Question

x + 1 & x + 2 & x + 4 \\ x + 3 & x + 5 & x + 8 \\ x + 7 & x + 10 & x + 14 \end{matrix} \right| =$$

• A. 2
• B. – 2
• C. $x^{2} - 2$
• D. None of these

Answer 1

Answer: (B)

– 2

Answer 2

$\Delta = \left| \begin{matrix}

• 1 & - 2 & x + 4 \
• 2 & - 3 & x + 8 \
• 3 & - 4 & x + 14 \end{matrix} \right|,$by$C_{1} \rightarrow C_{1} - C_{2} $$C_{2} \rightarrow C_{2} - C_{3} $

= $A^{3} = A^{2}A = \begin{bmatrix} 2 & 3 & 1 \\ 5 & 6 & 2 \\ 3 & 4 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 2 & 1 & 0 \end{bmatrix}$, by $C_{2} \rightarrow C_{2} - C_{1}$ $$C_{3} \rightarrow C_{3} + 4C_{1}$$

$$= - ( - x - 2 + x) + 1.( - 2x - 4 + 3x) + x(2 - 3)$$

= $2 + x - 4 - x = - 2$.

Trick : Put $C_{2}$. Then $\left| \begin{matrix} 2 & 3 & 5 \\ 4 & 6 & 9 \\ 8 & 11 & 15 \end{matrix} \right| = - 2$

Note : Since there is a option “None of these”, therefore we should check for one more different value of x. Put $x = - 1$.

0 & 1 & 3 \\ 2 & 4 & 7 \\ 6 & 9 & 13 \end{matrix} \right| = - 1(26 - 42) + 3(18 - 24) = - 2$$ Therefore answer is (2).