as sum of two vectors such that one is parallel to the vector 
and the other is perpendicular to 
.Solution: $\begin{array}{l} \vec{a}=6 \hat{\imath}-3 \hat{\jmath}-6 \hat{k} \\ \vec{b}=\hat{\imath}+\hat{\jmath}+\hat{k} \end{array}$ Let $\vec{a}$ be written as sum of $\vec{c}+\vec{d}$, where...
and 
, if 
and 
.Solution: $\begin{array}{l} \text { Given: } \vec{a}=2 \hat{\imath}-\hat{\jmath}+3 \hat{k} \\ \vec{b}=3 \hat{\imath}+\hat{\jmath}+2 \hat{k} \\ \vec{a}+\vec{b}=5 \hat{\imath}+5 \hat{k} \\...
of magnitude 
, making an angle 
with 
– axis, 
with 
– axis and an acute angle 
with z – axis.Solution: $\mathrm{I} \overrightarrow{\mathrm{a}} \mathrm{I}=5 \sqrt{2}$ Also it is given that $\alpha=\frac{\pi}{4}$ $\beta=\frac{\pi}{2}, \gamma=$ acute angle. So from this $\cos...
is equally inclined to the coordinate axes.Solution: $\begin{array}{l} \vec{a}=\cos \gamma=\frac{1}{\sqrt{3}} \\ \mathrm{I} \vec{a} \mathrm{I}=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3} \end{array}$ Since direction cosines or cosines of angle made by...
makes with the coordinate axes.Solution: Direction cosines gives the angle made by the vector with $x$-axis, $y-$ axis, z-axis respectively. $\begin{array}{l} \vec{a}=3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k} \\ \mathrm{I} \vec{a}...
is a unit vector such that 
, find 
.Solution: Given: $\mathrm{I} \vec{a} \mathrm{I}=1$ $\begin{array}{l} (\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=8 \\ \vec{x} \cdot \vec{x}+\vec{x} \cdot \vec{a}-\vec{a} \cdot \vec{x}-\vec{a} \cdot...
and 
then calculate the angle between 
and 
.Solution: $\vec{a}=\hat{\imath}+2 \hat{\jmath}-3 \hat{k}$ $\vec{b}=3 \hat{\imath}-\hat{\jmath}+2 \hat{k}$ $\begin{array}{l}2 \vec{a}+\vec{b} & =2(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})+(3...
and 
, when 
and 
and 
and 
.Solution: $\begin{array}{l} \text { i) } \vec{a}=\hat{\imath}-2 \hat{\jmath}+3 \hat{k} \\ \vec{b}=3 \hat{\imath}-2 \hat{\jmath}+\hat{k} \\ \mathrm{I} \vec{a}...
along the vector 
.Solution: $\begin{array}{l} \vec{a}=\hat{\imath}+\hat{\jmath}+\hat{k} \\ \vec{b}=\hat{\jmath} \end{array}$ $\mathrm{I} \vec{a} \mathrm{I}=\sqrt{3}=\sqrt{1^{2}+1^{2}+1^{2}}$ $\begin{array}{l}...
and 
. Find the projection of (i) 
on and (ii) on .Solution: If $\vec{a}=\mathrm{a} 1 \hat{\imath}+\mathrm{a} 2 \hat{\jmath}+\mathrm{a}_{3} \hat{k}$ and $\vec{b}=\mathrm{b}_{1} \hat{\imath}+\mathrm{b}_{2} \hat{\jmath}+\mathrm{b}_{3} \hat{k}$, then...
are mutually perpendicular unit vectors.Solution: $\begin{array}{l} \vec{a}=\frac{1}{7}(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k}) \\ \vec{b}=\frac{1}{7}(3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k}) \\ \vec{c}=\frac{1}{7}(6 \hat{\imath}+2...
and 
then find the value of 
so that and are orthogonal vectors.Solution: $\begin{array}{l} \vec{a}=\hat{\imath}-1 \hat{\jmath}+7 \hat{k} \\ \vec{b}=5 \hat{\imath}-1 \hat{\jmath}+\lambda \hat{k} \\ \vec{a}+\vec{b}=6 \hat{\imath}-2 \hat{\jmath}+(7+\lambda)...
and 
, show that is perpendicular to . 
and 
then show that and are orthogonal.Solution: If two vectors are perpendicular or orthogonal then their dot product is zero. $\begin{array}{l} \text { i) } \quad \vec{a}=\hat{\imath}+2 \hat{\jmath}-3 \hat{k} \\ \vec{b}=3...
when 
and 
and 
and 
Solution: If $\vec{a}=\mathrm{a} 1 \hat{\imath}+\mathrm{a} 2 \hat{\jmath}+\mathrm{a} 3 \hat{k}$ and $\vec{b}=\mathrm{b}_{1} \hat{\imath}+\mathrm{b}_{2} \hat{\jmath}+\mathrm{b}_{3} \hat{k}, \text {...
, where 
and 
are the points 
and 
respectively.Solution: $\begin{array}{l} \overrightarrow{O P}=\hat{\imath}+3 \hat{\jmath} \\ \overrightarrow{O Q}=4 \hat{\imath}+5 \hat{\jmath}+6 \hat{k} \\ \overrightarrow{P Q}=\overrightarrow{O...
and 
Solution: $\begin{array}{l} \overrightarrow{O A}=3 \hat{\imath}+2 \hat{\jmath}+6 \hat{k} \\ \overrightarrow{O B}=\hat{\imath}+4 \hat{\jmath}-2 \hat{k} \end{array}$ Let $\mathrm{C}$ be the mid-point...
which divides the line joining 
and 
in the ratio 
(i) internally (ii) externally.Solution: $\begin{array}{l} \overrightarrow{O A}=-2 \hat{\imath}+\hat{\jmath}+3 \hat{k} \\ \overrightarrow{O B}=3 \hat{\imath}+5 \hat{\jmath}-2 \hat{k} \end{array}$ (i) $\mathrm{R}$ divides...
and 
are and 
respectively. Find the position vector of a point 
which divides 
externally in the ratio 
. Also, show that 
is the mid-point of the line segment 
.Solution: Given: $\overrightarrow{O A}=(2 \vec{a}+\vec{b})$ $\overrightarrow{O B}=(\vec{a}-3 \vec{b})$ Position vector of $\mathrm{C}$ which divides $\mathrm{AB}$ in the ratio $1: 2$ externally is...
and 
(i) internally and (ii) externally in the ratio 
Solution: Given: $\overrightarrow{O A}=(2 \vec{a}-3 \vec{b})$ $\overrightarrow{O B}=(3 \vec{a}-2 \vec{b})$ (i) Let $\mathrm{P}$ be the point that divides $\mathrm{A}, \mathrm{B}$ internally in the...
and 
are the vertices of a rightangled triangle.Solution: $\begin{array}{l} \overrightarrow{O A}=\hat{\imath}-\hat{\jmath} \\ \overrightarrow{O B}=4 \hat{\imath}-3 \hat{\jmath}+\hat{k} \\ \overrightarrow{O C}=2 \hat{\imath}-4 \hat{\jmath}+5...
and of a 
be 
and 
, respectively, prove that is equilateral.Solution: $\begin{array}{l} \overrightarrow{O A}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k} \\ \overrightarrow{O B}=2 \hat{\imath}+3 \hat{\jmath}+\hat{k} \\ \overrightarrow{O C}=3...
and 
having position vectors 
and 
respectively, are collinear.Solution: $\begin{array}{l} \overrightarrow{O A}=\hat{\imath}+2 \hat{\jmath}+7 \hat{k} \\ \overrightarrow{O B}=2 \hat{\imath}+6 \hat{\jmath}+2 \hat{k} \\ \overrightarrow{O C}=3 \hat{\imath}+10...
and 
, find 
Solution: $\begin{array}{l} \vec{a}=\hat{\imath}-2 \hat{\jmath} \\ \vec{b}=2 \hat{\imath}-3 \hat{\jmath} \\ \vec{c}=2 \hat{\imath}+3 \hat{k} \\ \vec{a}+\vec{b}+\vec{c}=(\hat{\imath}-2...
