Let $\alpha \beta \gamma = 45$; $\alpha, \beta, \gamma \in R$. If $x(\alpha, 1, 2) + y(1, \beta, 2) + z(2, 3, \gamma) = (0, 0, 0)$ for some $x, y, z \in R$ such that $xyz \neq 0$,then $6\alpha + 4\beta + \gamma$ is equal to:

  • A
    $55$
  • B
    $56$
  • C
    $54$
  • D
    $31$

Explore More

Similar Questions

The number of $3 \times 3$ non-singular matrices,with four entries as $1$ and all other entries as $0$,is

If $P = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$,$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $Q = PAP^T$,then $P^T(Q^{2005})P$ is equal to

Let $a = \lim_{x \to 1} \left( \frac{x}{\ln x} - \frac{1}{x \ln x} \right)$,$b = \lim_{x \to 0} \frac{x^3 - 16x}{4x + x^2}$,$c = \lim_{x \to 0} \frac{\ln(1 + \sin x)}{x}$,and $d = \lim_{x \to -1} \frac{(x + 1)^3}{3(\sin(x + 1) - (x + 1))}$. Then the matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is:

Let $\alpha$ and $\beta$ be real numbers. Consider a $3 \times 3$ matrix $A$ such that $A^2 = 3A + \alpha I$. If $A^4 = 21A + \beta I$,then:

Let $S = \left\{ \begin{bmatrix} -1 & a \\ 0 & b \end{bmatrix} : a, b \in \{1, 2, 3, \ldots, 100\} \right\}$ and let $T_n = \{A \in S : A^{n(n+1)} = I\}$. Then the number of elements in $\bigcap_{n=1}^{100} T_n$ is

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