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. $\left|\begin{array}{ccc}68& 0& 0\\ 0& 68& 0\\ 0& 0& 68\end{array}\right|$
• C. $\left|\begin{array}{ccc}38& 0& 0\\ 0& 38& 0\\ 0& 0& 38\end{array}\right|$
  
• D. $\left|\begin{array}{ccc}32& 0& 0\\ 0& 32& 0\\ 0& 0& 32\end{array}\right|$

Answer 1

The correct option is B $\left|\begin{array}{ccc}68& 0& 0\\ 0& 68& 0\\ 0& 0& 68\end{array}\right|$

$A\left(adjA\right)=\left(adjA\right)A=det\left(A\right)I$

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

$=x(yz-8)-3(z-8)+2(2-2y)$

$=xyz-8x-3z+24+4-4y$

$=xyz-(8x+4y+3z)+28$

$=60-20+28$

$=68$

$I=\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$

$A\left(adjA\right)=68\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|=\left|\begin{array}{ccc}68& 0& 0\\ 0& 68& 0\\ 0& 0& 68\end{array}\right|$

Correct option is C