Question

Find the negative of determinant of matrix 'A' such that satisfying $\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]A=\left[\begin{array}{cc}1& 1\\ 0& -1\end{array}\right]$

Answer 1

Let $\frac{1}{-2}\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]A=\left[\begin{array}{cc}1& 1\\ 0& -1\end{array}\right]\to BA=C$

i.e, $B\frac{1}{-2}\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]\xi C=\left[\begin{array}{cc}1& 1\\ 0& -1\end{array}\right]$

Now, multiply both sides by $B^{-1},$ we get

$\Rightarrow B^{-1}\left(BA\right)=B^{-1}C$

$\Rightarrow A=B^{-1}C$ $[\therefore (B^{-1})A=IA=A]$

$=\frac{1}{\left|B\right|}\left[\begin{array}{cc}1& -3\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& -1\end{array}\right]$

$=\frac{1}{(1-0)}\left[\begin{array}{cc}1& -3\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& -1\end{array}\right]$

$=\left[\begin{array}{cc}1\times 1+0& (1\times 1)+[(-3)\times (-1)]\\ 0+0& 0-1\end{array}\right]$

$\therefore$ $A=\left[\begin{array}{cc}1& 4\\ 0& -1\end{array}\right]$

Hence, the determinant of $\left[\begin{array}{cc}1& 4\\ 0& -1\end{array}\right]=-1$