If $A = [x \quad y \quad z]$,$B = \begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}$,$C = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ and $(AB) \cdot C$ is an $m \times n$ order matrix,then:

  • A
    $m < n$
  • B
    $m+n=5$
  • C
    $m=n$
  • D
    $m > n$

Explore More

Similar Questions

If $A$ is a square matrix such that $A^2 = A$,then $(I + A)^2 - 3A =$ . . . . . . .

If $A+2B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix}$ and $2A-B = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix}$, then $\operatorname{tr}(A)-\operatorname{tr}(B) =$

Find $X$ and $Y$,if $X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]$ and $X-Y=\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]$.

Let $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$,$B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$,and $C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$. Find $A - B$.

Let $A = \begin{bmatrix} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0 & n \\ 0 & n & 0 \\ n & 0 & 0 \end{bmatrix}$. Then,$A^2 + B^2 + AB =$

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