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Sagot :
To determine the value of [tex]\( x \)[/tex] that makes the two matrices inverses of each other for the given matrices:
[tex]\[ A = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 1 & -2 \\ x & 5 \end{bmatrix}, \][/tex]
we need to find [tex]\( x \)[/tex] such that the product of the matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is the identity matrix [tex]\( I \)[/tex]:
[tex]\[ A \cdot B = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \][/tex]
Let's multiply the matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
[tex]\[ A \cdot B = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ x & 5 \end{bmatrix} \][/tex]
Perform the matrix multiplication:
[tex]\[ A \cdot B = \begin{bmatrix} (5 \cdot 1 + 2 \cdot x) & (5 \cdot -2 + 2 \cdot 5) \\ (2 \cdot 1 + 1 \cdot x) & (2 \cdot -2 + 1 \cdot 5) \end{bmatrix} = \begin{bmatrix} 5 + 2x & -10 + 10 \\ 2 + x & -4 + 5 \end{bmatrix} = \begin{bmatrix} 5 + 2x & 0 \\ 2 + x & 1 \end{bmatrix} \][/tex]
Now, we set the product equal to the identity matrix and solve for [tex]\( x \)[/tex]:
[tex]\[ \begin{bmatrix} 5 + 2x & 0 \\ 2 + x & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \][/tex]
From the matrix equality, we get the following system of equations:
[tex]\[ 5 + 2x = 1 \quad \text{(1)} \][/tex]
[tex]\[ 2 + x = 0 \quad \text{(2)} \][/tex]
First, solve equation (2) for [tex]\( x \)[/tex]:
[tex]\[ 2 + x = 0 \][/tex]
[tex]\[ x = -2 \][/tex]
Therefore, the value of [tex]\( x \)[/tex] that makes the two matrices inverses of each other is [tex]\( x = -2 \)[/tex].
To verify, substitute [tex]\( x = -2 \)[/tex] back into [tex]\( B \)[/tex] and check that [tex]\( A \cdot B = I \)[/tex]:
[tex]\[ B = \begin{bmatrix} 1 & -2 \\ -2 & 5 \end{bmatrix} \][/tex]
[tex]\[ A \cdot B = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 5 \cdot 1 + 2 \cdot (-2) & 5 \cdot -2 + 2 \cdot 5 \\ 2 \cdot 1 + 1 \cdot (-2) & 2 \cdot -2 + 1 \cdot 5 \end{bmatrix} = \begin{bmatrix} 5 - 4 & -10 + 10 \\ 2 - 2 & -4 + 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \][/tex]
Thus, when [tex]\( x = -2 \)[/tex], the matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are indeed inverses of each other.
[tex]\[ A = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 1 & -2 \\ x & 5 \end{bmatrix}, \][/tex]
we need to find [tex]\( x \)[/tex] such that the product of the matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is the identity matrix [tex]\( I \)[/tex]:
[tex]\[ A \cdot B = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \][/tex]
Let's multiply the matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
[tex]\[ A \cdot B = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ x & 5 \end{bmatrix} \][/tex]
Perform the matrix multiplication:
[tex]\[ A \cdot B = \begin{bmatrix} (5 \cdot 1 + 2 \cdot x) & (5 \cdot -2 + 2 \cdot 5) \\ (2 \cdot 1 + 1 \cdot x) & (2 \cdot -2 + 1 \cdot 5) \end{bmatrix} = \begin{bmatrix} 5 + 2x & -10 + 10 \\ 2 + x & -4 + 5 \end{bmatrix} = \begin{bmatrix} 5 + 2x & 0 \\ 2 + x & 1 \end{bmatrix} \][/tex]
Now, we set the product equal to the identity matrix and solve for [tex]\( x \)[/tex]:
[tex]\[ \begin{bmatrix} 5 + 2x & 0 \\ 2 + x & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \][/tex]
From the matrix equality, we get the following system of equations:
[tex]\[ 5 + 2x = 1 \quad \text{(1)} \][/tex]
[tex]\[ 2 + x = 0 \quad \text{(2)} \][/tex]
First, solve equation (2) for [tex]\( x \)[/tex]:
[tex]\[ 2 + x = 0 \][/tex]
[tex]\[ x = -2 \][/tex]
Therefore, the value of [tex]\( x \)[/tex] that makes the two matrices inverses of each other is [tex]\( x = -2 \)[/tex].
To verify, substitute [tex]\( x = -2 \)[/tex] back into [tex]\( B \)[/tex] and check that [tex]\( A \cdot B = I \)[/tex]:
[tex]\[ B = \begin{bmatrix} 1 & -2 \\ -2 & 5 \end{bmatrix} \][/tex]
[tex]\[ A \cdot B = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 5 \cdot 1 + 2 \cdot (-2) & 5 \cdot -2 + 2 \cdot 5 \\ 2 \cdot 1 + 1 \cdot (-2) & 2 \cdot -2 + 1 \cdot 5 \end{bmatrix} = \begin{bmatrix} 5 - 4 & -10 + 10 \\ 2 - 2 & -4 + 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \][/tex]
Thus, when [tex]\( x = -2 \)[/tex], the matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are indeed inverses of each other.
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