Matrix [ * 20 c 0& - 4 &1 4&0& - 5 - 1 &5&0 ]is - Vidyadip

MCQ

Matrix $\left[ {\begin{array}{*{20}{c}}0&{ - 4}&1\\4&0&{ - 5}\\{ - 1}&5&0\end{array}} \right]$is

A
Orthogonal
B
Idempotent
✓
Skew- symmetric
D
Symmetric

✓

Answer

Correct option: C.

Skew- symmetric

c
(c) It is skew-symmetric.

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

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 3 and 4 . determinant and metrices→STD 12 - 3 and 4 . determinant and metrices — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Two matrix $P =\left[\begin{array}{cc}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]$ and $Q ^{ T }=\left[\begin{array}{rrr}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]$, Find $P-Q$ :

Let $A = {1, 2, 3}.$ Then number of equivalence relations containing $(1, 2)$ is:

If $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=4,\big|\vec{\text{a}}.\vec{\text{b}}\big|=2,$ then $|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2=$

If $\frac{{dy}}{{dx}} = \frac{{xy + y}}{{xy + x}}$, then the solution of the differential equation is

If m and n are the order and degree of the differential equation $(\text{y}_{2})^{5}+\frac{4(\text{y}_{2})^{3}}{\text{y}^{3}}+\text{y}^{3}=\text{y}_{3}=\text{x}^{2}-1$, then

Choose the correct answer from the given four options:
The function $\text{f(x)}=4\sin^3\text{x}-6\sin^2\text{x}+12\sin\text{x}+100$ is strictly:

Let $a $ and $ b$ be two unit vectors inclined at an angle $\theta $, then $\sin \,(\theta /2)$ is equal to

For the $\text{LPP};$ maximise $z = x + 4y$ subject to the constraints $\text{x}+2\text{y}\leq2, \text{x}+2\text{y}\geq8,\text{x},\text{y}\geq0.$

The solution of $\frac{{{d^2}y}}{{d{x^2}}} = \cos x - \sin x$is

If $f(x)\, = \,\left\{ \begin{gathered}
\frac{{x - 1}}{2}\,,\,\,\,0 \leqslant x\, < 1 \hfill \\
1/2\,\,\,\,,\,\,\,1 \leqslant x\, < 2 \hfill \\
\end{gathered} \right.\,$

$g(x) = (2x + 1)(x - k) + 3,\,0 \leqslant x < \infty $ then $g(f(x)),$ will be continuous at $x = 1$ if $k$ is equal