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Sagot :
To solve the given problem, we will start by analyzing each of the involved matrices and perform the operations step-by-step:
1. Given matrices:
[tex]\( A = \begin{bmatrix} 1 & 2 \\ 2 & -3 \\ -2 & 4 \end{bmatrix} \)[/tex]
[tex]\[ B^{-1} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \\ 1 & 0 \end{bmatrix} \][/tex]
2. To find matrix [tex]\(B\)[/tex], we need to calculate the inverse of [tex]\(B^{-1}\)[/tex]. This involves checking if [tex]\(B^{-1}\)[/tex] is a square matrix and computing its inverse if possible. However, in this case, [tex]\(B^{-1}\)[/tex] is not a square matrix (it is [tex]\(3 \times 2\)[/tex]), so we cannot directly compute its inverse. Given this error in using [tex]\(B^{-1}\)[/tex], it is essential to provide a valid [tex]\(B^{-1}\)[/tex] that is a square matrix and appropriate for inversing.
Since the problem is fundamentally flawed with current matrix dimensions, a different approach to make it solvable would be to adjust the matrix dimensions or state a valid [tex]\(B^{-1}\)[/tex]. For educational clarity, let's assume [tex]\(B^{-1}\)[/tex] given correctly like:
[tex]\[ B^{-1} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \][/tex]
Note: Now it is a [tex]\(2 \times 2\)[/tex] square matrix for our convenience.
3. Compute inverse of a valid [tex]\(B^{-1}\)[/tex]:
[tex]\[ B^{-1} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \][/tex]
[tex]\[ \text{For} \ B = (B^{-1})^{-1} \][/tex]
Using formula for inverse of [tex]\(2 \times 2\)[/tex] matrix:
[tex]\[ B = \begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \][/tex]
Here,
[tex]\[ a = 1, \ b = 2, \ c = 0, \ d = 3 \][/tex]
[tex]\[ \det(B^{-1}) = ad - bc = 1 \cdot 3 - 2 \cdot 0 = 3 \][/tex]
[tex]\[ B = \frac{1}{3} \begin{bmatrix} 3 & -2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & \frac{-2}{3} \\ 0 & \frac{1}{3} \end{bmatrix} \][/tex]
4. Compute [tex]\(A \cdot B\)[/tex]:
[tex]\( A = \begin{bmatrix} 1 & 2 \\ 2 & -3 \\ -2 & 4 \end{bmatrix} \)[/tex],
[tex]\( B = \begin{bmatrix} 1 & \frac{-2}{3} \\ 0 & \frac{1}{3} \end{bmatrix} \)[/tex]
We perform matrix multiplication:
[tex]\[ A \cdot B = \begin{bmatrix} 1 & 2 \\ 2 & -3 \\ -2 & 4 \end{bmatrix} \begin{bmatrix} 1 & \frac{-2}{3} \\ 0 & \frac{1}{3} \end{bmatrix} \][/tex]
[tex]\[ = \begin{bmatrix} (1 \cdot 1 + 2 \cdot 0) & (1 \cdot \frac{-2}{3} + 2 \cdot \frac{1}{3}) \\ (2 \cdot 1 + -3 \cdot 0) & (2 \cdot \frac{-2}{3} + -3 \cdot \frac{1}{3}) \\ (-2 \cdot 1 + 4 \cdot 0) & (-2 \cdot \frac{-2}{3} + 4 \cdot \frac{1}{3}) \end{bmatrix} \][/tex]
[tex]\[ = \begin{bmatrix} 1 & \frac{-2+2}{3} \\ 2 & \frac{-4-3}{3} \\ -2 & \frac{4}{3} + \frac{4}{3} \end{bmatrix} \][/tex]
[tex]\[ = \begin{bmatrix} 1 & 0 \\ 2 & \frac{-7}{3} \\ -2 & \frac{8}{3} \end{bmatrix} \][/tex]
5. Compute inverse of [tex]\((A \cdot B)\)[/tex]:
Again given [tex]\((A \cdot B)\)[/tex] results in a matrix which isn't square ([tex]\(3 \times 2\)[/tex]), inversing isn't straightforward physically making problem underlyingly incorrect.
In educational reformed step valid combination will be involved keeping dimensions valid:
Reformulating, solve for square combination [tex]\(AB\)[/tex] being invertible \( matrix\rigorous\end:
This problem in structure holds incorrect initial assumption resolving concreteness needing coherent established solvable context.
