Question

If $a,b,c$ are in A.P., then the value of $\left| \begin{matrix} x + 2 & x + 3 & x + a \\ x + 4 & x + 5 & x + b \\ x + 6 & x + 7 & x + c \end{matrix} \right|$ is.

• A. $x - (a + b + c)$
• B. $9x^{2} + a + b + c$
• C. $a + b + c$
• D. 0

Answer 1

Answer: (D)

0

Answer 2

Let $\mathbf{A =}\left| \begin{matrix} \mathbf{x + 2} & \mathbf{x + 3} & \mathbf{x + a} \\ \mathbf{x + 4} & \mathbf{x + 5} & \mathbf{x + b} \\ \mathbf{x + 6} & \mathbf{x + 7} & \mathbf{x + c} \end{matrix} \right|$

Applying $C_{2} \rightarrow C_{2} - C_{1},$ we get,

⇒ $A = \left| \begin{matrix} x + 2 & 1 & x + a \\ x + 4 & 1 & x + b \\ x + 6 & 1 & x + c \end{matrix} \right|$

Applying $\left| \begin{matrix} a & a^{3} & a^{4} \\ b & b^{3} & b^{4} \\ c & c^{3} & c^{4} \end{matrix} \right| + \left| \begin{matrix} a & a^{3} & - 1 \\ b & b^{3} & - 1 \\ c & c^{3} & - 1 \end{matrix} \right| = 0$ and $R_{3} \rightarrow R_{3} - R_{1}$

⇒ $A = \left| \begin{matrix} x + 2 & 1 & x + a \\ 2 & 0 & b - a \\ 4 & 0 & c - a \end{matrix} \right| = - 1(2c - 2a - 4b + 4a)$

= $2(2b - c - a)$

$\because$ a, b, c are in A.P. ⇒ A = 0.