Question
If $\begin{bmatrix}\text{a }-\text{b}&2\text{a}+\text{c}\\2\text{a}-\text{b}&3\text{c}+\text{d} \end{bmatrix}=\begin{bmatrix}-1&5\\0&13 \end{bmatrix},$ fine the value of b.

Answer

$\begin{bmatrix}\text{a }-\text{b}&2\text{a}+\text{c}\\2\text{a}-\text{b}&3\text{c}+\text{d} \end{bmatrix}=\begin{bmatrix}-1&5\\0&13 \end{bmatrix}$
From the above matrices,
a - b = -1 ...(1)
2a - b = 0 ...(2)
Solving (1) and (2),
a = 1, b = 2
$\therefore$ b = 2

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 probability of student A passing an examination is $\frac{2}{9}$ and of student B passing is $\frac{5}{9}$. Assuming the two events: 'A passes', 'B passes' as independent, find the probability of:
Only A passing the examination.
For the following differential equation find the sum of its order and degree :
$
y=x\left(\frac{d y}{d x}\right)^3+\frac{d^2 y}{d x^2}
$
Evaluate:
$\int\frac{\text{e}^{6\log_\text{e}\text{x}}-\text{e}^{5\log_\text{e}\text{x}}}{\text{e}^{4\log_\text{e}\text{x}}-\text{e}^{3\log_\text{e}\text{x}}}\text{dx}$
Find the vector equation of a plane which is at a distance of 5 unit from the origin and which is normal to the vector $\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$
Find the integral of the function $\sin^3 x \cos^3 x$
Determine the order and degree of the following differential equations. state also whether they are linear or non linear.
$\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\frac{1}{\frac{\text{dy}}{\text{dx}}}=2$
If $\overrightarrow{\text{AO}}+\overrightarrow{\text{OB}}=\overrightarrow{\text{BO}}+\overrightarrow{\text{OC}}$, prove that A, B, C are collinear points.
Find both the maximum value and minimum value of $3{x^4} - 8{x^3} + 12{x^2} - 48x + 25$ on the interval [0, 3].
Find the sum of the following vectors: $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}},\vec{\text{b}}=2\hat{\text{i}}-3\hat{\text{j}},\vec{\text{c}}=2\hat{\text{i}}-3\hat{\text{k}}$.
If $\text{P}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$ and $\text{Q}=\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix},$ then prove that $\text{PQ}=\begin{bmatrix}\text{xa}&0&0\\0&\text{yb}&0\\0&0&\text{zc}\end{bmatrix}=\text{QP}.$