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

Toolbox:
• If a matrix $A=\begin{vmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{vmatrix}$ be a matrix of order $2\times 2$,then $\mid A\mid=\begin{vmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{vmatrix}=a_{11}a_{22}-a_{21}a_{12}$
Given $A=\begin{vmatrix}1 & 2\\4 & 2\end{vmatrix}$ then show that $\mid 2A\mid=4\mid A\mid.$

Let the determinant of $A=\begin{vmatrix}1 & 2\\4 & 2\end{vmatrix}$

To evaluate the value of the determinant let us multiply $a_{11}$ and $a_{22}$ and $a_{21}$ and $a_{12}$ and then subtract both.

$\mid A\mid=1\times 2-4\times 2$

$\qquad=2-8$

$\qquad=-6$

$4\mid A\mid =4\times -6$

$\qquad=-24$------(1)

$\mid 2A\mid=\begin{vmatrix}2 & 4\\8 & 4\end{vmatrix}$

After multiplying each element by 2

Hence following the above method to find the value of the determinant,

$\mid 2A\mid=2\times 4-8\times 4$

$\qquad=8-32$

$\qquad=-24$-----(2)

Hence from equ(1) and equ(2),we find that LHS=RHS.

(i.e)$\mid 2A\mid=4\mid A\mid$