$A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right],$ then show that $|3 A|=27|A|$.
$A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right],$ then show that $|3 A|=27|A|$.
The given matrix is $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right]$
It can be observed that in the first column, two entries are zero. Thus, we expand along the first column $(C_1 )$ for easier calculation.
$|A|=1\left|\begin{array}{ll}1 & 2 \\ 0 & 4\end{array}\right|-0\left|\begin{array}{ll}0 & 1 \\ 0 & 4\end{array}\right|+0\left|\begin{array}{ll}0 & 1 \\ 1 & 2\end{array}\right|=1(4-0)-0+0=4$
$\therefore 27|A|=27(4)=108......(i)$
${{\text{Now, }}3A = 3\left[ {\begin{array}{*{20}{l}}
1&0&1 \\
0&1&2 \\
0&0&4
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
3&0&3 \\
0&3&6 \\
0&0&{12}
\end{array}} \right]}$
${\therefore \,|3A| = 3\left| {\begin{array}{*{20}{l}}
3&6 \\
0&{12}
\end{array}} \right| - 0\left| {\begin{array}{*{20}{l}}
0&3 \\
0&{12}
\end{array}} \right| + 0\left| {\begin{array}{*{20}{l}}
0&3 \\
3&6
\end{array}} \right|}$
${\begin{array}{*{20}{l}}
{ = 3(36 - 0) = 3(36) = 108}
\end{array}}......(ii)$
From equations $( i )$ and $(ii)$, we have:
$|3 A|=27|A|$
Hence, the given result is proved.
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