# $\text{If A = } \begin{bmatrix} 1&0&1 \\ 0&1&2 \\ 0&0&4 \end{bmatrix}, \text{then show that } |\;3A\;| = 27|\;A\;|$

Toolbox:
• To evaluate a matrix of order $3\times 3$
• $\mid A\mid=\begin{vmatrix}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{vmatrix}$
• $\mid A \mid=a_{11}(a_{22}a_{33}-a_{32}a_{23})-a_{12}(a_{21}a_{33}-a_{31}a_{23}+a_{13}(a_{21}a_{32}-a_{31}a_{22})$
Given $A=\begin{vmatrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{vmatrix}$,then show that $\mid 3A\mid=27\mid A\mid.$

$\mid A \mid=a_{11}(a_{22}a_{33}-a_{32}a_{23})-a_{12}(a_{21}a_{33}-a_{31}a_{23}+a_{13}(a_{21}a_{32}-a_{31}a_{22})$

$\qquad=1(1\times 4-2\times 0)-0(0\times 4-2\times 0)+1(0\times 0-1\times 0)$

$\qquad=4$

27$\mid A\mid=27\times 4$

$\qquad=108$-----(1)

$\mid 3A\mid=\begin{vmatrix}3 & 0 & 3\\0 & 3 & 6\\0 & 0 &12\end{vmatrix}$(By multiplying each element by 3)

Hence the value of $\mid 3A\mid$ can be obtained by the above method

$\mid 3A\mid=3(3\times 12-6\times 0)-0(0\times 12-6\times 0)+3(0\times 0-1\times 0)$

$\qquad=3\times 36$

$\qquad=108$------(2)

Hence from equ(1) and equ(2) we find that

LHS=RHS.

(i.e)$\mid 3A\mid=27\mid A\mid$