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If $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right]$, then show that $|3 \mathbf{A}|=27|\mathbf{A}|$
Solution:
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Verified Answer
$3 A=3\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right]=\left[\begin{array}{ccc}3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12\end{array}\right]$
$\therefore \quad|3 \mathrm{~A}|=\left|\begin{array}{rrr}
3 & 0 & 3 \\
0 & 3 & 6 \\
0 & 0 & 12
\end{array}\right|$
Taking common 3 out 3 each from $\mathrm{R}_1, \mathrm{R}_2$ and $\mathrm{R}_3$ respectively.
$\begin{aligned}
&|3 \mathrm{~A}|=3 \times 3 \times 3 \times\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right|=27\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right| \\
&\Rightarrow|27 \mathrm{~A}|=27|\mathrm{~A}|
\end{aligned}$
$\therefore \quad|3 \mathrm{~A}|=\left|\begin{array}{rrr}
3 & 0 & 3 \\
0 & 3 & 6 \\
0 & 0 & 12
\end{array}\right|$
Taking common 3 out 3 each from $\mathrm{R}_1, \mathrm{R}_2$ and $\mathrm{R}_3$ respectively.
$\begin{aligned}
&|3 \mathrm{~A}|=3 \times 3 \times 3 \times\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right|=27\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right| \\
&\Rightarrow|27 \mathrm{~A}|=27|\mathrm{~A}|
\end{aligned}$
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