Question

If $\left| \begin{matrix} x + 1 & 3 & 5 \\ 2 & x + 2 & 5 \\ 2 & 3 & x + 4 \end{matrix} \right| = 0$, then x =

• A. 1, 9
• B. – 1, 9
• C. – 1, – 9
• D. 1, – 9

Answer 1

Answer: (D)

1, – 9

Answer 2

By $C_{1} \rightarrow C_{1} + C_{2} + C_{3}$,

we have $(9 + x)$ $\left| \begin{matrix} 1 & 3 & 5 \\ 1 & x + 2 & 5 \\ 1 & 3 & x + 4 \end{matrix} \right|$ = 0

$\Rightarrow (x + 9)$ $\left| \begin{matrix} 0 & 1 - x & 0 \\ 0 & - (1 - x) & 1 - x \\ 1 & 3 & x + 4 \end{matrix} \right| = 0$

$\Rightarrow (x + 9)$ $(1 - x)^{2}\left| \begin{matrix} 0 & 1 & 0 \\ 0 & - 1 & 1 \\ 1 & 3 & x + 4 \end{matrix} \right| = 0$

$\Rightarrow x = 1,1, - 9$, (Since the determinant = 1).