Question

If $\left|\begin{array}{ccc}{3}^2+k& 4^2& 3^2+3+k\\ 4^2+k& 5^2& 4^2+4+k\\ 5^2+k& 6^2& 5^2+5+k\end{array}\right|=0$, 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,
$\left|\begin{array}{ccc}{3}^2+k& 4^2& 3^2+k\\ 4^2+k& 5^2& 4^2+k\\ 5^2+k& 6^2& 5^2+k\end{array}\right|+\left|\begin{array}{ccc}{3}^2+k& 4^2& 3\\ 4^2+k& 5^2& 4\\ 5^2+k& 6^2& 5\end{array}\right|=0$
$\Rightarrow 0+\left|\begin{array}{ccc}9+k& 16& 3\\ 7& 9& 1\\ 9& 11& 1\end{array}\right||=0$
[Applying $R_3-R_2$ and $R_2-R_1$ in second det.]
$\Rightarrow \left|\begin{array}{ccc}9+k& 16& 3\\ 7& 9& 1\\ 2& 2& 0\end{array}\right|=0$
[Applying $R_3-R_2]$
$\Rightarrow \left|\begin{array}{ccc}9+k& 7-k& 3\\ 7& 2& 1\\ 2& 0& 0\end{array}\right|=0$
[ Applying $C_2-C_1]$
$\Rightarrow 2(7-k-6)=0$
$\Rightarrow k=1$