Question

State true or false:

If $A=$ $\left[\begin{array}{cc}2& 1\\ 0& -2\end{array}\right],$ $B=$ $\left[\begin{array}{cc}4& 1\\ -3& -2\end{array}\right]$ and $C=$ $\left[\begin{array}{cc}-3& 2\\ -1& 4\end{array}\right],$ then $A^2+AC-5B$ =$\left[\begin{array}{cc}17& 13\\ -12& -14\end{array}\right]$

• A. True
• B. False

Answer 1

The correct option is B False

Given

$A=\left(\begin{array}{cc}2& 1\\ 0& -2\end{array}\right),B=\left(\begin{array}{cc}4& 1\\ -3& -2\end{array}\right),C=\left(\begin{array}{cc}-3& 2\\ -1& 4\end{array}\right)$

$A^2=A.A=\left(\begin{array}{cc}2& 1\\ 0& -2\end{array}\right)\left(\begin{array}{cc}2& 1\\ 0& -2\end{array}\right)=\left(\begin{array}{cc}4& 0\\ 0& 4\end{array}\right)$

$AC=\left(\begin{array}{cc}2& 1\\ 0& -2\end{array}\right)\left(\begin{array}{cc}-3& 2\\ -1& 4\end{array}\right)=\left(\begin{array}{cc}-7& 8\\ 2& -8\end{array}\right)$

Taking LHS

$A^2+AC-5B=\left(\begin{array}{cc}4& 0\\ 0& 4\end{array}\right)+\left(\begin{array}{cc}-7& 8\\ 2& -8\end{array}\right)-\left(\begin{array}{cc}20& 5\\ -15& -10\end{array}\right)$

$=\left(\begin{array}{cc}-23& 3\\ -17& -6\end{array}\right)\ne RHS$

Hence $LHS\ne RHS$