Question

If A = $\left[\begin{array}{ccc}x& 5& 2\\ 2& y& 3\\ 1& 1& z\end{array}\right]$ ,xyz = 80, 3x + 2y + 10z = 20 and A adj A = kI, then k = _________________.

Answer 1

Given:
A = $\left[\begin{array}{ccc}x& 5& 2\\ 2& y& 3\\ 1& 1& z\end{array}\right]$
xyz = 80
3x + 2y + 10z = 20
A (adj A) = kI


Now,
$$\begin{gathered} A=\left[\begin{array}{ccc}x& 5& 2\\ 2& y& 3\\ 1& 1& z\end{array}\right] \\ \Rightarrow \left|A\right|=\left|\begin{array}{ccc}x& 5& 2\\ 2& y& 3\\ 1& 1& z\end{array}\right| \\ \Rightarrow \left|A\right|=x\left(yz-3\right)-2\left(5z-2\right)+1\left(15-2y\right) \\ \Rightarrow \left|A\right|=xyz-3x-10z+4+15-2y \\ \Rightarrow \left|A\right|=xyz-\left(3x+2y+10z\right)+19 \\ \Rightarrow \left|A\right|=80-\left(20\right)+19\left(\because xyz=80\mathrm{and}3x+2y+10z=20\right) \\ \Rightarrow \left|A\right|=60+19 \\ \Rightarrow \left|A\right|=79 \end{gathered}$$

As we know,
$$\begin{gathered} A\left(\mathrm{adj}A\right)=\left|A\right|I \\ \Rightarrow kI=79I \\ \Rightarrow k=79 \end{gathered}$$


Hence, k = 79.