.Solution: $\vec{A}=5 \hat{\imath}-\hat{\jmath}+2 \hat{k}$ Magnitude of required vector is 8 units $\mathrm{I} \vec{A} \mathrm{I}=\sqrt{5^{2}+1^{2}+2^{2}}=\sqrt{25+1+4}=\sqrt{30}$ $\begin{array}{l}...
.Solution: $\vec{A}=-2 \hat{\imath}+\hat{\jmath}+2 \hat{k}$ $\mathrm{I} \vec{A} \mathrm{I}=\sqrt{(-2)^{2}+1^{2}+2^{2}}=\sqrt{4+1+4}=\sqrt{9}=3$ Note: Any vector $\vec{X}$ is given by $\vec{X}=$...
and 
then find a unit vector parallel to .Solution: $\begin{array}{l} \vec{a}=\hat{\imath}+2 \hat{\jmath}-3 \hat{k} \\ \vec{b}=2 \hat{\imath}+4 \hat{\jmath}+9 \hat{k} \end{array}$ Then $\vec{a}+\vec{b}=(\hat{\imath}+2 \hat{\jmath}-3...
and 
then find a unit vector in the direction of .Solution: $\begin{array}{l} \vec{a}=3 \hat{\imath}+\hat{\jmath}-5 \hat{k} \\ \vec{b}=\hat{\imath}+2 \hat{\jmath}-\hat{k} \end{array}$ Then $\vec{a}-\vec{b}=(3 \hat{\imath}+\hat{\jmath}-5...
and 
then find the unit vector in the direction of .Solution: $\vec{a}=-\hat{\imath}+\hat{\jmath}-\hat{k}$ $\vec{b}=2 \hat{\imath}-\hat{\jmath}+2 \hat{k}$ Then $(\vec{a}+\vec{b})=(-\hat{\imath}+\hat{\jmath}-\hat{k})+(2 \hat{\imath}-\hat{\jmath}+2...
Solution: If $\vec{a}=\mathrm{a}_{1} \hat{\imath}+\mathrm{a}_{2} \hat{\jmath}+\mathrm{a}_{3} \hat{k}$, then Unit vector in the direction of $\vec{a}$ can be given by $\hat{a}=\frac{\vec{a}}{I...
Solution: A. $\vec{a}=\hat{\imath}+2 \hat{\jmath}+5 \hat{k}$ If $\vec{a}=\mathrm{a}_{1} \hat{\imath}+\mathrm{a}_{2} \hat{\jmath}+\mathrm{a}_{3} \hat{k}$, then $\mathrm{I} \vec{a}...
is 
Solution: $\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}^{2}$ Comparing with $\frac{d y}{d x}+P y=Q$ Integrating factor $=\mathrm{x}$ General solution is $\mathrm{yx}=\int...
is 
Solution: $\frac{d y}{d x}+y \operatorname{Cot} x=2 \operatorname{Cos} x$ Comparing with $\frac{d y}{d x}+P y=Q$ Integrating factor is $e^{\int \cot x d x}=\operatorname{Sin} x$ General solution is...
Solution: $\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y} \tan \mathrm{x}=\mathrm{Secx}$ Comparing with $\frac{d y}{d x}+P y=Q$ Integrating factor $e^{\int \tan x d x}=\operatorname{Sec} x$ General...
is 
Solution: $\frac{d y}{d x}=\frac{y}{x}+\sin \frac{y}{x}$ Let $\mathrm{y}=\mathrm{vx}$ $\begin{array}{l} \mathrm{dy} / \mathrm{dx}=\mathrm{v}+\mathrm{xdv} / \mathrm{d} \mathrm{x} \\...
is 
Solution: $\begin{array}{r} (x-y) d y+(x+y) d x=0 \\ \qquad \frac{d y}{d x}=\frac{x+y}{y-x} \end{array}$ Let $\mathrm{y}=\mathrm{vx}$ $\begin{array}{l} \mathrm{dy} /...
is 
Solution: $\begin{array}{r} 2 \mathrm{xydy}+\left(\mathrm{x}^{2} \quad \mathrm{y}^{2}\right) \mathrm{d} \mathrm{x}=0 \\ \frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y} \end{array}$ Let...
is 
Solution: $\mathrm{x} \frac{d y}{d x}=y+x \tan \frac{y}{x}$ Dividing both sides by $x$, we get, $\frac{d y}{d x}=\frac{y}{x}+\tan \frac{y}{x}$ Let $\mathrm{y}=\mathrm{vx}$ Differentiating both...
is. 
Solution: $\begin{array}{l} \mathrm{x}^{2} \frac{d y}{d x}=x^{2}+x y+y^{2} \\ \frac{d y}{d x}=1+\frac{y}{x}+\frac{y^{2}}{x^{2}} \end{array}$ Let $\mathrm{y}=\mathrm{vx}$ $\begin{array}{l}...
is 
Solution: $\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}$ Let $\mathrm{y}=\mathrm{vx}$ $\begin{array}{l} \mathrm{dy} / \mathrm{dx}=\mathrm{v}+\mathrm{xdv} / \mathrm{dx} \\ \frac{x^{2}...
Solution: $\begin{array}{r} \frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{1-\mathrm{x}^{2}} \sqrt{1-\mathrm{y}^{2}} \\ \frac{d y}{\sqrt{1-y^{2}}}=\sqrt{1-x^{2}} d x \end{array}$ $\operatorname{Let}...
is 
Solution: $\log \left(\frac{d y}{d x}\right)=(a x+b y)$ $\begin{array}{l} \frac{d y}{d x}=e^{a x+b y} \\ \frac{d y}{e^{b y}}=e^{a x} d x \end{array}$ On integrating on both sides we get...
is 
Solution: $\begin{array}{r} x \sqrt{1+y^{2}} \mathrm{dx}+\mathrm{y} \sqrt{1+x^{2}} \mathrm{dy}=0 \\ \frac{y d y}{\sqrt{1+y^{2}}}=\frac{-x d x}{\sqrt{1+x^{2}}} \end{array}$ Let $1+y^{2}=t$ and...
is 
Solution: $\left(1+x^{2}\right) d y-x y d x=0$ $\frac{d y}{y}=\frac{x}{1+x^{2}} d x$ Let $1+\mathrm{x}^{2}=\mathrm{t}$ $\begin{array}{r} 2 \mathrm{x} \mathrm{dx}=\mathrm{dt} \\ \frac{d y}{y}=\frac{d...
