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Sagot :
To determine which value of [tex]\( t \)[/tex] makes the two matrices inverses of each other, we need to verify if the product of the two matrices results in the identity matrix. We will consider the given matrices:
[tex]\[ A = \begin{pmatrix} -4 & 6 \\ 3 & -4 \end{pmatrix} \][/tex]
and
[tex]\[ B = \begin{pmatrix} 2 & 3 \\ 1.5 & 4 \end{pmatrix} \][/tex]
The problem specifies testing specific values for [tex]\( t \)[/tex]. However, let's first check the product of [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
Perform matrix multiplication:
[tex]\[ A \cdot B = \begin{pmatrix} -4 & 6 \\ 3 & -4 \end{pmatrix} \cdot \begin{pmatrix} 2 & 3 \\ 1.5 & 4 \end{pmatrix} \][/tex]
Calculate element-wise:
1. First row, first column:
[tex]\[ (-4 \cdot 2) + (6 \cdot 1.5) = -8 + 9 = 1 \][/tex]
2. First row, second column:
[tex]\[ (-4 \cdot 3) + (6 \cdot 4) = -12 + 24 = 12 \][/tex]
3. Second row, first column:
[tex]\[ (3 \cdot 2) + (-4 \cdot 1.5) = 6 - 6 = 0 \][/tex]
4. Second row, second column:
[tex]\[ (3 \cdot 3) + (-4 \cdot 4) = 9 - 16 = -7 \][/tex]
This forms the resultant matrix:
[tex]\[ A \cdot B = \begin{pmatrix} 1 & 12 \\ 0 & -7 \end{pmatrix} \][/tex]
For [tex]\( A \)[/tex] and [tex]\( B \)[/tex] to be inverses of each other, this product should be the identity matrix:
[tex]\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \][/tex]
Upon examining the resultant matrix, it is evident that:
[tex]\[ \begin{pmatrix} 1 & 12 \\ 0 & -7 \end{pmatrix} \][/tex]
is not the identity matrix.
Thus, matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are not inverses of each other under any normal circumstances. Hence, none of the provided options ([tex]\(-3, -2, 2, 3\)[/tex]) can make these matrices inverses of each other.
So, the final answer is:
[tex]\[ None \text{ of the values make the matrices inverses of each other.} \][/tex]
[tex]\[ A = \begin{pmatrix} -4 & 6 \\ 3 & -4 \end{pmatrix} \][/tex]
and
[tex]\[ B = \begin{pmatrix} 2 & 3 \\ 1.5 & 4 \end{pmatrix} \][/tex]
The problem specifies testing specific values for [tex]\( t \)[/tex]. However, let's first check the product of [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
Perform matrix multiplication:
[tex]\[ A \cdot B = \begin{pmatrix} -4 & 6 \\ 3 & -4 \end{pmatrix} \cdot \begin{pmatrix} 2 & 3 \\ 1.5 & 4 \end{pmatrix} \][/tex]
Calculate element-wise:
1. First row, first column:
[tex]\[ (-4 \cdot 2) + (6 \cdot 1.5) = -8 + 9 = 1 \][/tex]
2. First row, second column:
[tex]\[ (-4 \cdot 3) + (6 \cdot 4) = -12 + 24 = 12 \][/tex]
3. Second row, first column:
[tex]\[ (3 \cdot 2) + (-4 \cdot 1.5) = 6 - 6 = 0 \][/tex]
4. Second row, second column:
[tex]\[ (3 \cdot 3) + (-4 \cdot 4) = 9 - 16 = -7 \][/tex]
This forms the resultant matrix:
[tex]\[ A \cdot B = \begin{pmatrix} 1 & 12 \\ 0 & -7 \end{pmatrix} \][/tex]
For [tex]\( A \)[/tex] and [tex]\( B \)[/tex] to be inverses of each other, this product should be the identity matrix:
[tex]\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \][/tex]
Upon examining the resultant matrix, it is evident that:
[tex]\[ \begin{pmatrix} 1 & 12 \\ 0 & -7 \end{pmatrix} \][/tex]
is not the identity matrix.
Thus, matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are not inverses of each other under any normal circumstances. Hence, none of the provided options ([tex]\(-3, -2, 2, 3\)[/tex]) can make these matrices inverses of each other.
So, the final answer is:
[tex]\[ None \text{ of the values make the matrices inverses of each other.} \][/tex]
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