Question
Choose the correct answer from the given four options.
The matrix $\begin{bmatrix}1&0&0\\0&2&0\\0&0&4\end{bmatrix}$ is a:
  1. Identity matrix.
  2. Symmetric matrix.
  3. Skew-symmetric matrix.
  4. None of these.

Answer

  1. Symmetric matrix.

Solution:

We have $\text{A}=\begin{bmatrix}1&0&0\\0&2&0\\0&0&4\end{bmatrix}$

$\therefore\ \text{A}'=\text{A}$

So, the given matrix is a symmetric matrix.

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 $f(x)$ is a differentiable function such that $f:R \to r$ and $f\left( {\frac{1}{n}} \right) = 0\;\forall \;n \ge 1,n \in I$ then
$\frac{{\left[ {\left( {\vec a \times \vec b} \right) \times \left( {\vec b \times \vec c} \right)\left( {\vec b \times \vec c} \right) \times \left( {\vec c \times \vec a} \right)\left( {\vec c \times \vec a} \right) \times \left( {\vec a \times \vec b} \right)} \right]}}{{\left[ {\vec a \times \vec b\,\,\,\,\vec b \times \vec c\,\,\,\,\vec c \times \vec a} \right]}}$  is equal to
If $\vec x$ is a unit vector such that $\vec x \times \left( {\hat i - 2\hat j + \hat k} \right) =  - \hat i + \hat k$ , then $\vec x$ is 
If $y = f(x) = \frac{{x + 2}}{{x - 1}}$, then $x = $
If $A$ is square matrix such that $A^{2}=A$, then $(1+A)^{3}-7 A$ is equal to
A curve satisfying the initial condition, $y(1) = 0,$ satisfies the differential equation, $x \frac{{dy}}{{dx}} = y -x^2.$ The area bounded by the curve and the $x-$ axis is
Area bounded on the left by $y-$ axis, below by $x-$ axis, right by $x = \frac{\pi }{2}$ above left by $y = cosx$ and above right by $y = sinx$ is  
Let $f(\theta)=\sin \left(\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right)\right)$, where $-\frac{\pi}{4}<\theta<\frac{\pi}{4}$. Then the value of $\frac{d}{d(\tan \theta)}(f(\theta))$ is
If $a\,.\,b = b\,.\,c = c\,.\,a = 0$ , then what is the value of the scalar triple product $ [a b c] $ ?
If $f(x) = \left\{ \begin{array}{l}x\frac{{{e^{(1/x)}} - {e^{( - 1/x)}}}}{{{e^{(1/x)}} + {e^{( - 1/x)}}}},\,x \ne 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,\,x = 0\end{array} \right.$ then which of the following is true