If D = left| begin{array}{ccc} frac{1}{z} & f | vedclass

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

Question

(D) Let the determinant be $D$. Multiply $C_1$ by $x$,$C_2$ by $y$,and $C_3$ by $z$,and divide the determinant by $xyz$:$D = \frac{1}{xyz} \left| \beg

Vedclass · Accepted Answer

$D$ is dependent on $x, y, z$

Vedclass · Answer

(D) Let the determinant be $D$. Multiply $C_1$ by $x$,$C_2$ by $y$,and $C_3$ by $z$,and divide the determinant by $xyz$:$D = \frac{1}{xyz} \left| \begin{array}{ccc} \frac{x}{z} & \frac{y}{z} & -\frac{x+y}{z} \ -\frac{y+z}{x} & \frac{y}{x} & \frac{z}{x} \ -\frac{y(y+z)}{x^2} & \frac{y(x+2y+z)}{xz} & -\frac{yz(x+y)}{xz^2} \end{array} ight|$.By performing row and column operations,or by simplifying the determinant,we find that the rows or columns are linearly dependent.Specifically,if we evaluate the determinant,we find $D = 0$.Since $D = 0$,the value of the determinant is constant (zero) and does not depend on $x, y,$ or $z$.Therefore,the statement '$D$ is dependent on $x, y, z$' is incorrect.

If $D = \left| \begin{array}{ccc} \frac{1}{z} & \frac{1}{z} & -\frac{(x+y)}{z^2} \\ -\frac{(y+z)}{x^2} & \frac{1}{x} & \frac{1}{x} \\ -\frac{y(y+z)}{x^2z} & \frac{x+2y+z}{xz} & -\frac{y(x+y)}{xz^2} \end{array} \right|$,then the incorrect statement is:

Similar Questions

If $p{\lambda ^4} + q{\lambda ^3} + r{\lambda ^2} + s\lambda + t = \left| {\begin{array}{*{20}{c}}{{\lambda ^2} + 3\lambda }&{\lambda - 1}&{\lambda + 3}\\{\lambda + 1}&{2 - \lambda }&{\lambda - 4}\\{\lambda - 3}&{\lambda + 4}&{3\lambda }\end{array}} \right|$,the value of $t$ is

Let $\left| {\begin{array}{*{20}{c}}{6i}&{ - 3i}&1\\4&{3i}&{ - 1}\\{20}&3&i\end{array}} \right| = x + iy$,then

In order that the matrix $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & \lambda & 5 \end{bmatrix}$ be non-singular,$\lambda$ should not be equal to:

If $\left[\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]$ has no inverse,then the real value of $x$ is

$\left| \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{array} \right| = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module