MCQ
The matrix $\text{A}=\begin{bmatrix}1&0&0\\0&2&0\\0&0&4\end{bmatrix}$ is:
  • A
    Identity matrix.
  • B
    Symmetric matrix.
  • C
    Skew-symmetric matrix.
  • Diagonal matrix.

Answer

Correct option: D.
Diagonal matrix.
A matrix is called Diagonal matrix if all the elements, except those in the leading diagonal, are zero.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If A and B are two events such that $\text{A}\subset\text{B}$ and $\text{P}(\text{B})\neq0,$ then which of the following is correct?
The eqution of the plane through the line $x + y + 3 = 0 = 2x - y + 3z + 1$ and parallel to the line $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$ is:
$y=x(x-3)^2$ decreases for the values of $x$ given by
The function $y=2 x^2-\ln |x|, x \neq 0$ decreases when $x \in$
If $\alpha=\tan^{-1}\Big(\frac{\sqrt3\text{x}}{2\text{y}-\text{x}}\Big),\beta=\tan^{-1}\Big(\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}\Big),$ then $\alpha-\beta=$
For $\mathrm{x} \in \mathbb{R}$, let $\mathrm{y}(\mathrm{x})$ be a solution of the differential equation $\left(x^2-5\right) \frac{d y}{d x}-2 x y=-2 x\left(x^2-5\right)^2$ such that $y(2)=7$
A line $'l'$ passing through origin is perpendicular to the lines  $l_{1}: \overrightarrow{ r }=(3+ t ) \hat{ i }+(-1+2 t ) \hat{ j }+(4+2 t ) \hat{ k }$ ; $l_{2}: \overrightarrow{ r }=(3+2 s ) \hat{ i }+(3+2 s ) \hat{ j }+(2+ s ) \hat{ k }$ . If the co-ordinates of the point in the first octant on ${ }^{\prime} l_{2}^{\prime}$ at a distance of $\sqrt{17}$ from the point of intersection of $^{\prime} l^{\prime}$ and ${ }^{\prime} l_{1}^{\prime}$ are $( a , b , c ),$ then $18( a+ b + c )$ is equal to ........ .
If $\text{A}=\begin{bmatrix} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix},$ then det $\text{(adj (adj A))}$ is :
If the solution of the differential equation $\frac{d y}{d x}+e^{x}\left(x^{2}-2\right) y=\left(x^{2}-2 x\right)\left(x^{2}-2\right) e^{2 x} \quad$ satisfies $y(0)=0$, then the value of $y(2)$ is
The straigth line $\frac{\text{x}-3}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-1}{0}$ is: