 & -2 & 1 \end{pmatrix} = \begin{pmatrix} 3/8 & 2/8 & -1/8 \\ -6/8 & 4/8 & 2/8 \\ 5/8 & -2/8 & 1/8 \end{pmatrix}$

$A^{-1} = \begin{pmatrix} 3/8 & 1/4 & -1/8 \\ -3/4 & 1/2 & 1/4 \\ 5/8 & -1/4 & 1/8 \end{pmatrix}$

3. Solve the system of equations:

The system of equations can be written in matrix form as $AX = B$, where $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ and $B = \begin{pmatrix} -3 \\ 1 \\ 2 \end{pmatrix}$.

To solve for $X$, we use the inverse matrix: $X = A^{-1}B$.

$X = \begin{pmatrix} 3/8 & 1/4 & -1/8 \\ -3/4 & 1/2 & 1/4 \\ 5/8 & -1/4 & 1/8 \end{pmatrix} \begin{pmatrix} -3 \\ 1 \\ 2 \end{pmatrix}$

$x = (3/8)(-3) + (1/4)(1) + (-1/8)(2) = -9/8 + 1/4 - 2/8 = -9/8 + 2/8 - 2/8 = -9/8$

$y = (-3/4)(-3) + (1/2)(1) + (1/4)(2) = 9/4 + 1/2 + 2/4 = 9/4 + 2/4 + 2/4 = 13/4$

$z = (5/8)(-3) + (-1/4)(1) + (1/8)(2) = -15/8 - 1/4 + 2/8 = -15/8 - 2/8 + 2/8 = -15/8$

3. Final Answer

The solution to the system of equations is:

$x = -9/8$

$y = 13/4$

$z = -15/8$