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 $a, b, c$ are distinct and $\left| \begin{array}{ccc} a & a^2 & a^3 - 1 \\ b & b^2 & b^3 - 1 \\ c & c^2 & c^3 - 1 \end{array} \right| = 0$,then

The value of the determinant $\left| \begin{array}{ccc} 0 & b^3 - a^3 & c^3 - a^3 \\ a^3 - b^3 & 0 & c^3 - b^3 \\ a^3 - c^3 & b^3 - c^3 & 0 \end{array} \right|$ is equal to:

If $\alpha, \beta, \gamma$ are the roots of $\left|\begin{array}{ccc} 1-x & -2 & 1 \\ -2 & 4-x & -2 \\ 1 & -2 & 1-x \end{array}\right|=0$,then $\alpha \beta+\beta \gamma+\gamma \alpha=$

The area of a triangle is $4$ sq. units,and its vertices are $(-2, 0)$,$(0, 4)$,and $(0, k)$. Find the value of $k$.

If $a \ne 6, b, c$ satisfy $\left| \begin{array}{ccc} a & 2b & 2c \\ 3 & b & c \\ 4 & a & b \end{array} \right| = 0$,then $abc = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module