Question

Given \[ A = \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix} \] find \( AB \). Hence, solve the system of linear equations: \[ x - y + z = 4, \] \[ x - 2y - 2z = 9, \] \[ 2x + y + 3z = 1. \]

Hint:

For solving systems using matrices, use \( AX = B \Rightarrow X = A^{-1} B \).

Answer 1

To solve the problem, we are given two matrices \( A \) and \( B \). First, we need to find the product \( AB \), and then use it to solve the system of linear equations.

1. Writing the System in Matrix Form:
The system of equations can be written in the matrix form \( BX = C \), where:

\( B = \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad C = \begin{bmatrix} 4 \\ 9 \\ 1 \end{bmatrix} \)

2. Calculating \( AB \):
We compute the matrix multiplication \( AB \), where:
\[ A = \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix} \] Now computing the product \( AB \):
\[ AB = \begin{bmatrix} -4(1) + 4(1) + 4(2) & -4(-1) + 4(-2) + 4(1) & -4(1) + 4(-2) + 4(3) \\ -7(1) + 1(1) + 3(2) & -7(-1) + 1(-2) + 3(1) & -7(1) + 1(-2) + 3(3) \\ 5(1) + (-3)(1) + (-1)(2) & 5(-1) + (-3)(-2) + (-1)(1) & 5(1) + (-3)(-2) + (-1)(3) \end{bmatrix} \] Simplifying:
\[ AB = \begin{bmatrix} -4 + 4 + 8 & 4 -8 + 4 & -4 -8 + 12 \\ -7 + 1 + 6 & 7 -2 + 3 & -7 -2 + 9 \\ 5 -3 -2 & -5 + 6 -1 & 5 + 6 -3 \end{bmatrix} = \begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{bmatrix} \]

3. Observing the Result:
We find that \( AB = 8I \), where \( I \) is the identity matrix. This means that \( B^{-1} = \frac{1}{8}A \).

4. Solving the System:
We now solve \( BX = C \). Multiply both sides by \( B^{-1} \):

\[ X = B^{-1}C = \frac{1}{8}AC \]
We compute \( AC \), where:
\[ C = \begin{bmatrix} 4 \\ 9 \\ 1 \end{bmatrix} \] \[ AC = \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} \begin{bmatrix} 4 \\ 9 \\ 1 \end{bmatrix} = \begin{bmatrix} -4(4) + 4(9) + 4(1) \\ -7(4) + 1(9) + 3(1) \\ 5(4) + (-3)(9) + (-1)(1) \end{bmatrix} = \begin{bmatrix} -16 + 36 + 4 \\ -28 + 9 + 3 \\ 20 - 27 - 1 \end{bmatrix} = \begin{bmatrix} 24 \\ -16 \\ -8 \end{bmatrix} \] \[ X = \frac{1}{8} \begin{bmatrix} 24 \\ -16 \\ -8 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \\ -1 \end{bmatrix} \]

Final Answer:
The solution to the system is: \( x = 3, \, y = -2, \, z = -1 \)