is 
Solution: $\frac{d y}{d x}+y \log y \operatorname{Cot} x=0$ Let $\log \mathrm{y}=\mathrm{t}$ On differentiating we get $\begin{array}{l} \frac{1}{y} d y=d t \\ \frac{d t}{t}=-\operatorname{Cot} x d...
is 
Solution: $\begin{array}{l} x \operatorname{Cosydy}=\left(x e^{x} \log x+e^{x}\right) d x \\ \operatorname{Cosydy}=\frac{x \operatorname{exlogx}+e x}{x} d x \end{array}$ On integrating on both sides...
is 
Solution: $\cos x(1+\cos y) d x-\sin y(1+\sin x) d y=0$ Let $1+\cos y=t$ and $1+\sin x=u$ On differentiating both equations, we get $-\sin y d y=d t$ and $\cos x d x=d u$ Substitute this in the...
is 
Solution: $\begin{array}{l} \frac{d y}{d x}=\frac{-2 x y}{x^{2}+1} \\ \frac{d y}{y}=\frac{-2 x d x}{x^{2}+1} \end{array}$ Let $\mathrm{x}^{2}+1=\mathrm{t}$ On differentiating on both sides we get $2...
is 
Solution: $\begin{array}{l} \frac{d y}{d x}=\frac{1-\operatorname{Cos} x}{1+\operatorname{Cos} x} \\ \frac{d y}{d x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}, \\ \frac{d y}{d x}=\tan...
is 
Solution: $\frac{\mathrm{dy}}{\mathrm{dx}}+\sqrt{\frac{1-\mathrm{y}^{2}}{1-\mathrm{x}^{2}}}=0$ $\frac{-d y}{\sqrt{1-y^{2}}}=\frac{d x}{\sqrt{1-x^{2}}}$ On integrating on both sides, we get $-\sin...
is 
Solution: $\begin{array}{l} \frac{d y}{d x}=e^{x+y}+x^{2} e^{y} \\ \left(e^{-y}\right) d y=\left(e^{x}+e^{2}\right) d x \end{array}$ On integrating on both sides, we get $\begin{array}{l}...
is 
Solution: $\begin{array}{l} \frac{d y}{d x}=1-x+y-x y \\ \frac{d y}{d x}=1-x+y(1-x) \\ \frac{d y}{d x}=(1+y)(1-x) \\ \frac{d y}{1+y}=(1-x) d x \end{array}$ On integrating on both sides, we get $\log...
is. 
Solution: $\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}$ On integrating on both sides, we get $\begin{array}{l} \tan ^{-1} y=\tan ^{-1} x+c \\ \frac{y-x}{1+y x}=\mathrm{c} \\...
is 
Solution: $x \frac{d y}{d x}=\cot y$ Separating the variables, we get, $\begin{array}{c} \frac{d y}{\cot y}=\frac{d x}{x} \\ \tan y d y=\frac{d x}{x} \end{array}$ Integrating both sides, we get,...
is 
Solution: $\begin{array}{l} \text { Given } x d y+y d x=0 \\ x d y=-y d x \end{array}$ $-\frac{d y}{y}=\frac{d x}{d x}$ On integrating on both sides we get, $\begin{array}{l} -\log y=\log x+c \\...
is 
Solution: $\begin{array}{l} \left(e^{x}+1\right) y d y=(y+1) e^{x} d x \\ \frac{y d y}{y+1}=\frac{e^{x} d x}{\left(e^{x}+1\right)} \end{array}$ On differentiating on both sides we get $e^{x} d x=d...
is 
Solution: $\begin{array}{c} \frac{d y}{d x}=e^{x+y} \\ \frac{d y}{d x}=e^{x} e^{y} \\ e^{-y} d y=e^{x} d x \end{array}$ On integrating on both sides, we get $\begin{array}{c}...
, given that when 
, then 
.Solution: $\left(1+y^{2}\right) d x+\left(x-e^{-\tan ^{-1} y}\right) d y=0\dots (1)$ General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=$ Qis given by, $y...
Solution: $(x+y+1) \frac{d y}{d x}=1-\ldots(1)$ General solution for the differential equation in the form of $\frac{d y}{d x}+\overline{\mathrm{Py}}=$ Qis given by, $y \cdot(I . F .)=\int Q \cdot(I...
Solution: $(\mathrm{x}+\mathrm{y}) \frac{d y}{d x}=1-\ldots(1)$ General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ is given by, $y \cdot\left(I ....
Solution: $y d x+\left(x-y^{2}\right) d y=0\dots (1)$ General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ General solution is given by, $y...
and the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 
Find the equation of the curve.Solution: General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=$ Qis given by, $y \cdot(I . F .)=\int Q \cdot(I . F .) d x+c$ Where, integrating factor, $I . F...
is equal to the sum of the coordinates of the point. Find the equation of the curve.Solution: General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=$ Qis given by, $y \cdot(I . F .)=\int Q \cdot(I . F .) d x+c$ Where, integrating factor, $\text...
, given that 
when Solution: $\frac{d y}{d x}+\operatorname{ytan} \mathrm{x}=2 \mathrm{x}+x^{2} \tan x\dots(1)$ General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$...
, given that when 
Solution: $x \frac{d y}{d x}-y=\log x-\ldots(1)$ General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ is given by, $y \cdot(I . F .)=\int Q \cdot(I...
, given that when Solution: $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=4 x^{2}-\ldots(1)$ General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ is given by, $y...
, given that , when Solution: $\frac{d y}{d x}+2 y=e^{-2 x} \cdot \operatorname{Sin} x\dots(1)$ General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ General solution is...
, given that when Solution: General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}\dots(1)$ General solution is given by, $y \cdot(I . F .)=\int Q \cdot(I . F .) d x+c$...
, given that when 
Solution: $\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y} \cdot \operatorname{Cot} \mathrm{x}=4 \mathrm{x} \operatorname{Cosec} \mathrm{x}^{--}-\mathrm{c}-\mathrm{c}-\mathrm{c}(1)\dots (1)$ General...
, given that when 
Solution: $\mathrm{x} \frac{d y}{d x}+y=x^{3}\dots(1)$ General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ is given by, $y \cdot(I . F .)=\int Q...
Solution: $\frac{d y}{d x}+2 y \tan x=\sin x\dots(1)$ To solve (1) we will use following formula $\int \tan x d x=\log _{\mid} \sec x \mid$ $\begin{array}{l} \operatorname{alog} \mathrm{b}=\log...
Solution: $\frac{d y}{d x}+y \cot x=\sin 2 x-\dots (1)$ General solution: For the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ General solution is given by, $y...
Solution: $\frac{d y}{d x}=\operatorname{ytan} x-2 \sin x-\ldots(1)$ General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ General solution is given...
Solution: $x \frac{d y}{d x}-y=2 x^{2} \sec x \dots (1)$ General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ General solution is given by, $y...
Solution: $\frac{d y}{d x}+2 \mathrm{y}(\cot x)=3 x^{2} \operatorname{cosec}^{2} x\dots (1)$ General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$...
Solution: $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=\cot x-\ldots(1)$ General solution for the differential equation in the form of $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ General solution is...
Solution: $\frac{d y}{d x}+y=\cos x-\sin x-\cdots \dots(1)$ General solution for the differential equation in the form of $\frac{d y}{d x}+P y=Q$ is given by, $y \cdot(I . F .)=\int Q \cdot(I . F .)...
