If $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right],$ then show that $|2 A|=4|A|$.
If $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right],$ then show that $|2 A|=4|A|$.
The given matrix is $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right]$
$\therefore 2 A=2\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right]=\left[\begin{array}{ll}2 & 4 \\ 8 & 4\end{array}\right]$
$\begin{aligned} L H S:|2 A| &=\left|\begin{array}{ll}2 & 4 \\ 8 & 4\end{array}\right| \\ &=2 \times 4-4 \times 8 \\ &=8-32=-24 \end{aligned}$
Now, $|A|=\left[\begin{array}{ll}1 & 2 \\ 4 & 2\end{array}\right]=1 \times 2-2 \times 4=2 \times 8=-6$
$R H S: 4|A|=4 \times(-6)=-24$
$\therefore L. H. S.=\therefore \mathrm{R.} \mathrm{H.} \mathrm{S}$
Similar Questions
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$x + \left( {\sin \,\alpha } \right)y + \left( {\cos \,\alpha } \right)z = 0$
$x + \left( {\cos \,\alpha } \right)y + \left( {\sin \alpha } \right)z = 0$
$x - \left( {\sin \,\alpha } \right)y - \left( {\cos \alpha } \right)z = 0$
has a non-trivial solution for only one value of $\alpha $ lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$
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$\left| {\begin{array}{*{20}{c}}
{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha } \\
{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha } \\
{\cos {\mkern 1mu} \alpha }&{ - \sin {\mkern 1mu} \alpha }&{ - \cos {\mkern 1mu} \alpha }
\end{array}} \right| = 0$
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Statement $-1$ : The system of linear equations
$x + \left( {\sin \,\alpha } \right)y + \left( {\cos \,\alpha } \right)z = 0$
$x + \left( {\cos \,\alpha } \right)y + \left( {\sin \alpha } \right)z = 0$
$x - \left( {\sin \,\alpha } \right)y - \left( {\cos \alpha } \right)z = 0$
has a non-trivial solution for only one value of $\alpha $ lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$
Statement $-2$ : The equation in $\alpha $
$\left| {\begin{array}{*{20}{c}}
{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha } \\
{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha } \\
{\cos {\mkern 1mu} \alpha }&{ - \sin {\mkern 1mu} \alpha }&{ - \cos {\mkern 1mu} \alpha }
\end{array}} \right| = 0$
has only one solution lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$
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