begin{aligned} & text { If }left|begin{array} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the determinant $\Delta = \left|\begin{array}{ccc}n^2 & (n+1)^2 & (n+2)^2 \ (n+1)^2 & (n+2)^2 & (n+3)^2 \ (n+2)^2 & (n+3)^2 & (n+4)^2\end{

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Given the determinant $\Delta = \left|\begin{array}{ccc}n^2 & (n+1)^2 & (n+2)^2 \ (n+1)^2 & (n+2)^2 & (n+3)^2 \ (n+2)^2 & (n+3)^2 & (n+4)^2\end{array}ight|$.Applying row operations $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_2$, we get:$\Delta = \left|\begin{array}{ccc}n^2 & (n+1)^2 & (n+2)^2 \ 2n+1 & 2n+3 & 2n+5 \ 2 & 2 & 2\end{array}ight|$.Applying $R_2 ightarrow R_2 - R_1$ again, we get:$\Delta = \left|\begin{array}{ccc}n^2 & 2n+1 & 2 \ 2n+1 & 2 & 0 \ 2 & 0 & 0\end{array}ight| = 2(0 - 4) = -8$.Now, given the second determinant equation:$\left|\begin{array}{ccc}1 & -4 & 7 \ -2 & 3 & -5 \ 3 & x & -3\end{array}ight| = 2\Delta + 1 = 2(-8) + 1 = -15$.Expanding the determinant:$1(-9 + 5x) + 4(6 + 15) + 7(-2x - 9) = -15$.$-9 + 5x + 84 - 14x - 63 = -15$.$-9x + 12 = -15$.$-9x = -27$.$x = 3$.

Similar Questions

If matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,then $|A|^{-1}$ is equal to

Find the equation of the line joining $(1, 2)$ and $(3, 6)$ using determinants.

$\Delta = \left| \begin{array}{ccc} a & a+b & a+b+c \\ 3a & 4a+3b & 5a+4b+3c \\ 6a & 9a+6b & 11a+9b+6c \end{array} \right|$ where $a = i, b = \omega, c = \omega^2$,then $\Delta$ is equal to

The value of $\left| \begin{array}{ccc} 1 & 1 & 1 \\ bc & ca & ab \\ b+c & c+a & a+b \end{array} \right|$ is

At what value of $x$ will $\left| \begin{array}{ccc} x + \omega^2 & \omega & 1 \\ \omega & \omega^2 & 1 + x \\ 1 & x + \omega & \omega^2 \end{array} \right| = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module