If $A=\left[\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right],$ find $|A|$.
Let $A=\left[\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right]$
By expanding along the first row, we have:
$A=\left[\begin{array}{lll}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{array}\right]$
$|A|=1\left|\begin{array}{cc}1 & -3 \\ 4 & -9\end{array}\right|-1\left|\begin{array}{cc}2 & -3 \\ 5 & -9\end{array}\right|-2\left|\begin{array}{cc}2 & 1 \\ 5 & 4\end{array}\right|$
$=1(-9+12)-1(-18+15)-2(8-5)$
$=1(3)-1(-3)-2(3)$
$=3+3-6$
$=6-6$
$=0$
If $|A|$ denotes the value of the determinant of the square matrix $A$ of order $3$ , then $ |-2A|=$
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{{A_1}{\kern 1pt} }&{{B_1}}&{{C_1}} \\
{{A_2}}&{{B_2}}&{{C_2}} \\
{{A_3}}&{{B_3}}&{{C_3}}
\end{array}} \right|$ ; then $\Delta $ is divisible by
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