The rank of the matrix $A=\begin{bmatrix} 1 & -1 & 0 & -2 \\ -4 & 4 & 0 & 8 \\ -2 & 1 & 2 & 4 \end{bmatrix}$ is

  • A
    $1$
  • B
    $0$
  • C
    $3$
  • D
    $2$

Explore More

Similar Questions

If $y = \sin(px)$ and $y_n$ is the $n^{th}$ derivative of $y$,then $\left| \begin{array}{ccc} y & y_1 & y_2 \\ y_3 & y_4 & y_5 \\ y_6 & y_7 & y_8 \end{array} \right|$ is equal to:

Difficult
View Solution

If $\Delta(x) = \begin{vmatrix} x-2 & (x-1)^2 & x^3 \\ x-1 & x^2 & (x+1)^3 \\ x & (x+1)^2 & (x+2)^3 \end{vmatrix}$, then the coefficient of $x$ in $\Delta(x)$ is:

If $f(x) = \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \\ 3 + 2 \cos^4 x & 2 \sin^4 x & \sin^2 2x \\ 2 \cos^4 x & 3 + 2 \sin^4 x & \sin^2 2x \end{vmatrix}$,then $\frac{1}{5} f'(0)$ is equal to:

If $A = \begin{vmatrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{vmatrix}$ and $B = \begin{vmatrix} x & 1 \\ 1 & x \end{vmatrix}$,then $\frac{dA}{dx}$ is equal to

If $f: N \to Z$ is defined by $f(n) = \det \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3(2k+1) & 3(k+2)+1 \end{vmatrix}$, where $k \in N$ and $\sum_{n=1}^k f(n) = 98$, then $k$ is equal to:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo