To solve the given system of equations using the inverse of a matrix, we start by writing the system in matrix form:
1. [tex]\( -x + y - 8 = 0 \)[/tex]
2. [tex]\( 4x + 3y - 12 = 0 \)[/tex]
3. [tex]\( x - 7y + z = 15 \)[/tex]
Rewriting the system in the form [tex]\( AX = B \)[/tex]:
The coefficient matrix [tex]\( A \)[/tex] is:
[tex]\[
A = \begin{pmatrix}
-1 & 1 & 0 \\
4 & 3 & 0 \\
1 & -7 & 1
\end{pmatrix}
\][/tex]
The constant matrix [tex]\( B \)[/tex] is:
[tex]\[
B = \begin{pmatrix}
8 \\
12 \\
15
\end{pmatrix}
\][/tex]
The variable matrix [tex]\( X \)[/tex] we're solving for is:
[tex]\[
X = \begin{pmatrix}
x \\
y \\
z
\end{pmatrix}
\][/tex]
Using the formula [tex]\( X = A^{-1}B \)[/tex], we first find the inverse of matrix [tex]\( A \)[/tex], denoted [tex]\( A^{-1} \)[/tex], and then multiply it by matrix [tex]\( B \)[/tex].
Given the already known result for [tex]\( X = A^{-1}B \)[/tex]:
[tex]\[
X = \begin{pmatrix}
-1.7142857142857142 \\
6.285714285714286 \\
60.714285714285715
\end{pmatrix}
\][/tex]
The approximate values for [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex] are:
[tex]\[
x \approx -1.714 \approx -\frac{12}{7}
\][/tex]
[tex]\[
y \approx 6.286 \approx \frac{44}{7}
\][/tex]
[tex]\[
z \approx 60.714 \approx \frac{425}{7}
\][/tex]
Comparing these approximate values with the given answer choices, we match [tex]\( x = -\frac{12}{7} \)[/tex], [tex]\( y = \frac{44}{7} \)[/tex], and [tex]\( z = \frac{425}{7} \)[/tex]. Therefore, the correct answer is:
C. [tex]\( x = -\frac{12}{7}, y = \frac{44}{7} \)[/tex], and [tex]\( z = \frac{425}{7} \)[/tex]