Solution: $\frac{d y}{d x}+2 y=\sin x- \dots(1)$ General solution for the differential equation in the form of is given by, $y \cdot(I . F .)=\int Q \cdot(I . F .) d x+c$ Where, integrating factor,...
Solution: $x d y+\left(y-x^{3}\right) d x=0\dots (1)$ General solution for the differential equation in the form of is given by,$\frac{d y}{d x}+P y=Q$ $y \cdot(I . F .)=\int Q \cdot(I . F .) d x+c$...
Solution: $x d y-\left(y+2 x^{2}\right) d x=0\dots (1)$ General solution for the differential equation in the form of $\frac{d y}{d x}+P y=Q$ is given by $y \cdot(I . F .)=\int Q \cdot(I . F \cdot)...
Solution: $\left(\mathrm{y}+3 \mathrm{x}^{2}\right) \frac{d x}{d y}=\mathrm{x}-\cdots \dots(1)$ General solution for the differential equation in the form of is given by,$\frac{d y}{d x}+P y=Q$ $y...
Solution: $\frac{d y}{d x}+\frac{4 x}{x^{2}+1} y+\frac{1}{\left(1+x^{2}\right)^{2}}=0\dots (1)$ General solution for the differential equation in the form of $\frac{d y}{d x}+P y=Q$ is given by, $y...
Solution: $(1+x) \frac{d y}{d x}-y=e^{3 x}(1+x)^{2}\dots (1)$ General solution for the differential equation in the form of is given by $\frac{d y}{d x}+P y=Q$ $y \cdot(I . F .)=\int Q \cdot(I . F...
Solution: $\begin{array}{r} x \frac{d y}{d x}+2 y=x^{2} \log x-\ldots(1) \\ \qquad \frac{d(\log x)}{d x}=\frac{1}{x} \end{array}$ General solution for the differential equation in the form of...
Solution: $x \frac{d y}{d x}+y=x \log x\dots (1)$ To solve (1) we will use following formula $\begin{array}{l} \int \frac{1}{x} d x=\log x \\ a^{\log _{a} b=b} \\ \qquad \int u \cdot v d x=u . \int...
Solution: $\frac{d y}{d x}-y \tan x=e^{x} \sec x-\ldots(1)$ To solve (1) we will use following formula $\begin{array}{l} \quad \int \tan x d x=\log (\sec x) \\ a \log b=\log b^{a} \\ a^{\log _{a}...
Solution: Given Differential Equation : $\frac{\text { dy }}{\mathrm{dx}}+\frac{1}{\mathrm{x}} \cdot \mathrm{y}=\mathrm{x}^{2} \ldots \ldots \ldots \mathrm{eq}(1)$ Formula : i) $\int \frac{1}{x} d...
Solution: $\frac{d y}{d x}+8 y=5 e^{-3 x} \ldots(1)$ To solve (1) we will use following formula $\begin{array}{l} \int 1 d x=x \\ \int e^{k x} d x=\frac{e^{k x}}{k} \end{array}$ General solution for...
Solution: $\frac{d y}{d x}+3 y=e^{-2 x}\dots \dots(1)$ To solve (1) we will use following formula $\begin{array}{l} \int 1 d x=x \\ \int e^{k x} d x=\frac{e^{k x}}{k} \end{array}$ General solution...
Solution: $\frac{d y}{d x}+2 y=6 e^{x}$ To solve (1) we will use following formula $\begin{array}{l} \int 1 d x=x \\ \int e^{k x} d x=\frac{e^{k x}}{k} \end{array}$ General solution for the...
Solution: $\left(x^{2}+1\right) \frac{d y}{d x}-2 x y=\left(x^{2}+1\right)\left(x^{2}+2\right)^{-\ldots}-\ldots(1)$ To solve (1) we will use following formula $\begin{array}{l} \int...
Solution: $\left(1-x^{2}\right) \frac{d y}{d x}=x y=a x-\ldots(1)$ To solve (1) we will use following formula $\begin{array}{r} \int \frac{f^{\prime}(X)}{f(x)} d x=\log f(x)+\mathrm{C} \\ a \log...
Solution: $\left(1-x^{2}\right) \frac{d y}{d x}+x y=x \sqrt{1-x^{2}}\dots \dots (1)$ To solve (1) we will use following formula $\begin{array}{l} \int \frac{f^{\prime}(x)}{f(x)} d x=\log f(x) \\...
Solution: $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=\frac{1}{\left(1+x^{2}\right)}\dots \dots(1)$ To solve (1) we will use following formula $\begin{array}{l} \int \frac{f^{\prime}(x)}{f(x)} d...
Solution: $x \frac{d y}{d x}-y=x+1\dots \dots (1)$ To solve (1) we will use following formula $\begin{array}{l} \int \frac{1}{x} d x=\log x \\ \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \\ a \log b=\log...
Solution: $x \frac{d y}{d x}-y+2 x^{3}\dots \dots(1)$ To solve (1) we will use following formula $\begin{array}{c} \int \frac{1}{x} d x=\log x \\ a \log b=\log b^{a} \end{array}$ $a^{\log _{a}...
Solution: $x \frac{d y}{d x}+y=3 x^{2}-3\dots \dots (1)$ To solve (1) we will use following formula $\begin{array}{l} \int \frac{1}{x} d x=\log x \\ a^{\log _{a} b}=\log b \end{array}$ General...
Solution: To solve the given equation we will use following formula $\begin{array}{l} \int \frac{1}{x} d x=\log \mathrm{x}+\mathrm{C} \\ \int x^{n} d x=\frac{x^{n+1}}{n+1}+\mathrm{c} \\ a \log...
Solution: $x \frac{d y}{d x}+2 y=x^{2}\dots \dots (1)$ To solve (1) we will use following formula $\begin{array}{l} \int \frac{1}{x} d x=\log x \\ \int x^{n} d x=\frac{x^{n}+1}{n+1}+c \\ a^{\log...
Solution: Given Differential Equation : $\frac{\text { dy }}{\mathrm{dx}}+\frac{1}{\mathrm{x}} \cdot \mathrm{y}=\mathrm{x}^{2} \ldots \ldots \ldots \mathrm{eq}(1)$ Formula : i) $\int \frac{1}{x} d...
Solution: The transpose of the matrix is an operation of making interchange of elements by the rule on positioned element $a_{j i}$ shifted to new position $a_{j i}$. The symmetric matrix is defined...
![A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7da3aec34a65e6e0aeb82a3f6a232300_l3.png "Rendered by QuickLaTeX.com")
and 
, find 
.Solution: We have $A=\left(\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right)$. Now addition/subtraction of two matrices is possible if order of both the matrices are same and multiplication...
Solution: We have $A$ and $B$ are symmetric matrices. Therefore $A^{T}=A$ and $B^{T}=B$ The transpose of the matrix is an operation of making interchange of elements by the rule on positioned...
![A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b5c8c8d44a123b24c17830efc625a77b_l3.png "Rendered by QuickLaTeX.com")
, show that 
.Solution: We have $A=\left(\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right)$. The transpose of the matrix is an operation of making interchange of...
![\mathrm{A}=\left[\begin{array}{ll}4 & 2 \\ 1 & 3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bd81dbbffa9684064602a99b2f36c226_l3.png "Rendered by QuickLaTeX.com")
and ![\mathrm{B}=\left[\begin{array}{cc}-2 & 1 \\ 3 & 2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aefeccc89d0278dab4e14dc06c63bae5_l3.png "Rendered by QuickLaTeX.com")
, find a matrix 
such that 
Solution: We have $A=\left(\begin{array}{ll}4 & 2 \\ 1 & 3\end{array}\right), B=\left(\begin{array}{cc}-2 & 1 \\ 3 & 2\end{array}\right)$ and $3 A-2 B+X=0$ We can have the addition...
