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Sagot :
To solve for [tex]\( X \)[/tex] in the matrix equation [tex]\( A \cdot X + B = C \)[/tex], we can follow these steps:
1. Rearrange the Equation:
First, we need to isolate [tex]\( A \cdot X \)[/tex]. So, we rearrange the equation as follows:
[tex]\[ A \cdot X = C - B \][/tex]
2. Calculate [tex]\( C - B \)[/tex]:
We need to subtract matrix [tex]\( B \)[/tex] from matrix [tex]\( C \)[/tex]:
[tex]\[ C = \begin{pmatrix} -42 & -20 \\ 5 & 4 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} -7 & -9 \\ 4 & -1 \end{pmatrix} \][/tex]
[tex]\[ C - B = \begin{pmatrix} -42 & -20 \\ 5 & 4 \end{pmatrix} - \begin{pmatrix} -7 & -9 \\ 4 & -1 \end{pmatrix} = \begin{pmatrix} -42 - (-7) & -20 - (-9) \\ 5 - 4 & 4 - (-1) \end{pmatrix} = \begin{pmatrix} -35 & -11 \\ 1 & 5 \end{pmatrix} \][/tex]
So,
[tex]\[ C - B = \begin{pmatrix} -35 & -11 \\ 1 & 5 \end{pmatrix} \][/tex]
3. Solve for [tex]\( X \)[/tex] Using the Inverse of [tex]\( A \)[/tex]:
To solve for [tex]\( X \)[/tex], we need to multiply both sides of [tex]\( A \cdot X = C - B \)[/tex] by the inverse of [tex]\( A \)[/tex] (denoted as [tex]\( A^{-1} \)[/tex]):
[tex]\[ X = A^{-1} \cdot (C - B) \][/tex]
4. Calculate the Inverse of [tex]\( A \)[/tex]:
First, let's find the inverse of matrix [tex]\( A \)[/tex]:
[tex]\[ A = \begin{pmatrix} -3 & -4 \\ 1 & 0 \end{pmatrix} \][/tex]
The inverse of [tex]\( A \)[/tex], denoted by [tex]\( A^{-1} \)[/tex], is given by:
[tex]\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \][/tex]
where [tex]\(\text{det}(A)\)[/tex] is the determinant of [tex]\( A \)[/tex] and [tex]\(\text{adj}(A)\)[/tex] is the adjugate of [tex]\( A \)[/tex].
The determinant of [tex]\( A \)[/tex] is:
[tex]\[ \text{det}(A) = (-3)(0) - (1)(-4) = 4 \][/tex]
The adjugate of [tex]\( A \)[/tex] is:
[tex]\[ \text{adj}(A) = \begin{pmatrix} 0 & -1 \\ 4 & -3 \end{pmatrix}^{\top} = \begin{pmatrix} 0 & 4 \\ -1 & -3 \end{pmatrix} \][/tex]
Thus,
[tex]\[ A^{-1} = \frac{1}{4} \cdot \begin{pmatrix} 0 & 4 \\ -1 & -3 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -0.25 & -0.75 \end{pmatrix} \][/tex]
5. Multiply [tex]\( A^{-1} \)[/tex] by [tex]\( C - B \)[/tex]:
Now, we calculate [tex]\( X \)[/tex]:
[tex]\[ X = A^{-1} \cdot (C - B) = \begin{pmatrix} 0 & 1 \\ -0.25 & -0.75 \end{pmatrix} \cdot \begin{pmatrix} -35 & -11 \\ 1 & 5 \end{pmatrix} \][/tex]
We perform the matrix multiplication:
[tex]\[ X = \begin{pmatrix} 0 \cdot (-35) + 1 \cdot 1 & 0 \cdot (-11) + 1 \cdot 5 \\ -0.25 \cdot (-35) + -0.75 \cdot 1 & -0.25 \cdot (-11) + -0.75 \cdot 5 \end{pmatrix} = \begin{pmatrix} 1 & 5 \\ 8 & -1 \end{pmatrix} \][/tex]
Therefore, the value of [tex]\( X \)[/tex] is:
