The roots of the equation left| begin{array}{ | vedclass

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

Question

(B) Given the determinant equation: $\left| \begin{array}{ccc} x-1 & 1 & 1 \ 1 & x-1 & 1 \ 1 & 1 & x-1 \end{array} ight| = 0$Applying the column o

Vedclass · Accepted Answer

$-1, 2$

Vedclass · Answer

(B) Given the determinant equation: $\left| \begin{array}{ccc} x-1 & 1 & 1 \ 1 & x-1 & 1 \ 1 & 1 & x-1 \end{array} ight| = 0$Applying the column operation $C_1 	o C_1 + C_2 + C_3$:$\left| \begin{array}{ccc} x+1 & 1 & 1 \ x+1 & x-1 & 1 \ x+1 & 1 & x-1 \end{array} ight| = 0$Taking $(x+1)$ common from $C_1$:$(x+1) \left| \begin{array}{ccc} 1 & 1 & 1 \ 1 & x-1 & 1 \ 1 & 1 & x-1 \end{array} ight| = 0$Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$(x+1) \left| \begin{array}{ccc} 1 & 1 & 1 \ 0 & x-2 & 0 \ 0 & 0 & x-2 \end{array} ight| = 0$Expanding along $C_1$:$(x+1) [1 \cdot (x-2)(x-2) - 0 + 0] = 0$$(x+1)(x-2)^2 = 0$Thus,the roots are $x = -1$ and $x = 2$.

The roots of the equation $\left| \begin{array}{ccc} x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{array} \right| = 0$ are

Similar Questions

If $\left|\begin{array}{ccc}\cos (A+B) & -\sin (A+B) & \cos 2 B \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B\end{array}\right|=0$, then $B$ is equal to

If $2\left|\begin{array}{ll}\sin ( A + B ) & \cos ( A + B ) \\ \cos ( A - B ) & \sin ( A - B )\end{array}\right|+\sqrt{3}= 0$,then $A =$ . . . . . . .

$\left| {\begin{array}{ccc} 1 + i & 1 - i & i \\ 1 - i & i & 1 + i \\ i & 1 + i & 1 - i \end{array}} \right| = $

$\left|\begin{array}{ccc}x & 3x+2 & 2x-1 \\ 2x-1 & 4x & 3x+1 \\ 7x-2 & 17x+6 & 12x-1\end{array}\right|=0$ is true for

Let $A = \begin{bmatrix} x+2 & 3x \\ 3 & x+2 \end{bmatrix}$ and $B = \begin{bmatrix} x & 0 \\ 5 & x+2 \end{bmatrix}$. Then all solutions of the equation $\det(AB) = 0$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module