Question

If $x,y,z$ are in arithmetic progression with common difference $d,x\ne 3d,$ and the determinant of the matrix $\left[\begin{array}{ccc}3& 4\sqrt{2}& x\\ 4& 5\sqrt{2}& y\\ 5& k& z\end{array}\right]$ is zero, then the value of $k^2$ is:

• A. $6$
• B. $36$
• C. $72$
• D. $12$

Answer 1

The correct option is C $72$

Given $\left|\begin{array}{ccc}3& 4\sqrt{2}& x\\ 4& 5\sqrt{2}& y\\ 5& k& z\end{array}\right|=0$

Applying $R_1\to R_1+R_3-2R_2$
$\Rightarrow \left|\begin{array}{ccc}0& 4\sqrt{2}+k-10\sqrt{2}& 0\\ 4& 5\sqrt{2}& y\\ 5& k& z\end{array}\right|=0(\because 2y=x+z)$

$\Rightarrow (k-6\sqrt{2})(4z-5y)=0$
$\Rightarrow k=6\sqrt{2}$ (or) $4z=5y$
If $4z=5y\Rightarrow 4(x+2d)=5(x+d)$
$\Rightarrow x=3d($Not possible $)$
So, $k^2=72$