![A=\left[\begin{array}{cc}2 & -3 \\ 4 & 5\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-40f9b0b32323b8d7e2a6115865036362_l3.png "Rendered by QuickLaTeX.com")
and ![B=\left[\begin{array}{cc}-1 & 2 \\ 0 & 3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2386be89f405fcc8381ad1375da5f2f9_l3.png "Rendered by QuickLaTeX.com")
, find a matrix such that 
.Solution: We have $A=\left(\begin{array}{cc}2 & -3 \\ 4 & 5\end{array}\right), B=\left(\begin{array}{cc}-1 & 2 \\ 0 & 3\end{array}\right)$ and $A+2 B+X=0$. We can have the addition...
![\mathrm{A}=\left[\begin{array}{ll}4 & 5 \\ 1 & 8\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8158b32a662df332e0a5eb4fe7a2edb9_l3.png "Rendered by QuickLaTeX.com")
, show that 
is symmetricSolution: We have $\left(\begin{array}{ll}4 & 5 \\ 1 & 8\end{array}\right)$. The transpose of the matrix is an operation of making interchange of elements by the rule on positioned element...
and for which ![\left[\begin{array}{cc} \mathrm{x} & \mathrm{y} \\ 3 \mathrm{y} & \mathrm{x} \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \end{array}\right]=\left[\begin{array}{l} 3 \\ 5 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0863b7d782cc25f6aa50594877777cc3_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $\left(\begin{array}{cc}x & y \\ 3 y & x\end{array}\right)\left(\begin{array}{l}1 \\ 2\end{array}\right)=\left(\begin{array}{l}3 \\ 5\end{array}\right)$. Use the...
and for which ![\left[\begin{array}{cc} 2 & -3 \\ 1 & 1 \end{array}\right]\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \end{array}\right]=\left[\begin{array}{l} 1 \\ 3 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4bea77a5649bdf186a1bac7467012af0_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $\left(\begin{array}{cc}2 & -3 \\ 1 & 1\end{array}\right)\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{l}1 \\ 3\end{array}\right)$. Use the...
![A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-237a953de1ebc9d655318073b827c503_l3.png "Rendered by QuickLaTeX.com")
then find the least value of 
for which 
.Solution: We have $A=\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right)$ Use the addition rule and get $A+A^{T}=I_{2}$ as follow:...
![\mathrm{A}=\left[\begin{array}{cr}1 & -5 \\ -3 & 2 \\ 4 & -2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b75e51d5c2d1c5e218878039b0ef52a7_l3.png "Rendered by QuickLaTeX.com")
and ![\mathrm{B}=\left[\begin{array}{cc}3 & 1 \\ 2 & -1 \\ -2 & 3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4438109c3e36aa09efaebe02d6551a19_l3.png "Rendered by QuickLaTeX.com")
, find the matrix such that 
is a zero matrixSolution: We have $A=\left(\begin{array}{cc}1 & -5 \\ -3 & 2 \\ 4 & -2\end{array}\right) ; B=\left(\begin{array}{cc}3 & 1 \\ 2 & -1 \\ -2 & 3\end{array}\right) .$ and...
![\begin{array}{l} \cos \theta \cdot\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]+\sin \theta \\ {\left[\begin{array}{cc} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{array}\right]=I} \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9f6e84acdf2d6d135e00de04b3a005f2_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $\cos \theta\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right)+\sin \theta\left(\begin{array}{cc}\sin \theta & -\cos...
Solution: We have $A=\operatorname{diag}(3-25)=\left(\begin{array}{ccc}3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 5\end{array}\right) ; B=$...
![\left[\begin{array}{cc}x & 6 \\ -1 & 2 w\end{array}\right]+\left[\begin{array}{cc}4 & x+y \\ z+w & 3\end{array}\right]=3\left[\begin{array}{cc}x & y \\ z & w\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3458f4e6f17d7e715a86e43d08489614_l3.png "Rendered by QuickLaTeX.com")
, find the values of 
Solution: We have $\left(\begin{array}{cc}x & 6 \\ -1 & 2 w\end{array}\right)+\left(\begin{array}{cc}4 & x+y \\ z+w & 3\end{array}\right)=3\left(\begin{array}{cc}x & y \\ z &...
![\left[\begin{array}{cc}x & 3 x-y \\ 2 x+z & 3 y-w\end{array}\right]=\left[\begin{array}{ll}3 & 2 \\ 4 & 7\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-611c97c74d8c38eb224393d8f387948c_l3.png "Rendered by QuickLaTeX.com")
, find the values of 
.Solution: We have $\left(\begin{array}{cc}x & 3 x-y \\ 2 x+z & 3 y-w\end{array}\right)=\left(\begin{array}{ll}3 & 2 \\ 4 & 7\end{array}\right)$. Now from the equality of matrices we...
![x \cdot\left[\begin{array}{l}2 \\ 3\end{array}\right]+y \cdot\left[\begin{array}{c}-1 \\ 1\end{array}\right]=\left[\begin{array}{r}10 \\ 5\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b06993b29db939642e38c580586c51db_l3.png "Rendered by QuickLaTeX.com")
, find the values of and .Solution: We have $x\left(\begin{array}{l}2 \\ 3\end{array}\right)+y\left(\begin{array}{c}-1 \\ 1\end{array}\right)=\left(\begin{array}{c}10 \\ 5\end{array}\right)$. Use the addition rule and get...
and , if ![2\left[\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e3b199653bed94af618cb5f0d4f0ec35_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $2\left(\begin{array}{ll}1 & 3 \\ 0 & x\end{array}\right)+\left(\begin{array}{ll}y & 0 \\ 1 & 2\end{array}\right)=\left(\begin{array}{ll}5 & 6 \\ 1 &...
![\left[\begin{array}{cc}x+2 y & -y \\ 3 x & 4\end{array}\right]=\left[\begin{array}{cc}-4 & 3 \\ 6 & 4\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-33d06ec59a67edd192829f442472121d_l3.png "Rendered by QuickLaTeX.com")
, find the values of and Solution: We have $\left(\begin{array}{cc}x+2 y & -y \\ 3 x & 4\end{array}\right)=\left(\begin{array}{cc}-4 & 3 \\ 6 & 4\end{array}\right)$ Now from the equality of matrices we can...
matrix whose elements are given by 
Solution: We have $a_{i j}=\frac{1}{2}|-3 i+j|^{2}$ Now $\begin{array}{l} a_{11}=\frac{|-3(1)+1|}{2}=1, a_{12}=\frac{|-3(1)+2|}{2}=\frac{9}{2}, a_{13}=\frac{|-3(1)+3|}{2}=\frac{9}{2} \\...
matrix whose elements are given by 
Solution: We have $a_{i j}=\frac{1}{2}(i-2 j)^{2}$ Now $\begin{array}{l} a_{11}=\frac{(1-2(1))^{2}}{2}=\frac{1}{2}, a_{12}=\frac{(1-2(2))^{2}}{2}=\frac{9}{2} \\ a_{21}=\frac{(2-2(1))^{2}}{2}=0...
