MCQ
If $\left[\begin{array}{cc}2 a+b & a-2 b \\ 5 c-d & 4 c+3 d\end{array}\right]=\left[\begin{array}{cc}4 & -3 \\ 11 & 24\end{array}\right]$, then value of $a+b-c+2 d$ is
  • $8$
  • B
    $10$
  • C
    $4$
  • D
    $-8$

Answer

Correct option: A.
$8$
From the definition of equality of two matrices, we have
$2 a+b=4 ....(i)$
$5 c-d=11 .....(iii)$
$a-2 b=-3.....(ii)$
$4 c+3 d=24......(iv)$
Solving $(i)$ and $(ii)$, we get
$5 a=5 \Rightarrow a=1, b=2$
Solving $(iii)$ and $(iv)$, we get
$19 c=57 \Rightarrow c=3, d=4$
$\therefore a+b-c+2 d=1+2-3+8=8$

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 $\int\limits_0^x {f\left( t \right)} dt = {x^2} + \int\limits_x^1 {{t^2}f\left( t \right)dt} $, then $f'(1/2)$ is
$\int {\frac{{x\,\,dx}}{{{x^2} + 4x + 5}} = } $
A die is thrown three times. Getting a $3$ or a $6$ is considered success. Then the probability of at least two successes is
Let $f(x)=\frac{x^2-6 x+5}{x^2-5 x+6}$.

Match the conditions / expressions in Column $I$ with statements in Column $II$ and indicate your answers by darkening the appropriate bubbles in $4 \times 4$ matrix given in the $ORS$.

Column $I$ Column $II$
$(A)$ If $-1 < x < 1$, then $f$ ( $x$ ) satisfies $(p)$ $ 0 < $ f (x) $ < 1$
$(B)$ If $1 < x < 2$, then $f(x)$ satisfies $(q)$ $\mathrm{f}(\mathrm{x}) < 0$
$(C)$ If $3 < x < 5$, then $f(x)$ satisfies $(r)$ $ \mathrm{f}(\mathrm{x}) > 0$
$(D)$ If $x > 5$, then $f(x)$ satisfies $(s)$ $ f (\mathrm{x}) < 1$
The solution of the differential equation $\frac{{dy}}{{dx}} + \frac{{1 + {x^2}}}{x} = 0$ is
Choose the correct answers from the given four options:
If $\text{y}=\sqrt{\sin\text{x}+\text{y}},$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:
If the points whose position, vectors are $3i - 2j - k,$ $2i + 3j - 4k,$ $ - i + j + 2k$and $4i + 5j + \lambda k$ lie on a plane, then $\lambda = $
A curve passes through the point $\left(1, \frac{\pi}{6}\right)$. Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x}+\sec \left(\frac{y}{x}\right)$, $x>0$. Then the equation of the curve is
Integrating factor of the differential equation $\left(1-x^2\right) \frac{d y}{d x}-x y=1$ is
The function $f(x) = 2x^3- 15x^2 + 36x + 4$ is maximum at $x =$