Question

Given 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

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

$\begin{array}{c}Given,\\ A=\left[\begin{array}{ccc}x& 3& 2\\ 1& y& 4\\ 2& 2& z\end{array}\right]\\ Now,\\ \left|A\right|=\left|\begin{array}{ccc}x& 3& 2\\ 1& y& 4\\ 2& 2& z\end{array}\right|\\ =x\left(yz-8\right)-3\left(z-8\right)+2\left(2-2y\right)\\ =xyz-8x-3z+24+4-4y\\ =60+28-20\\ \left|A\right|=68\\ Hence,A.adj\left(A\right)=\left[\begin{array}{ccc}68& 0& 0\\ 0& 68& 0\\ 0& 0& 68\end{array}\right]\end{array}$

Hence, this is the answer.