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If $A=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right], B=(\operatorname{adj} A)$ and $C=5 A$, then $\frac{|C|}{|\operatorname{adj} B|}$ is equal to
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125
We have,
$$
\begin{aligned}
A &=\left[\begin{array}{lll}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array}\right] \\
\text { Now, }|A| &=\left[\begin{array}{lll}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array} \mid\right.\\
&=3(-3+4)+3(2-0)+4(-2-0) \\
&=3+6-8=1 \\
\text { Now, } \frac{|C|}{|\operatorname{adj} B|} &=\frac{|5 A|}{|B|^{3-1}}=\frac{5^{3}|A|}{|B|^{2}}=\frac{5^{3}|A|}{|\operatorname{adj} A|^{2}} \\
&=\frac{5^{3}|A|}{\left(|A|^{2}\right)^{2}}=\frac{5^{3}|A|}{|A|^{4}}=\frac{5^{3}}{|A|^{3}} \\
&=\frac{5^{3}}{1^{3}}=5^{3}
\end{aligned}
$$
$$
\begin{aligned}
A &=\left[\begin{array}{lll}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array}\right] \\
\text { Now, }|A| &=\left[\begin{array}{lll}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array} \mid\right.\\
&=3(-3+4)+3(2-0)+4(-2-0) \\
&=3+6-8=1 \\
\text { Now, } \frac{|C|}{|\operatorname{adj} B|} &=\frac{|5 A|}{|B|^{3-1}}=\frac{5^{3}|A|}{|B|^{2}}=\frac{5^{3}|A|}{|\operatorname{adj} A|^{2}} \\
&=\frac{5^{3}|A|}{\left(|A|^{2}\right)^{2}}=\frac{5^{3}|A|}{|A|^{4}}=\frac{5^{3}}{|A|^{3}} \\
&=\frac{5^{3}}{1^{3}}=5^{3}
\end{aligned}
$$
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