MCQ
If $A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{lr}4 & 0 \\ 1 & -2 \\ 0 & 3\end{array}\right]$ then. $A B=$_________ .
  • A
    $\left[\begin{array}{rr}4 & 0 \\ 1 & -2 \\ 0 & 3\end{array}\right]$
  • B
    $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
  • C
    $\left[\begin{array}{rr}4 & 0 \\ 1 & -2\end{array}\right]$
  • Does not exist

Answer

Correct option: D.
Does not exist
D

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

Let $\vec{a}=\hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+2 \hat{k}$. Then the number of vectors $\vec{b}$ such that $\vec{b} \times \vec{c}=\vec{a}$ and $|\vec{b}| \in\{1,2, \ldots ., 10\}$ is
$\int_{}^{} {{a^{3x + 3}}dx} = $
The number of distinct real values of $\lambda $ on for which the lines $\frac{{x - 1}}{1} = \frac{{y - 2}}{2} = \frac{{z + 3}}{{{\lambda ^2}}}$ and $\frac{{x - 3}}{1} = \frac{{y - 2}}{{{\lambda ^2}}} = \frac{{z - 1}}{2}$ are coplanar is
${d \over {dx}}\left[ {{{{e^{ax}}} \over {\sin (bx + c)}}} \right] = $
Choose the correct answer from the given four options.

If $\text{A}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then A2 is equal to:

  1. $\begin{bmatrix}0&1\\1&0\end{bmatrix}$

  2. $\begin{bmatrix}1&0\\1&0\end{bmatrix}$

  3. $\begin{bmatrix}0&1\\0&1\end{bmatrix}$

  4. $\begin{bmatrix}1&0\\0&1\end{bmatrix}$

The solution of the differention equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^{2}+\text{xy}+\text{y}^{2}}{\text{x}^{2}}$ is:
  1. $\tan^{-1}\big(\frac{\text{x}}{\text{y}}\big)-\log\text{y}+\text{C}$ 
  2. $\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)-\log\text{x}+\text{C}$
  3. $\tan^{-1}\big(\frac{\text{x}}{\text{y}}\big)=\log\text{x}+\text{C}$
  4. $\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)=\log\text{y}+\text{C}$
Choose the correct answer in Exercises:

If $\frac{\text{d}}{\text{dx}}\text{f}\text{(x)}=4\text{x}^3-\frac{3}{\text{x}^4}$such that $\text{f}(2)=0.$Then $\text{f}\text{(x)}$ is

  1. $\text{x}^4+\frac{1}{\text{x}^3}-\frac{129}{8}$

  2. $\text{x}^3+\frac{1}{\text{x}^4}+\frac{129}{8}$

  3. $\text{x}^4+\frac{1}{\text{x}^3}+\frac{129}{8}$

  4. $\text{x}^3+\frac{1}{\text{x}^4}-\frac{129}{8}$

The position vectors of four points  $ P, Q, R, S$  are $2a + 4c,\,$ $5a + 3\sqrt 3 \,b + 4c,$ $ - 2\sqrt 3 b + c$ and $2a + c$ respectively, then
Twenty metres of wire is available for fencing off a flowerbed in the form of a circular sector. Then the maximum area (in sq. m) of the flower bed is :
Let $\vec a = 3\hat i + 2\hat j + 2\hat k$ and $\vec b = \hat i + 2\hat j - 2\hat k$ be two vectors. If a vector perpendicular to both the vectors $\vec a + \vec b$ and $\vec a - \vec b$ has the magnitude $12$ then one such vector is