Wheeler Rowe
10/20/2023 · High School
i) \( (X A)^{-1}=A^{-1} \mathrm{~B} \)
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Step-by-step Solution
To analyze the equation \( (X A)^{-1} = A^{-1} B \), we can start by recalling the properties of matrix inverses. Specifically, for any invertible matrices \( M \) and \( N \), the inverse of their product is given by:
\[
(M N)^{-1} = N^{-1} M^{-1}
\]
In your case, we have \( M = X \) and \( N = A \). Therefore, we can apply this property to the left-hand side of the equation:
\[
(X A)^{-1} = A^{-1} X^{-1}
\]
Now, we can set this equal to the right-hand side of your equation:
\[
A^{-1} X^{-1} = A^{-1} B
\]
To simplify this equation, we can multiply both sides by \( A \) (assuming \( A \) is invertible):
\[
X^{-1} = B
\]
Finally, if we take the inverse of both sides, we find:
\[
X = B^{-1}
\]
Thus, the equation \( (X A)^{-1} = A^{-1} B \) implies that \( X = B^{-1} \) under the assumption that \( A \) is invertible.
Quick Answer
The equation \( (X A)^{-1} = A^{-1} B \) implies \( X = B^{-1} \) if \( A \) is invertible.
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