![\left[\begin{array}{ccc} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f03cb7f5858669bd60950918832693a7_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{ccc}-1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary...
![\left[\begin{array}{ccc} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9b986c3bb2233699ef7411fc62105e4b_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{ccc}1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5\end{array}\right)$. To get the inverse we will proceed by augmented matrix with elementary...
![\left[\begin{array}{ccc} 3 & -1 & -2 \\ 2 & 0 & -1 \\ 3 & -5 & 0 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6f98ac5a8bdc0a4c68b59ad5fff13f27_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{ccc}3 & -1 & -2 \\ 2 & 0 & -1 \\ 3 & -5 & 0\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary...
![\left[\begin{array}{lll} 3 & 0 & 2 \\ 1 & 5 & 9 \\ 6 & 4 & 7 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c2d4be16f7e50993b9d9ddb636e32da5_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{lll}3 & 0 & 2 \\ 1 & 5 & 9 \\ 6 & 4 & 7\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row...
![\left[\begin{array}{ccc} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9e97cd7932741e795ff74dfcc62f2080_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row...
![\left[\begin{array}{ll} 4 & 0 \\ 2 & 5 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-00e067d8b3d687152537a9d78c193c7f_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{ll}4 & 0 \\ 2 & 5\end{array}\right)$. To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...
![\left[\begin{array}{cc} 2 & -3 \\ 4 & 5 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6d086db9247d6fa2dd93801088be8f5f_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{cc}2 & -3 \\ 4 & 5\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...
![\left[\begin{array}{rr} 2 & 5 \\ -3 & 1 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4e7a3750c7e99f5875f5fe5e581bc3bc_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{cc}2 & 5 \\ -3 & 1\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...
![\left[\begin{array}{cc} 1 & 2 \\ 2 & -1 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-918a8e2a33fbf14a8141e6d365aa3df9_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...
![\left[\begin{array}{ll} 1 & 2 \\ 3 & 7 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6a1968e1a662e8c4373a2fec80dd99d1_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{ll}1 & 2 \\ 3 & 7\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...
, show that 
.Solution: Given that $A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \theta & \cos \alpha\end{array}\right]$. We wil find $A$ $A^{\prime}=\left[\begin{array}{cc}\cos \alpha...
and , verify that 
: ![A=\left[\begin{array}{rrr}-1 & 2 & -3 \\ 4 & -5 & 6\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4b8a6371313bb6ec237c7ac8ff4e7f9d_l3.png "Rendered by QuickLaTeX.com")
and ![B=\left[\begin{array}{cc}3 & -4 \\ 2 & 1 \\ -1 & 0\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d5b55b58779f0457b2c521b7bc825595_l3.png "Rendered by QuickLaTeX.com")
Solution: Take $\mathrm{C}=\mathrm{AB}$ $\begin{array}{l} C=\left[\begin{array}{ccc} -1 & 2 & -3 \\ 4 & -5 & 6 \end{array}\right]\left[\begin{array}{cc} 3 & -4 \\ 2 & 1 \\ -1...
and , verify that 
: ![\mathrm{A}=\left[\begin{array}{c} -1 \\ 2 \\ 3 \end{array}\right] \text { and } \mathrm{B}=[-2 \cdot 1-4]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-86cfd44f3d05fdc26f7ec6f9a464ad20_l3.png "Rendered by QuickLaTeX.com")
Solution: Take $C=A B$ $\begin{array}{l} C=\left[\begin{array}{c} -1 \\ 2 \\ 3 \end{array}\right]\left[\begin{array}{lll} -2 & -1 & -4 \end{array}\right] \\...
and , verify that : ![A=\left[\begin{array}{rr} 3 & -1 \\ 2 & -2 \end{array}\right] \text { and } B=\left[\begin{array}{rr} 1 & -3 \\ 2 & -1 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3f4483ea07db74116ba0bc5d55287e5b_l3.png "Rendered by QuickLaTeX.com")
Solution: Take $C=A B$ $\begin{array}{l} C=\left[\begin{array}{rr} 3 & -1 \\ 2 & -2 \end{array}\right]\left[\begin{array}{ll} 1 & -3 \\ 2 & -1 \end{array}\right] \\...
and , verify that : ![A=\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right] \text { and } B=\left[\begin{array}{ll} 1 & 4 \\ 2 & 5 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c3667d21de9c5134248177485655ffeb_l3.png "Rendered by QuickLaTeX.com")
Solution: Take $\mathrm{C}=\mathrm{A} 8$ $\begin{array}{l} C=\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]\left[\begin{array}{ll} 1 & 4 \\ 2 & 5 \end{array}\right] \\...
![A=\left[\begin{array}{lll}3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7aaa077b9058e56f42a554ca8fcb171f_l3.png "Rendered by QuickLaTeX.com")
as sum af two matrices such that and is symmetric and the other is skew-symmetric.Solution: Given that $A=\left[\begin{array}{lll}3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7\end{array}\right]$, to express as sum of symmetric matrix $P$ and skew symmetric matrix...
![\mathrm{A}=\left[\begin{array}{rr}3 & -4 \\ 1 & -1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4c224184d619670975e11f1ad1098530_l3.png "Rendered by QuickLaTeX.com")
as the sum of a symmetric matrix and a skew-symmetric matrix.Solution: Given that $\mathrm{A}=\left[\begin{array}{rr}3 & -4 \\ 1 & -1\end{array}\right]$,to express as the sum of symmetric matrix $\mathrm{P}$ and skew symmetric matrix Q. $A=P+Q$ Where...
![\mathrm{A}=\left[\begin{array}{cc}2 & 3 \\ -1 & 4\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0327e38dfafec712de874ce9c4347484_l3.png "Rendered by QuickLaTeX.com")
as the sum of a symmetric matrix and a skew-symmetric matrix.Solution: Given that $A=\left[\begin{array}{cc}2 & 3 \\ -1 & 4\end{array}\right]$, As for a symmetric matrix $A^{\prime}=A$ hence $\begin{array}{l} A+A^{\prime}=2 A \\...
![\mathrm{A}=\left[\begin{array}{ccc}0 & \mathrm{a} & \mathrm{b} \\ -\mathrm{a} & 0 & \mathrm{c} \\ -\mathrm{b} & -\mathrm{c} & 0\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-46d4fca5a1e4ccd3daf4e4683a818d9d_l3.png "Rendered by QuickLaTeX.com")
is skew-symmetric.Solution: We have $A=\left(\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right)$. The transpose of the matrix is an operation of making interchange of...
![A=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-16f776216ed4662f25a9fa186c592e2f_l3.png "Rendered by QuickLaTeX.com")
, show that 
is skew-symmetric.Solution: We have $A=\left(\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right)$ The transpose of the matrix is an operation of making interchange of elements by the rule on positioned...
![A=\left[\begin{array}{ll}4 & 1 \\ 5 & 8\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2cd8607c366b843d16c49d92af8d1f55_l3.png "Rendered by QuickLaTeX.com")
, show that is symmetric.Solution: We have $A=\left(\begin{array}{ll}4 & 1 \\ 5 & 8\end{array}\right)$. The transpose of the matrix is an operation of making interchange of elements by the rule on positioned element...
