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If the product of the matrix $B=\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]$
with a matrix $A$ has the inverse $C=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]$, then $A^{-1}$ equals
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with a matrix $A$ has the inverse $C=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]$, then $A^{-1}$ equals
Solution:
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Verified Answer
The correct answer is:
$\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]$
$(B A)^{-1}=C$ (given)
or $A^{-1} B^{-1}=C$
or $A^{-1}\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]^{-1}=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]$
Multiply by $B$ on both sides, we get
$A^{-1}\left(B^{-1} B\right)=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]$
or
$$
A^{-1}=\left[\begin{array}{ccc}
-3 & -5 & -5 \\
0 & 9 & 2 \\
2 & 14 & 6
\end{array}\right]_{3 \times 3}
$$
or $A^{-1} B^{-1}=C$
or $A^{-1}\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]^{-1}=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]$
Multiply by $B$ on both sides, we get
$A^{-1}\left(B^{-1} B\right)=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]$
or
$$
A^{-1}=\left[\begin{array}{ccc}
-3 & -5 & -5 \\
0 & 9 & 2 \\
2 & 14 & 6
\end{array}\right]_{3 \times 3}
$$
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