Question

If $A=\left[\begin{array}{cc}1& -2\\ 3& 0\end{array}\right],B=\left[\begin{array}{cc}-1& 4\\ 2& 3\end{array}\right]$ and $ABC=\left[\begin{array}{cc}4& 8\\ 3& 7\end{array}\right]$, then $C$ equals

• A. $\frac{1}{66}\left[\begin{array}{cc}72& -32\\ 57& -29\end{array}\right]$
• B. $\frac{1}{66}\left[\begin{array}{cc}-54& -110\\ 3& 11\end{array}\right]$
• C. $\frac{1}{66}\left[\begin{array}{cc}-54& 110\\ 3& -11\end{array}\right]$
• D. $\frac{1}{66}\left[\begin{array}{cc}-72& 32\\ -57& 29\end{array}\right]$

Answer 1

The correct option is B $\frac{1}{66}\left[\begin{array}{cc}-54& -110\\ 3& 11\end{array}\right]$

$A=\left[\begin{array}{cc}1& -2\\ 3& 0\end{array}\right],B=\left[\begin{array}{cc}-1& 4\\ 2& 3\end{array}\right]$ and $ABC=\left[\begin{array}{cc}4& 8\\ 3& 7\end{array}\right]$
$ABC=\left[\begin{array}{cc}4& 8\\ 3& 7\end{array}\right]$
$\Rightarrow \left(AB{)}^{-1}\right(AB)C=(AB{)}^{-1}\left[\begin{array}{cc}4& 8\\ 3& 7\end{array}\right]$
$\Rightarrow C=B^{-1}{A}^{-1}\left[\begin{array}{cc}4& 8\\ 3& 7\end{array}\right]$ -----(1)
$B^{-1}=\frac{1}{11}\left[\begin{array}{cc}-3& 4\\ 2& 1\end{array}\right]$
$A^{-1}=\frac{1}{6}\left[\begin{array}{cc}0& 2\\ -3& 1\end{array}\right]$
$\Rightarrow B^{-1}{A}^{-1}=\frac{1}{66}\left[\begin{array}{cc}-12& -2\\ -3& 5\end{array}\right]$
From (1)
$C=\frac{1}{66}\left[\begin{array}{cc}-54& -110\\ 3& 11\end{array}\right]$
Hence, option B is correct.