Solve this

Solution:

Given:

$A=\left[\begin{array}{lll}x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z\end{array}\right]$

$x y z=80$

$3 x+2 y+10 z=20$

$A(\operatorname{adj} A)=k l$

Now,

$A=\left[\begin{array}{lll}x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z\end{array}\right]$

$\Rightarrow|A|=\left|\begin{array}{lll}x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z\end{array}\right|$

$\Rightarrow|A|=x(y z-3)-2(5 z-2)+1(15-2 y)$

$\Rightarrow|A|=x y z-3 x-10 z+4+15-2 y$

$\Rightarrow|A|=x y z-(3 x+2 y+10 z)+19$

$\Rightarrow|A|=80-(20)+19 \quad(\because x y z=80$ and $3 x+2 y+10 z=20)$

$\Rightarrow|A|=60+19$

$\Rightarrow|A|=79$

As we know,

$A(\operatorname{adj} A)=|A| I$

$\Rightarrow k I=79 I$

$\Rightarrow k=79$

Hence, $k=\underline{79}$.