Question

Matrix $A=\left[\begin{array}{ccc}x& 3& 2\\ 1& y& 4\\ 2& 2& z\end{array}\right]$. If $xyz=60$ and $8x+4y+3z=20$, then $A\left(adjA\right)$ is equal to

• A. $\left[\begin{array}{ccc}64& 0& 0\\ 0& 64& 0\\ 0& 0& 64\end{array}\right]$
• B. $\left[\begin{array}{ccc}88& 0& 0\\ 0& 88& 0\\ 0& 0& 88\end{array}\right]$
• C. $\left[\begin{array}{ccc}68& 0& 0\\ 0& 68& 0\\ 0& 0& 68\end{array}\right]$
• D. $\left[\begin{array}{ccc}34& 0& 0\\ 0& 34& 0\\ 0& 0& 34\end{array}\right]$

Answer 1

Answer: (C)

$\left[\begin{array}{ccc}68& 0& 0\\ 0& 68& 0\\ 0& 0& 68\end{array}\right]$

Given:

$A=\left[\begin{array}{ccc}x& 3& 2\\ 1& y& 4\\ 2& 2& z\end{array}\right]$

$xyz=60$ and $8x+4y+3z=20$

$|A|=\left|\begin{array}{ccc}x& 3& 2\\ 1& y& 4\\ 2& 2& z\end{array}\right|$

$\Rightarrow |A|=x(yz-8)-3(z-8)+2(2-2y)$
$\Rightarrow |A|=xyz-8x-4y-3z+28=68$

We know,

$A\left(adjA\right)=|A|I$, where $I$ is the Identity matrix.
$\therefore A\left(adjA\right)=\left[\begin{array}{ccc}68& 0& 0\\ 0& 68& 0\\ 0& 0& 68\end{array}\right]$

Hence, option C.