Question

If the matrix $A=\left[\begin{array}{ccc}x& 3& 2\\ 1& y& 4\\ 2& 2& z\end{array}\right],xyz=60$ and $8x+4y+3z=10$, 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. $\left[\begin{array}{ccc}35& 0& 0\\ 0& 35& 0\\ 0& 0& 35\end{array}\right]$

Answer 1

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

$\because A\cdot dj(A)=\left| A \right| I\quad $$\therefore \left| A \right| =xyz-8x-3(z-8)+2(2-2y)$$=60-20+28=68\quad $$\quad \therefore A\cdot adj(A)=68\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 68 & 0 & 0 \\ 0 & 68 & 0 \\ 0 & 0 & 68 \end{bmatrix}$∵A⋅dj(A

$\because A\cdot dj\left(A\right)=\left|A\right|I$
$\therefore \left|A\right|=xyz-8x-3(z-8)+2(2-2y)$
$=60-20+28=68$
$\therefore A\cdot adj\left(A\right)=68\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}68& 0& 0\\ 0& 68& 0\\ 0& 0& 68\end{array}\right]$