MCQ
If $A=\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$ and $B=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$ two matrix, then find AB ?
  • $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
  • B
    $\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right]$
  • C
    $\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
  • D
    $\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]$

Answer

Correct option: A.
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
(A)
$AB =\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]=\left[\begin{array}{cc}4-3 & 6-6 \\ -2+2 & -3+4\end{array}\right]$ $=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

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

The function $\text{f(x)}=\frac{4-\text{x}^2}{4\text{x}-\text{x}^3}$
  1. Discontinuous at only one point.
  2. Discontinuous exactly at two points.
  3. Discontinuous exactly at three points.
  4. None of these.
The order of the differential equation whose general solution is given by
$
y=\left(C_1+C_2\right) \cos \left(x+C_3\right)-C_4 e^{x+C_5}
$
where $C_1, C_2, C_3, C_4, C_5$ are arbitrary constants, is
Let $A=\left[\begin{array}{ccc}2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2\end{array}\right]$. If $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} 2 A))|=(16)^{ n }$, then $n$ is equal to
$\int\limits_{ - \,a}^a {\,f\,(x)\,dx} $=
Time period is a:
  1. Vector quantity
  2. Scalar quantity
  3. Neither scalar nor vector
  4. None of these
If $\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}^\text{k}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then the least positive integral value of k is:
  1. 3
  2. 4
  3. 6
  4. 7
If $y = {t^{4/3}} - 3{t^{ - 2/3}}$, then $\frac{{dy}}{{dt}}=$
If $\vec a = 3\vec j + 4\vec k$ , $\vec b = 2\vec i + \vec k$ and $\vec c$ , $\vec d$ are respectively the component of $\vec a$ parallel & perpendicular to $\vec b$ ,then $\left[ {\left( {\vec a \times \vec c} \right) \times \left( {\vec c \times \vec d} \right)\,\left( {\vec c \times \vec d} \right) \times \left( {\vec d \times \vec a} \right)\left( {\vec d \times \vec a} \right) \times \left( {\vec a \times \vec c} \right)} \right]$ equals
Objective function of a L.P.P. is
If $\left| {\begin{array}{*{20}{c}}
1&1&1\\
a&b&c\\
{{a^2}}&{{b^2}}&{{c^2}}
\end{array}} \right| = 5$ , then $\left| {\begin{array}{*{20}{c}}
{b{c^2} - {b^2}c}&{{a^2}c - a{c^2}}&{a{b^2} - b{a^2}}\\
{{b^2} - {c^2}}&{{c^2} - {a^2}}&{{a^2} - {b^2}}\\
{c - b}&{a - c}&{b - a}
\end{array}} \right|$ is equal to