Question

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

Answer 1

The correct option is C $A=\left[\begin{array}{ccc}68& 0& 0\\ 0& 68& 0\\ 0& 0& 68\end{array}\right]$

Given, $A=\left[\begin{array}{ccc}x& 3& 2\\ 1& y& 4\\ 2& 2& z\end{array}\right]$
$|A|=xyz-(8x+4y+3z)+28$
$\Rightarrow |A|=68$
Now, $A\left(adjA\right)=|A|I$
$=68I$
$=\left[\begin{array}{ccc}68& 0& 0\\ 0& 68& 0\\ 0& 0& 68\end{array}\right]$