Question

The value of the determinant of $\begin{bmatrix}x+1&x+2&x+4\\\ x+3 &x+5&x+8\\\ x+7 &x+10&x+14\end{bmatrix}$ is :

• A. -2
• B. $x^2 + 2$
• C. 2
• D. 3

Answer 1

Answer: (A)

-2

Answer 2

Let $A = \begin{vmatrix}x+1&x+2&x+4\\\ x+3 &x+5&x+8\\\ x+7 &x+10&x+14\end{vmatrix}$ Operate $R_{2} \to R_{2} -R_{1} ,$ and $R_{3} \to R_{3} - R_{1}$ So, $A = \begin{vmatrix}x+1&x+2&x+4\\\ 2 &3&4\\\ 6&8&10\end{vmatrix}$ Again operate, $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$ Thus, $A = \begin{vmatrix}x+1&1&3\\\ 2 &1&2\\\ 6&2&4\end{vmatrix}$ Expanding by first row, we get A = (x + 1) (4 - 4) - 1 (8 - 12) + 3 (4 - 6) = 4 - 6 = - 2.