Question

If A=$\begin{bmatrix}1&0&1\\\0&1&2\\\0&0&4\end{bmatrix}$,then show that$\mid A\mid=27\mid A \mid$

Answer 1

The given matrix is A=$\begin{bmatrix}1&0&1\\\0&1&2\\\0&0&4\end{bmatrix}$
It can be observed that in the first column, two entries are zero.
Thus, we expand along
the first column (C1) for easier calculation.
$\mid A \mid=1\begin{vmatrix}1&2&\\\0&4\end{vmatrix}-0\begin{vmatrix}0&1\\\0&4\end{vmatrix}+0\begin{vmatrix}0&1\\\1&2\end{vmatrix}=1(4-0)-0+0=4$
so 27IAI=27(4)=108 ....(1)

Now 3A=3$\begin{bmatrix}1&1&1\\\0&1&2\\\0&0&4\end{bmatrix}$[=$\begin{bmatrix}3&0&3\\\0&3&6\\\0&0&12\end{bmatrix}$[303 036 0012]
so I3AI=$3\begin{vmatrix}3&6\\\0&12\end{vmatrix}-0\begin{vmatrix}0&3\\\0&12\end{vmatrix}+0\begin{vmatrix}0&3\\\3&6\end{vmatrix}$

=3(36-0)=3x36=108 ...(2)
From equations (1) and (2), we have:
I3AI=27IAI

Hence, the given result is proved