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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]
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]
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