[tex]\[ X = \begin{pmatrix} 1 & 5 \\ 8 & -1 \end{pmatrix} \][/tex]
1. Rearrange the Equation:
First, we need to isolate [tex]\( A \cdot X \)[/tex]. So, we rearrange the equation as follows:
[tex]\[ A \cdot X = C - B \][/tex]
2. Calculate [tex]\( C - B \)[/tex]:
We need to subtract matrix [tex]\( B \)[/tex] from matrix [tex]\( C \)[/tex]:
[tex]\[ C = \begin{pmatrix} -42 & -20 \\ 5 & 4 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} -7 & -9 \\ 4 & -1 \end{pmatrix} \][/tex]
[tex]\[ C - B = \begin{pmatrix} -42 & -20 \\ 5 & 4 \end{pmatrix} - \begin{pmatrix} -7 & -9 \\ 4 & -1 \end{pmatrix} = \begin{pmatrix} -42 - (-7) & -20 - (-9) \\ 5 - 4 & 4 - (-1) \end{pmatrix} = \begin{pmatrix} -35 & -11 \\ 1 & 5 \end{pmatrix} \][/tex]
So,
[tex]\[ C - B = \begin{pmatrix} -35 & -11 \\ 1 & 5 \end{pmatrix} \][/tex]
3. Solve for [tex]\( X \)[/tex] Using the Inverse of [tex]\( A \)[/tex]:
To solve for [tex]\( X \)[/tex], we need to multiply both sides of [tex]\( A \cdot X = C - B \)[/tex] by the inverse of [tex]\( A \)[/tex] (denoted as [tex]\( A^{-1} \)[/tex]):
[tex]\[ X = A^{-1} \cdot (C - B) \][/tex]
4. Calculate the Inverse of [tex]\( A \)[/tex]:
First, let's find the inverse of matrix [tex]\( A \)[/tex]:
[tex]\[ A = \begin{pmatrix} -3 & -4 \\ 1 & 0 \end{pmatrix} \][/tex]
The inverse of [tex]\( A \)[/tex], denoted by [tex]\( A^{-1} \)[/tex], is given by:
[tex]\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \][/tex]
where [tex]\(\text{det}(A)\)[/tex] is the determinant of [tex]\( A \)[/tex] and [tex]\(\text{adj}(A)\)[/tex] is the adjugate of [tex]\( A \)[/tex].
The determinant of [tex]\( A \)[/tex] is:
[tex]\[ \text{det}(A) = (-3)(0) - (1)(-4) = 4 \][/tex]
The adjugate of [tex]\( A \)[/tex] is:
[tex]\[ \text{adj}(A) = \begin{pmatrix} 0 & -1 \\ 4 & -3 \end{pmatrix}^{\top} = \begin{pmatrix} 0 & 4 \\ -1 & -3 \end{pmatrix} \][/tex]
Thus,
[tex]\[ A^{-1} = \frac{1}{4} \cdot \begin{pmatrix} 0 & 4 \\ -1 & -3 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -0.25 & -0.75 \end{pmatrix} \][/tex]
5. Multiply [tex]\( A^{-1} \)[/tex] by [tex]\( C - B \)[/tex]:
Now, we calculate [tex]\( X \)[/tex]:
[tex]\[ X = A^{-1} \cdot (C - B) = \begin{pmatrix} 0 & 1 \\ -0.25 & -0.75 \end{pmatrix} \cdot \begin{pmatrix} -35 & -11 \\ 1 & 5 \end{pmatrix} \][/tex]
We perform the matrix multiplication:
[tex]\[ X = \begin{pmatrix} 0 \cdot (-35) + 1 \cdot 1 & 0 \cdot (-11) + 1 \cdot 5 \\ -0.25 \cdot (-35) + -0.75 \cdot 1 & -0.25 \cdot (-11) + -0.75 \cdot 5 \end{pmatrix} = \begin{pmatrix} 1 & 5 \\ 8 & -1 \end{pmatrix} \][/tex]
Therefore, the value of [tex]\( X \)[/tex] is:
[tex]\[ X = \begin{pmatrix} 1 & 5 \\ 8 & -1 \end{pmatrix} \][/tex]
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