Question

If $A=\left|\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 0& 0& 4\end{array}\right|,$ then show that $|3A|=27|A|$

Answer 1

Given $A=\left|\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 0& 0& 4\end{array}\right|$
To show: $|3A|=27|A|$
Taking L.H.S.
$|3A|=\left|\begin{array}{c}\left(3\right)\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 0& 0& 4\end{array}\right]\end{array}\right|=\left[\begin{array}{ccc}3& 0& 3\\ 0& 3& 6\\ 0& 0& 12\end{array}\right]$
$=3\left|\begin{array}{cc}3& 6\\ 0& 12\end{array}\right|-0\left|\begin{array}{cc}0& 6\\ 0& 12\end{array}\right|+3\left|\begin{array}{cc}0& 3\\ 0& 0\end{array}\right|$
$=3\left(3\right(12)-0(6\left)\right)-0\left(0\right(12)-0(6\left)\right)+3\left(0\right(0)-0(3\left)\right)$
$=3(36-0)-0\left(0\right)+3\left(0\right)$
$=3\left(36\right)+0+0=108$
$\therefore L.H.S=|3A|=108$

Taking R.H.S.
$27|A|=\left(27\right)\left|\begin{array}{c}\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 0& 0& 4\end{array}\right]\end{array}\right|$
$=\left(27\right)(1\left|\begin{array}{cc}1& 2\\ 0& 4\end{array}\right|-0\left|\begin{array}{cc}0& 2\\ 0& 4\end{array}\right|+1\left|\begin{array}{cc}0& 1\\ 0& 0\end{array}\right|)$
$=\left(27\right)\left(1\right(1\left(4\right)-0\left(2\right))-0(0\left(4\right)-0\left(2\right))+1(0-0\left(1\right)\left)\right)$
$=\left(27\right)\left(1\right(4-0)-0(0)+1(0\left)\right)$
$=27\left(4\right)$
$=108$
$\therefore L.H.S.=R.H.S.$
Hence Proved