Question

Let$A=\left[\begin{array}{cc}2& -1\\ 3& 4\end{array}\right],B=\left[\begin{array}{cc}5& 2\\ 7& 4\end{array}\right]$ and $C=\left[\begin{array}{cc}2& 5\\ 3& 8\end{array}\right]$. Find a matrix D such that $CD-AB=0$
$\left[5\right]$

Answer 1

Since $A,B,C$ are all square matrics of order $2$, and $CD-AB$ is well defined, D must be a square matrix of order $2$.

Let $D=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ then $CD-AB=0$ gives

$\left[\begin{array}{cc}2& 5\\ 3& 8\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]-\left[\begin{array}{cc}2& -1\\ 3& 4\end{array}\right]\left[\begin{array}{cc}5& 2\\ 7& 4\end{array}\right]=0$

or $\left[\begin{array}{cc}2a+5c& 2b+5d\\ 3a+8c& 3b+8d\end{array}\right]-\left[\begin{array}{cc}3& 0\\ 43& 22\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$\left[2\right]$

or $\left[\begin{array}{cc}2a+5c-3& 2b+5d\\ 3a+8c-43& 3b+8d-22\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$\left[1\right]$

By equating the corresponding elements of matrices, we get
$2a+5c-3=0$ ...(i)
$3a+8c-43=0$ ...(ii)
$2b+5d=0$ ...(iii)
and $3b+8d-22=0$ ...(iv)
$\left[1\right]$

Solving (i) and (ii), we get $a=-191,c=77$
Solving (iii) and (iv), we get $b=-110,d=44$
$\left[1\right]$

Therefore $D=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}-191& -110\\ 77& 44\end{array}\right]$