Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
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]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.