Let's solve the matrix equation [tex]\( A X + B = C \)[/tex] for [tex]\( X \)[/tex].
Given matrices:
[tex]\[
A = \begin{pmatrix}
-3 & -4 \\
1 & 0
\end{pmatrix},
\quad B = \begin{pmatrix}
-7 & -9 \\
4 & -1
\end{pmatrix},
\quad C = \begin{pmatrix}
-42 & -20 \\
5 & 4
\end{pmatrix}
\][/tex]
We need to isolate [tex]\( X \)[/tex]. First, we rearrange the equation:
[tex]\[ A X + B = C \][/tex]
[tex]\[ A X = C - B \][/tex]
[tex]\[ X = A^{-1} (C - B) \][/tex]
### Step 1: Calculate [tex]\( C - B \)[/tex]
[tex]\[
C - B = \begin{pmatrix}
-42 & -20 \\
5 & 4
\end{pmatrix} - \begin{pmatrix}
-7 & -9 \\
4 & -1
\end{pmatrix} = \begin{pmatrix}
-42 - (-7) & -20 - (-9) \\
5 - 4 & 4 - (-1)
\end{pmatrix}
\][/tex]
[tex]\[
C - B = \begin{pmatrix}
-35 & -11 \\
1 & 5
\end{pmatrix}
\][/tex]
### Step 2: Calculate the inverse of matrix [tex]\( A \)[/tex]
The inverse of a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( A = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix} \)[/tex] is given by:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix}
d & -b \\
-c & a
\end{pmatrix}
\][/tex]
For matrix [tex]\( A \)[/tex]:
[tex]\[
A = \begin{pmatrix}
-3 & -4 \\
1 & 0
\end{pmatrix}
\][/tex]
Let's calculate the determinant [tex]\( \text{det}(A) \)[/tex]:
[tex]\[ \text{det}(A) = (-3 \times 0) - (1 \times -4) = 0 + 4 = 4 \][/tex]
Since the determinant is non-zero, [tex]\( A \)[/tex] is invertible.
[tex]\[
A^{-1} = \frac{1}{4} \begin{pmatrix}
0 & 4 \\
-1 & -3
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
-0.25 & -0.75
\end{pmatrix}
\][/tex]
### Step 3: Multiply [tex]\( A^{-1} \)[/tex] by [tex]\( C - B \)[/tex]
[tex]\[
X = A^{-1} (C - B) = \begin{pmatrix}
0 & 1 \\
-0.25 & -0.75
\end{pmatrix} \begin{pmatrix}
-35 & -11 \\
1 & 5
\end{pmatrix}
\][/tex]
Calculating this matrix multiplication:
[tex]\[
X_{11} = 0 \cdot (-35) + 1 \cdot 1 = 1
\][/tex]
[tex]\[
X_{12} = 0 \cdot (-11) + 1 \cdot 5 = 5
\][/tex]
[tex]\[
X_{21} = -0.25 \cdot (-35) + (-0.75) \cdot 1 = 8.75 - 0.75 = 8
\][/tex]
[tex]\[
X_{22} = -0.25 \cdot (-11) + (-0.75) \cdot 5 = 2.75 - 3.75 = -1
\][/tex]
So, the matrix [tex]\( X \)[/tex] is:
[tex]\[
X = \begin{pmatrix}
1 & 5 \\
8 & -1
\end{pmatrix}
\][/tex]
Thus, the correct value of [tex]\( X \)[/tex] is:
[tex]\[
X = \begin{pmatrix}
1 & 5 \\
8 & -1
\end{pmatrix}
\][/tex]
The corresponding choice is:
[tex]\[
X = \begin{pmatrix}
1 & 5 \\
8 & -1
\end{pmatrix}
\][/tex]