![P=\left[\begin{array}{rr}3 & 4 \\ 2 & -1 \\ 0 & 5\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-25a68e7dd1de3e3d940408c0f29c6ce9_l3.png "Rendered by QuickLaTeX.com")
and ![P=\left[\begin{array}{cc}7 & -5 \\ -4 & 0 \\ 2 & 6\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0609e63fb51547a4b3ca4db5d7e945bf_l3.png "Rendered by QuickLaTeX.com")
, verify that 
Solution: We have $P=\left(\begin{array}{cc}3 & 4 \\ 2 & -1 \\ 0 & 5\end{array}\right)$ and $Q=\left(\begin{array}{cc}7 & -5 \\ -4 & 0 \\ 2 & 6\end{array}\right)$. The...
![A=\left[\begin{array}{rrr}3 & 2 & -1 \\ -5 & 0 & -6\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c32e17bbb774b334da0bf3fafe18b27e_l3.png "Rendered by QuickLaTeX.com")
and ![B=\left[\begin{array}{ccc}-4 & -5 & -2 \\ 3 & 1 & 8\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-733024809307a4305930fcb87f2b032f_l3.png "Rendered by QuickLaTeX.com")
, verify that 
.Solution: We have $A=\left(\begin{array}{ccc}3 & 2 & -1 \\ -5 & 0 & -6\end{array}\right)$ and $B=\left(\begin{array}{ccc}-4 & -5 & -2 \\ 3 & 1 & 8\end{array}\right)$....
![A=\left[\begin{array}{ccc}2 & -3 & 5 \\ 0 & 7 & -4\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6ac0f76ea62d4473fa0142e884721096_l3.png "Rendered by QuickLaTeX.com")
, verify that 
.Solution: We have $A=\left(\begin{array}{ccc}2 & -3 & 5 \\ 0 & 7 & -4\end{array}\right)$. The transpose of the matrix is an operation of making interchange of elements by the rule on...
![\left[\begin{array}{cc}2 & 3 \\ 5 & 7\end{array}\right]\left[\begin{array}{cc}1 & -3 \\ -2 & 4\end{array}\right]=\left[\begin{array}{cc}-4 & 6 \\ -9 & \mathrm{x}\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6910ac8e55e169c0e6b215dd371a238c_l3.png "Rendered by QuickLaTeX.com")
, find the value of Solution: We have $A=\left(\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right), B=\left(\begin{array}{cc}1 & -3 \\ -2 & 4\end{array}\right)$ and $C=\left(\begin{array}{cc}-4 & 6 \\...
![A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4b294a697ebb148d82263ae46fb13478_l3.png "Rendered by QuickLaTeX.com")
and ![B=\left[\begin{array}{cc}0 & 4 \\ -1 & 7\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-170fe1f79db53b97fc71e66a945d55c8_l3.png "Rendered by QuickLaTeX.com")
, find 
.Solution: We have $A=\left(\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right), B=\left(\begin{array}{cc}0 & 4 \\ -1 & 7\end{array}\right)$. (i) Let's compute first $A^{2}$ $A^{2}=A...
Solution: We have $\boldsymbol{A B}=\boldsymbol{A} \boldsymbol{C}$ but $\boldsymbol{B} \neq \boldsymbol{C}$. WE need to find: $\boldsymbol{A}, \boldsymbol{B}$. Let's take...
Solution: We have $\boldsymbol{A} \neq \boldsymbol{O}, \boldsymbol{B} \neq \boldsymbol{O}, \boldsymbol{A B}=\boldsymbol{O}$ and $\boldsymbol{B A} \neq \boldsymbol{O}$. We need to find:...
![A=\left[\begin{array}{cc}3 & 4 \\ -4 & -3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fec9588c04898858cde19179ff79aa8f_l3.png "Rendered by QuickLaTeX.com")
, find , where 
.Solution: We have $A=\left(\begin{array}{cc}3 & 4 \\ -4 & -3\end{array}\right)$ and equation $f(x)=x^{2}-5 x+7$. (i) Let us compute first $A^{2}$ $A^{2}=A A=\left(\begin{array}{cc} 3 & 4...
and 
for which ![\left[\begin{array}{cc} a & b \\ -a & 2 b \end{array}\right]\left[\begin{array}{c} 2 \\ -1 \end{array}\right]=\left[\begin{array}{l} 5 \\ 4 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-54ffd2f53dfe0a81dce111e7e3dda667_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}\boldsymbol{a} & \boldsymbol{b} \\ -\boldsymbol{a} & \boldsymbol{2} \boldsymbol{b}\end{array}\right),...
![\left[\begin{array}{lll}1 & x & 1\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & 2 & 5\end{array}\right]\left[\begin{array}{c}1 \\ -2 \\ 3\end{array}\right]=\mathrm{O}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4941f73e6b8cc12c72d34e0f4983af39_l3.png "Rendered by QuickLaTeX.com")
, find Solution: We have $A=\left(\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & 2 & 5\end{array}\right), B=\left(\begin{array}{lll}1 & x & 1\end{array}\right)$ and...
, show that ![A^{2}=\left[\begin{array}{cc}\cos 2 \alpha & \sin 2 \alpha \\ -\sin 2 \alpha & \cos 2 \alpha\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-004ae2a0cb0effd2df50d6aef24065f4_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right)$ and to show $A^{2}=$ $\left(\begin{array}{cc}\cos ^2 \alpha & \sin...
![F(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fe54b977730cd78576884611e79ac060_l3.png "Rendered by QuickLaTeX.com")
, show that 
Solution: We have $\boldsymbol{F}(\boldsymbol{X})=\left(\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right)$ and to show...
![\left[\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right]=\left[\begin{array}{cc}0 & -4 \\ 10 & 3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-870749f112d49ec389561d6cc400b796_l3.png "Rendered by QuickLaTeX.com")
.Solution: We have $\boldsymbol{B}=\left(\begin{array}{ll}\mathbf{2} & \mathbf{3} \\ \mathbf{4} & \mathbf{5}\end{array}\right)$ and $\boldsymbol{C}=\left(\begin{array}{cc}\mathbf{0} &...
![A=\left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-866d9f588aa87affb3349473fc083592_l3.png "Rendered by QuickLaTeX.com")
, find the value of a and such that 
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ll}\mathbf{3} & \mathbf{2} \\ \mathbf{1} & \mathbf{1}\end{array}\right)$. To find $\boldsymbol{a}, \boldsymbol{b}$ such that...
![A=\left[\begin{array}{ll}3 & 1 \\ 7 & 5\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0988384a97f81e0966a9037a51b5ab9b_l3.png "Rendered by QuickLaTeX.com")
, find and such that 
.Solution: We have $\boldsymbol{A}=\left(\begin{array}{ll}\mathbf{3} & \mathbf{1} \\ \mathbf{7} & \mathbf{5}\end{array}\right)$. To find $\boldsymbol{x}, \boldsymbol{y}$ such that...
and , when Solution: We have $A=\left(\begin{array}{cc}2 & -3 \\ 1 & 1\end{array}\right), B=\left(\begin{array}{l}1 \\ 3\end{array}\right)$ and $X=\left(\begin{array}{l}x \\ y\end{array}\right)$. To...
![A=\left[\begin{array}{cc}-1 & 2 \\ 3 & 1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-abd4fa7d6a73d83d97ab6e13a9b5a009_l3.png "Rendered by QuickLaTeX.com")
, find , where 
Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}-1 & 2 \\ 3 & 1\end{array}\right)$. Now addition/subtraction of two matrices is possible if order of both the matrices are same and...
