Question

If $\begin{vmatrix}3^{2}+ k & 4^{2} & 3^{2}+3+ k \\ 4^{2}+ k & 5^{2} & 4^{2}+4+ k \\ 5^{2}+ k & 6^{2} & 5^{2}+5+ k \end{vmatrix}=0$, then the value of $k$ is

• A. 0
• B. -1
• C. 2
• D. 1

Answer 1

Answer: (D)

Breaking the given determinant into two determinants, we get,
$\begin{vmatrix}3^{2}+ k & 4^{2} & 3^{2}+ k \\ 4^{2}+ k & 5^{2} & 4^{2}+ k \\ 5^{2}+ k & 6^{2} & 5^{2}+ k \end{vmatrix} +\begin{vmatrix}3^{2}+ k & 4^{2} & 3 \\ 4^{2}+ k & 5^{2} & 4 \\ 5^{2}+ k & 6^{2} & 5\end{vmatrix}=0$
$\Rightarrow 0+\begin{vmatrix}9+ k & 16 & 3 \\ 7 & 9 & 1 \\ 9 & 11 & 1\end{vmatrix}|=0$
[Applying $R _{3}- R _{2}$ and $R _{2}- R _{1}$ in second det.]
$\Rightarrow \begin{vmatrix}9+ k & 16 & 3 \\ 7 & 9 & 1 \\ 2 & 2 & 0\end{vmatrix}=0$
[Applying $R_{3}- R_{2}]$
$\Rightarrow \begin{vmatrix}9+ k & 7- k & 3 \\ 7 & 2 & 1 \\ 2 & 0 & 0\end{vmatrix}=0$
[ Applying $\left.C _{2}- C _{1}\right]$
$\Rightarrow 2(7-k-6)=0$
$ \Rightarrow k=1$