If $A\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right]$, then show that $|3 A|=27|A|$.
$A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right]$
$\Rightarrow 3 A=\left[\begin{array}{ccc}3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12\end{array}\right]$ [Multiplying each element of A by 3 ]
$\Rightarrow|3 A|=(-1)^{1+1} 3(36-0)+(-1)^{1+2} 0(0-0)+(-1)^{1+3} 3(0-0)=3(36-0)-0(0-0)+3(0-0) \quad\left[\right.$ Expanding along $\left.R_{1}\right]$
$=3 \times 36=108$ $\ldots(1)$
$\Rightarrow|A|=(-1)^{1+1} 1(4-0)+(-1)^{1+2} 0(0-0)+(-1)^{1+3} 1(0-0)=1(4-0)-0(0-0)+1(0-0)=4 \quad\left[\right.$ Expanding along $\left.R_{1}\right]$ $\Rightarrow 27|A|=27 \times 4=108 \quad \ldots(2)$
$\therefore|3 A|=27|A| \quad[$ From eqs. (1) and (2) $]$