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 ..... .
- ✓$72$
- B$12$
- C$36$
- D$6$
$R _{2} \rightarrow R _{1}+ R _{3}-2 R _{2}$
$\Rightarrow\left|\begin{array}{ccc}3 & 4 \sqrt{2} & x \\ 0 & k-6 \sqrt{2} & 0 \\ 5 & k & z\end{array}\right|=0$
$\Rightarrow(k-6 \sqrt{2})(3 z-5 x)=0$
if $3 z-5 x=0 \Rightarrow 3(x+2 d)-5 x=0$
$\Rightarrow x=3 d$ (Not possible)
$\Rightarrow k =6 \sqrt{2} \quad \Rightarrow k ^{2}=72$
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$STATEMENT -1$ : $\mathrm{P}\left(\mathrm{H}_{\mathrm{i}} \mid \mathrm{E}\right)>\mathrm{P}\left(\mathrm{E} \mid \mathrm{H}_{\mathrm{i}}\right) \cdot \mathrm{P}\left(\mathrm{H}_{\mathrm{i}}\right)$ for $\mathrm{i}=1,2, \ldots, \mathrm{n}$ because
$STATEMENT$ $-2: \sum_{1=1}^{\mathrm{n}} \mathrm{P}\left(\mathrm{H}_{\mathrm{i}}\right)=1$
$A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|$. Then $2 A_{10}-A_8$