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Q. If $A=\begin{bmatrix}8 & 27 & 125 \\ 2 & 3 & 5 \\ 1 & 1 & 1\end{bmatrix}$, then the value of $A^{2}$ is equal to

KEAMKEAM 2016Determinants

Solution:

$A=\begin{vmatrix} 8 & 27 & 125 \\ 2 & 3 & 5 \\ 1 & 1 & 1\end{vmatrix} $
$ =8(3-5)-27(2-5)+125(2-3)$
$= 8(-2)-27(-3)+125(-1) $
$ =-16+81-125 $
$ =-60$
$ \therefore A^{2} =(-60)^{2}=3600 $