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:
\[
x - y + z = 4,
\]
\[
x - 2y - 2z = 9,
\]
\[
2x + y + 3z = 1.
\]
Show Hint
Solution and Explanation
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: \( x = 3, \, y = -2, \, z = -1 \)
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