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

We have, $A=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 0& 0& 4\end{array}\right]$
$\therefore 3A=\left[\begin{array}{ccc}3& 0& 3\\ 0& 3& 6\\ 0& 0& 12\end{array}\right]$
Clearly both $A$ and $3A$ are upper triangular matrices
Thus $|A|=\text{Product of diagonal element of matrix A}=1\times 1\times 4=4$
& $|3A|=\text{Product of diagonal element of matrix 3A}=3\times 3\times 12=27\times 4=27|A|$
Hence $|3A|=27|A|$ proved.