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Sagot :
To find the matrix [tex]\( B \)[/tex] given that [tex]\( A \)[/tex] and [tex]\( AB \)[/tex] are provided, we need to solve for [tex]\( B \)[/tex]. Here are the steps to do so:
1. Given Matrices:
[tex]\[ A = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right] \][/tex]
2. The matrix product [tex]\( AB \)[/tex] is given by:
[tex]\[ AB = \left[\begin{array}{rrr} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array}\right] \][/tex]
3. To find [tex]\( B \)[/tex], we use the equation [tex]\( AB = C \)[/tex] (where [tex]\( C \)[/tex] is known):
[tex]\[ A B = C \][/tex]
We can isolate [tex]\( B \)[/tex] by multiplying both sides by the inverse of [tex]\( A \)[/tex]:
[tex]\[ B = A^{-1} C \][/tex]
4. Calculate [tex]\( A^{-1} \)[/tex]:
The inverse of [tex]\( A \)[/tex] for diagonal matrices is simply the reciprocal of each diagonal element. Since elements of [tex]\( A \)[/tex] are [tex]\(\{1, -1, 1\}\)[/tex], the inverse is:
[tex]\[ A^{-1} = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right] \][/tex]
5. Multiply [tex]\( A^{-1} \)[/tex] and [tex]\( C \)[/tex] to find [tex]\( B \)[/tex]:
[tex]\[ A^{-1} C = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right] \left[\begin{array}{rrr} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array}\right] = \left[ \begin{array}{rrr} 1 \cdot (-1) + 0 \cdot 0 + 0 \cdot 0 & 0 & 0 \\ 0 & (-1) \cdot (-1) + 0 \cdot 0 + 0 \cdot 0 & 0 \\ 0 & 0 & 1 \cdot (-1) + 0 \cdot 0 + 0 \cdot 0 \end{array}\right] \][/tex]
[tex]\[ B = \left[\begin{array}{rrr} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right] \][/tex]
Thus, the matrix [tex]\( B \)[/tex] that satisfies the given equation [tex]\( AB = C \)[/tex] is:
[tex]\[ B = \left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{array}\right] \][/tex]
The correct answer is:
B. [tex]\(\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{array}\right]\)[/tex]
1. Given Matrices:
[tex]\[ A = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right] \][/tex]
2. The matrix product [tex]\( AB \)[/tex] is given by:
[tex]\[ AB = \left[\begin{array}{rrr} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array}\right] \][/tex]
3. To find [tex]\( B \)[/tex], we use the equation [tex]\( AB = C \)[/tex] (where [tex]\( C \)[/tex] is known):
[tex]\[ A B = C \][/tex]
We can isolate [tex]\( B \)[/tex] by multiplying both sides by the inverse of [tex]\( A \)[/tex]:
[tex]\[ B = A^{-1} C \][/tex]
4. Calculate [tex]\( A^{-1} \)[/tex]:
The inverse of [tex]\( A \)[/tex] for diagonal matrices is simply the reciprocal of each diagonal element. Since elements of [tex]\( A \)[/tex] are [tex]\(\{1, -1, 1\}\)[/tex], the inverse is:
[tex]\[ A^{-1} = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right] \][/tex]
5. Multiply [tex]\( A^{-1} \)[/tex] and [tex]\( C \)[/tex] to find [tex]\( B \)[/tex]:
[tex]\[ A^{-1} C = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right] \left[\begin{array}{rrr} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array}\right] = \left[ \begin{array}{rrr} 1 \cdot (-1) + 0 \cdot 0 + 0 \cdot 0 & 0 & 0 \\ 0 & (-1) \cdot (-1) + 0 \cdot 0 + 0 \cdot 0 & 0 \\ 0 & 0 & 1 \cdot (-1) + 0 \cdot 0 + 0 \cdot 0 \end{array}\right] \][/tex]
[tex]\[ B = \left[\begin{array}{rrr} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right] \][/tex]
Thus, the matrix [tex]\( B \)[/tex] that satisfies the given equation [tex]\( AB = C \)[/tex] is:
[tex]\[ B = \left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{array}\right] \][/tex]
The correct answer is:
B. [tex]\(\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{array}\right]\)[/tex]
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