MCQ
Find the value of x, if $\begin{bmatrix}2&5\\3&\text{x}\end{bmatrix}=\begin{bmatrix}\text{x}&-1\\5&3\end{bmatrix}$ is:
- ✓20
- B-20
- C30
- D-30
✓
Answer
Correct option: A.
20
$\begin{bmatrix}2&5\\3&\text{x}\end{bmatrix}=\begin{bmatrix}\text{x}&-1\\5&3\end{bmatrix}$
$\Rightarrow2\text{x}-15=3\text{x}+5 $
$\Rightarrow\text{x}=-20$
$\Rightarrow2\text{x}-15=3\text{x}+5 $
$\Rightarrow\text{x}=-20$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
$\left| {\,\begin{array}{*{20}{c}}5&3&{ - 1}\\{ - 7}&x&{ - 3}\\9&6&{ - 2}\end{array}\,} \right| = 0$, then $ x$ is equal to
→$\int\frac{\text{x}^3}{\text{x}+1}$ is equal to:
→If $A=\left[\begin{array}{rr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$ then $A ^{-1}= ?$
→If ${x^x}{y^y}{z^z} = c$, then ${{\partial z} \over {\partial x}} = $
→Choose the correct answer from the given four options.
The feasible solution for a LPP shown in Fig. 12.12. Let z = 3x - 4y be objective functio. (Maximum value of Z + Minimum value of Z) is equal to:
The feasible solution for a LPP shown in Fig. 12.12. Let z = 3x - 4y be objective functio. (Maximum value of Z + Minimum value of Z) is equal to:
If $\int {\frac{{dx}}{{{x^3}{{\left( {1 + {x^6}} \right)}^{2/3}}}} = xf\left( x \right){{\left( {1 + {x^6}} \right)}^{\frac{1}{3}}} + C} $ where $C$ is a constant of integration, then the function $f(x)$ is equal to
→Area lying first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the line $x = 0$ and $x = 2,$ is:
→The solution of the differential equation $\frac{d y}{d x}=\frac{1+y^2}{1+x^2}$ is
→The value of $\int_2^3 {\frac{{x + 1}}{{{x^2}(x - 1)}}dx} $ is
→Let $f(x)=2 \cos ^{-1} x+4 \cot ^{-1} x-3 x^{2}-2 x+10, x \in[-$ $1,1]$. If $[ a , b ]$ is the range of the function then $4 a -$ $b$ is equal to
→