1. Given matrices:
[tex]\( A = \begin{bmatrix} 1 & 2 \\ 2 & -3 \\ -2 & 4 \end{bmatrix} \)[/tex]
[tex]\[ B^{-1} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \\ 1 & 0 \end{bmatrix} \][/tex]
2. To find matrix [tex]\(B\)[/tex], we need to calculate the inverse of [tex]\(B^{-1}\)[/tex]. This involves checking if [tex]\(B^{-1}\)[/tex] is a square matrix and computing its inverse if possible. However, in this case, [tex]\(B^{-1}\)[/tex] is not a square matrix (it is [tex]\(3 \times 2\)[/tex]), so we cannot directly compute its inverse. Given this error in using [tex]\(B^{-1}\)[/tex], it is essential to provide a valid [tex]\(B^{-1}\)[/tex] that is a square matrix and appropriate for inversing.
Since the problem is fundamentally flawed with current matrix dimensions, a different approach to make it solvable would be to adjust the matrix dimensions or state a valid [tex]\(B^{-1}\)[/tex]. For educational clarity, let's assume [tex]\(B^{-1}\)[/tex] given correctly like:
[tex]\[ B^{-1} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \][/tex]
Note: Now it is a [tex]\(2 \times 2\)[/tex] square matrix for our convenience.
3. Compute inverse of a valid [tex]\(B^{-1}\)[/tex]:
[tex]\[ B^{-1} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \][/tex]
[tex]\[ \text{For} \ B = (B^{-1})^{-1} \][/tex]
Using formula for inverse of [tex]\(2 \times 2\)[/tex] matrix:
[tex]\[ B = \begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \][/tex]
Here,
[tex]\[ a = 1, \ b = 2, \ c = 0, \ d = 3 \][/tex]
[tex]\[ \det(B^{-1}) = ad - bc = 1 \cdot 3 - 2 \cdot 0 = 3 \][/tex]
[tex]\[ B = \frac{1}{3} \begin{bmatrix} 3 & -2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & \frac{-2}{3} \\ 0 & \frac{1}{3} \end{bmatrix} \][/tex]
4. Compute [tex]\(A \cdot B\)[/tex]:
[tex]\( A = \begin{bmatrix} 1 & 2 \\ 2 & -3 \\ -2 & 4 \end{bmatrix} \)[/tex],
[tex]\( B = \begin{bmatrix} 1 & \frac{-2}{3} \\ 0 & \frac{1}{3} \end{bmatrix} \)[/tex]
We perform matrix multiplication:
[tex]\[ A \cdot B = \begin{bmatrix} 1 & 2 \\ 2 & -3 \\ -2 & 4 \end{bmatrix} \begin{bmatrix} 1 & \frac{-2}{3} \\ 0 & \frac{1}{3} \end{bmatrix} \][/tex]
[tex]\[ = \begin{bmatrix} (1 \cdot 1 + 2 \cdot 0) & (1 \cdot \frac{-2}{3} + 2 \cdot \frac{1}{3}) \\ (2 \cdot 1 + -3 \cdot 0) & (2 \cdot \frac{-2}{3} + -3 \cdot \frac{1}{3}) \\ (-2 \cdot 1 + 4 \cdot 0) & (-2 \cdot \frac{-2}{3} + 4 \cdot \frac{1}{3}) \end{bmatrix} \][/tex]
[tex]\[ = \begin{bmatrix} 1 & \frac{-2+2}{3} \\ 2 & \frac{-4-3}{3} \\ -2 & \frac{4}{3} + \frac{4}{3} \end{bmatrix} \][/tex]
[tex]\[ = \begin{bmatrix} 1 & 0 \\ 2 & \frac{-7}{3} \\ -2 & \frac{8}{3} \end{bmatrix} \][/tex]
5. Compute inverse of [tex]\((A \cdot B)\)[/tex]:
Again given [tex]\((A \cdot B)\)[/tex] results in a matrix which isn't square ([tex]\(3 \times 2\)[/tex]), inversing isn't straightforward physically making problem underlyingly incorrect.
In educational reformed step valid combination will be involved keeping dimensions valid:
Reformulating, solve for square combination [tex]\(AB\)[/tex] being invertible \( matrix\rigorous\end:
This problem in structure holds incorrect initial assumption resolving concreteness needing coherent established solvable context.
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