If \( \begin{vmatrix}2&3&2\\x&x&x\\4&9&1\end{vmatrix}+\begin{vmatrix}3&2\\4&5\end{vmatrix}=0 \), then the value of \( x \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → −7/3
The second row of the first determinant has the common factor x, which comes out of the determinant:
\( \displaystyle \begin{vmatrix}2&3&2\\x&x&x\\4&9&1\end{vmatrix}=x\begin{vmatrix}2&3&2\\1&1&1\\4&9&1\end{vmatrix} \)
\( \displaystyle =x\left[2(1-9)-3(1-4)+2(9-4)\right] \)
\( \displaystyle =x\left[-16+9+10\right] \)
\( \displaystyle =3x \)
The second determinant is
\( \displaystyle \begin{vmatrix}3&2\\4&5\end{vmatrix}=15-8 \)
\( \displaystyle =7 \)
So
\( \displaystyle 3x+7=0 \)
\( \displaystyle x=-\frac{7}{3} \)