<p>Find the value of <strong>k</strong> such that the determinant <br/>\[\begin{vmatrix} 3^2 + k & 4^2 & 3^2 + 3 + k \\ 4^2 + k & 5^2 & 4^2 + 4 + k \\ 5^2 + k & 6^2 & 5^2 + 5 + k \end{vmatrix} = 0\]</p>
Step-by-Step Solution
Key Concept: Use column and row operations to simplify the determinant, then expand to find the value of k.
<p><strong>Step 1:</strong> Apply column operation $C_3 \to C_3 - C_1$:</p><p>$$\begin{vmatrix} 3^2 + k & 4^2 & 3 \\ 4^2 + k & 5^2 & 4 \\ 5^2 + k & 6^2 & 5 \end{vmatrix} = 0$$</p><p><strong>Step 2:</strong> Apply row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:</p><p>$$\begin{vmatrix} 9 + k & 16 & 3 \\ 7 & 9 & 1 \\ 16 & 20 & 2 \end{vmatrix} = 0$$</p><p><strong>Step 3:</strong> Expand the determinant:</p><p>$(9 + k)(18 - 20) - 16(14 - 16) + 3(140 - 144) = 0$</p><p>$(9 + k)(-2) - 16(-2) + 3(-4) = 0$</p><p>$-18 - 2k + 32 - 12 = 0$</p><p>$2k = 2$</p><p>$\therefore k = 1$</p>
Correct Answer: 1