If x, y, z are all different and not equal to | vedclass

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

Question

(D) Given the determinant equation: $\left|\begin{array}{ccc}1+x & 1 & 1 \ 1 & 1+y & 1 \ 1 & 1 & 1+z\end{array}ight|=0$ Divide the first row by $x

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(D) Given the determinant equation: $\left|\begin{array}{ccc}1+x & 1 & 1 \ 1 & 1+y & 1 \ 1 & 1 & 1+z\end{array}ight|=0$ Divide the first row by $x$,the second row by $y$,and the third row by $z$: $xyz \left|\begin{array}{ccc}\frac{1}{x}+1 & \frac{1}{x} & \frac{1}{x} \ \frac{1}{y} & \frac{1}{y}+1 & \frac{1}{y} \ \frac{1}{z} & \frac{1}{z} & \frac{1}{z}+1\end{array}ight|=0$ Since $x, y, z 
eq 0$,we can divide by $xyz$: $\left|\begin{array}{ccc}\frac{1}{x}+1 & \frac{1}{x} & \frac{1}{x} \ \frac{1}{y} & \frac{1}{y}+1 & \frac{1}{y} \ \frac{1}{z} & \frac{1}{z} & \frac{1}{z}+1\end{array}ight|=0$ Apply the operation $C_1 ightarrow C_1 + C_2 + C_3$: $\left|\begin{array}{ccc}1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} & \frac{1}{x} & \frac{1}{x} \ 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} & \frac{1}{y}+1 & \frac{1}{y} \ 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} & \frac{1}{z} & \frac{1}{z}+1\end{array}ight|=0$ Factor out $(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$: $(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \left|\begin{array}{ccc}1 & \frac{1}{x} & \frac{1}{x} \ 1 & \frac{1}{y}+1 & \frac{1}{y} \ 1 & \frac{1}{z} & \frac{1}{z}+1\end{array}ight|=0$ Since $x, y, z$ are distinct,the determinant part is non-zero. Therefore,$1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} = 0$ $x^{-1}+y^{-1}+z^{-1} = -1$

If $x, y, z$ are all different and not equal to zero and $\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z\end{array}\right|=0$,then the value of $x^{-1}+y^{-1}+z^{-1}$ is equal to

Similar Questions

If $A = \begin{bmatrix} x & 2 & 1 \\ 2 & x & 1 \\ 2 & 1 & 0 \end{bmatrix}$ and $\det(A^3) = 125$,then $x =$

The values of $\alpha$,for which $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$,lie in the interval

If $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ and $|A^3| = 27$,then $\alpha = $

If $\left|\begin{array}{ll}2017 & 2018 \\ 2019 & 2020\end{array}\right|+\left|\begin{array}{ll}2021 & 2022 \\ 2023 & 2024\end{array}\right|=2 k$,then $k^3=$ . . . . . .

For $0 < \theta < \frac{\pi}{2}$,if $A = \begin{bmatrix} 1 & -\cos \theta & -1 \\ \cos \theta & 1 & -\cos \theta \\ 1 & \cos \theta & 1 \end{bmatrix}$,then which of the following is true regarding $\operatorname{det}(A)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module