Question

If $x,y,z$ are in arithmetic progression with common difference $d,x\ne 3d$, and the determinant of the matrix

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

is zero, then the value of $k^2$ is:

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

Answer 1

The correct option is C

$72$

Explanation for the correct option.

If $x,y,z$ are in arithmetic progression, then $y-x=z-y=d$.

In the given matrix, let us take

$R_1\to R_1-R_2\mathrm{and}{R}_2\to R_2-R_3$, so we will get

$\left[\begin{array}{ccc}-1& -\sqrt{2}& -d\\ -1& \left(5\sqrt{2}-k\right)& -d\\ 5& k& z\end{array}\right]=0$

Now, by $R_1\to R_1-R_2$, we get

$\left[\begin{array}{ccc}0& \left(-6\sqrt{2}+k\right)& 0\\ -1& \left(5\sqrt{2}-k\right)& -d\\ 5& k& z\end{array}\right]=0$

Now, the determinant will be

$\begin{array}{rcl}0\left[\left(5\sqrt{2}-k\right)z+dk\right]-\left(k-6\sqrt{2}\right)\left[-z+5d\right]+0\left[-k-5\left(5\sqrt{2}-k\right)\right]& =& 0\\ \Rightarrow -\left(k-6\sqrt{2}\right)\left[-z+5d\right]& =& 0\\ \Rightarrow \left(k-6\sqrt{2}\right)\left[-z+5d\right]& =& 0\end{array}$

Now, as $x,y,z$ are in arithmetic progression and $x\ne 3d$, that means $y\ne 4d$ and $z\ne 5d$.

$$\begin{gathered} \begin{array}{rcl}& \Rightarrow & k-6\sqrt{2}=0\\ \Rightarrow k& =& 6\sqrt{2}\end{array} \\ \Rightarrow \begin{array}{rcl}& & k^2=72\end{array} \end{gathered}$$

Hence, option C is correct.