Question

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

Answer 1

Given $A=\left[\begin{array}{cc}2& -1\\ 3& 4\end{array}\right],B=\left[\begin{array}{cc}5& 2\\ 7& 4\end{array}\right],C=\left[\begin{array}{cc}2& 5\\ 3& 8\end{array}\right]$

Since $A,B,C$ are all square matrices of order $2,$ and $CD–AB=O$, $D$ 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

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

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

$\Rightarrow \left[\begin{array}{cc}2a+5c& 2b+5d\\ 3a+8c& 3b+8d\end{array}\right]-\left[\begin{array}{cc}10-7& 4-4\\ 15+28& 6+16\end{array}\right]=0$

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

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

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

Since matrices are equal,

Hence,

$2a+5c-3=0$ $\dots \left(1\right)$

$3a+8c-43=0$ $\dots \left(2\right)$

$2b+5d=0$ $\dots \left(3\right)$

$3b+8d-22=0$ $\dots \left(4\right)$

Solving equations

Solving equation $\left(1\right)$

$2a+5c-3=0$

$\Rightarrow 2a+5c=3$

$\Rightarrow 2a=3-5c$

$\Rightarrow a=\left(\frac{3-5c}{2}\right)$

Putting values of a in equation $\left(2\right)$

$3a+8c-43=0$

$\Rightarrow 3\left(\frac{3-5c}{2}\right)+8c-43=0$

$\Rightarrow \frac{3(3-5c)+2(8c-43)}{2}=0$

$\Rightarrow 9-15c+16c-86=0$

$\Rightarrow -15c+16c-86+9=0$

$\Rightarrow c-77=0$

$\Rightarrow c=77$

Putting value of $c=77$ in equation $\left(1\right)$

$\Rightarrow 2a+5\times 77-3=0$

$\Rightarrow 2a+385-3=0$

$\Rightarrow 2a=-382$

$\Rightarrow a=\frac{-382}{2}$

$\Rightarrow a=-191$

From equation $\left(3\right)$

$2b+5d=0$

$\Rightarrow 2b=-5d$

$\Rightarrow b=\left(\frac{-5}{2}\right)d$

Putting values of $b$ in equation $\left(4\right)$

$\Rightarrow 3\left(\frac{-5}{2}\right)d+8d-22=0$

$\Rightarrow \frac{-15d}{2}+8d-22=0$

$\Rightarrow \frac{-15d+16d-44}{2}=0$

$\Rightarrow d-44=0$

$\Rightarrow d=44$

Putting value of $d$ in equation $\left(3\right)$

$\Rightarrow 2b+5\times 44=0$

$\Rightarrow 2b+220=0$

$\Rightarrow 2b=-220$

$\Rightarrow b=\frac{-220}{2}$

$\Rightarrow b=-110$

Hence, $a=-191,b=-110,c=77,d=44$.

Therefore, Matrix $D$ is $\left[\begin{array}{cc}-191& -110\\ 77& 44\end{array}\right]$