Question

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

• A. $\left[\begin{array}{ccc}-3& -5& 5\\ 0& 9& 14\\ 2& 2& 9\end{array}\right]$
• B. $\left[\begin{array}{ccc}-3& 5& 5\\ 0& 0& 9\\ 2& 14& 16\end{array}\right]$
• C. $\left[\begin{array}{ccc}-3& -5& -5\\ 0& 0& 2\\ 2& 14& 6\end{array}\right]$
• D. $\left[\begin{array}{ccc}-3& -5& -5\\ 0& 9& 2\\ 2& 14& 6\end{array}\right]$

Answer 1

Answer: (D)

$(BA{)}^{-1}=C$ (given)
or $A^{-}1B^{-1}=C$
or $A^{-1}{\left[\begin{array}{ccc}-3& -3& -5\\ 0& 9& 2\\ 2& 14& 6\end{array}\right]}^{-1}=\left[\begin{array}{ccc}1& 0& 1\\ 1& 1& 3\\ 2& 0& 2\end{array}\right]$
Multiply by B on both sides, we get
$A^{-1}(B^{-1}B)=\left[\begin{array}{ccc}-1& 0& 1\\ 1& 1& 3\\ 2& 0& 2\end{array}\right]\left[\begin{array}{ccc}2& 6& 4\\ 1& 0& 1\\ -1& 1& -1\end{array}\right]$
or $A^{-1}={\left[\begin{array}{ccc}-3& -5& -5\\ 0& 9& 2\\ 2& 14& 6\end{array}\right]}_{3\times 3}$