![A=\left[\begin{array}{rr}3 & -2 \\ 4 & -2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dcf4081f0dd51325845f0b232bbdf385_l3.png "Rendered by QuickLaTeX.com")
, find 
so that 
.Solution: We have $\boldsymbol{A}=\left(\begin{array}{ll}\mathbf{3} & -\mathbf{2} \\ \mathbf{4} & -\mathbf{2}\end{array}\right)$. Now addition/subtraction of two matrices is possible if...
![\mathrm{A}=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a90c945b5fcc948c6a2940f9dbfc3dcf_l3.png "Rendered by QuickLaTeX.com")
satisfies the equation 
.Solution: We have $\boldsymbol{A}=\left(\begin{array}{ll}\mathbf{2} & \mathbf{3} \\ \mathbf{1} & \mathbf{2}\end{array}\right)$. Now addition of two matrices is possible if order of both the...
![A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ba6b294e8b773dce761b0691a57562ee_l3.png "Rendered by QuickLaTeX.com")
, show that 
.Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}\mathbf{3} & \mathbf{1} \\ -\mathbf{1} & \mathbf{2}\end{array}\right)$. Now addition of two matrices is possible if order of both the...
![A=\left[\begin{array}{cc}2 & -2 \\ -3 & 4\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-27e32a8a28ce812aeb94edeb354a68bd_l3.png "Rendered by QuickLaTeX.com")
then find 
Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}\mathbf{2} & \mathbf{- 2} \\ -\mathbf{3} & \mathbf{4}\end{array}\right)$. Now addition of two matrices is possible if order of both...
![A=\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4aecc6df45ed27297ab2d5657e1884ca_l3.png "Rendered by QuickLaTeX.com")
and , find .Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}\mathbf{2} & \mathbf{- 1} \\ \mathbf{3} & \mathbf{2}\end{array}\right)$ and $\boldsymbol{B}=\left(\begin{array}{cc}\mathbf{0} &...
![A=\left[\begin{array}{ccc}4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2d60fb53e459506aacf730b3fd4b58ff_l3.png "Rendered by QuickLaTeX.com")
, show that 
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ccc}\mathbf{4} & -\mathbf{1} & -\mathbf{4} \\ \mathbf{3} & \mathbf{0} & -\mathbf{4} \\ \mathbf{3} & -\mathbf{1} &...
![A=\left[\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9fe04bd131b7b3438ccdb0c839265f03_l3.png "Rendered by QuickLaTeX.com")
, show that 
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ccc}\mathbf{2} & \mathbf{- 2} & -\mathbf{4} \\ \mathbf{- 1} & \mathbf{3} & \mathbf{4} \\ \mathbf{1} & -\mathbf{2} &...
, when ![\mathrm{A}=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right], \mathrm{B}=\left[\begin{array}{cc} 2 & 0 \\ 1 & -3 \end{array}\right] \text { and } \mathrm{C}=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-65c5d9f1c3d09f9b98677f8ffcad25dc_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ll}\mathbf{1} & \mathbf{2} \\ \mathbf{3} & \mathbf{4}\end{array}\right), \boldsymbol{B}=\left(\begin{array}{cc}\mathbf{2} &...
: ![\mathrm{A}=\left[\begin{array}{ccc} 2 & 3 & -1 \\ 3 & 0 & 2 \end{array}\right], \mathrm{B}=\left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right] \text { and } \mathrm{C}=\left[\begin{array}{ll} 1 & -2 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f267062030e6a4b1a9e0ae96655b75b9_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ccc}\mathbf{2} & \mathbf{3} & \mathbf{- 1} \\ \mathbf{3} & \mathbf{0} & \mathbf{2}\end{array}\right),...
: ![\mathrm{A}=\left[\begin{array}{lll} 1 & 2 & 5 \\ 0 & 1 & 3 \end{array}\right], \mathrm{B}=\left[\begin{array}{lll} 2 & 3 & 0 \\ 1 & 0 & 4 \\ 1 & -1 & 2 \end{array}\right] \text { and } \mathrm{C}=\left[\begin{array}{l} 1 \\ 4 \\ 5 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-28076805d7b71c6daa66bcab8ce41845_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{lll}1 & 2 & 5 \\ 0 & 1 & 3\end{array}\right), B=\left(\begin{array}{ccc}2 & 3 & 0 \\ 1 & 0 & 4 \\ 1 & -1 &...
![A=\left[\begin{array}{ccc}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-38b9629a3ecd89b9f213d5de336a84f0_l3.png "Rendered by QuickLaTeX.com")
and ![B=\left[\begin{array}{lll}a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2}\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-95fc7f6ea73a57dfb76fa56240b6d2f5_l3.png "Rendered by QuickLaTeX.com")
. show that 
is a zero matrix.Solution: We have $\boldsymbol{A}=\left(\begin{array}{ccc}\mathbf{0} & \boldsymbol{c} & -\boldsymbol{b} \\ -\boldsymbol{c} & \mathbf{0} & \boldsymbol{a} \\ \boldsymbol{b} &...
in each of the following cases: ![A=\left[\begin{array}{ccc} 1 & 3 & -1 \\ 2 & 2 & -1 \\ 3 & 0 & -1 \end{array}\right] \text { and } B=\left[\begin{array}{ccc} -2 & 3 & -1 \\ -1 & 2 & -1 \\ -6 & 9 & -4 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7a2b2d708af06466b10968e7314747a5_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ccc}\mathbf{1} & \mathbf{3} & -\mathbf{1} \\ \mathbf{2} & \mathbf{2} & -\mathbf{1} \\ \mathbf{3} & \mathbf{0} &...
in each of the following cases: ![\mathrm{A}=\left[\begin{array}{lll} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{ccc} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f2cf553262923f440cdb2486ff51fdb6_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{lll}1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2\end{array}\right)$ and $B=\left(\begin{array}{ccc}10 & -4 & -1 \\ -11 & 5 & 0...
in each of the following cases: ![A=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \text { and } B=\left[\begin{array}{cc} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-de32b04501d491874e21effd827b2a59_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $A=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right)$ and $B=\left(\begin{array}{cc}\cos \phi & -\sin \phi \\ \sin \phi...
in each of the following cases: ![\mathrm{A}=\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{ccc} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4ffeaf0e1df4be603ccf10fb6d6f40b6_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $\boldsymbol{A}=\left(\begin{array}{lll}\mathbf{1} & \mathbf{2} & \mathbf{3} \\ \mathbf{0} & \mathbf{1} & \mathbf{0} \\ \mathbf{1} & \mathbf{1} &...
in each of the following cases : ![A=\left[\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}\right] \text { and } B=\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7f5c45bafaa8a499f178625171ba32fc_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}\mathbf{5} & -\mathbf{1} \\ \mathbf{6} & \mathbf{7}\end{array}\right)$ and $\boldsymbol{B}=\left(\begin{array}{ll}\mathbf{2} &...
and 
, which ever exists when ![A=\left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{array}\right] \text { and } B=\left[\begin{array}{rrr} 1 & 0 & 1 \\ -1 & 2 & 1 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-279e1d695b832080e5f60bbaf01191b3_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $\boldsymbol{B}=\left(\begin{array}{ccc}\mathbf{1} & \mathbf{0} & \mathbf{1} \\ -\mathbf{1} & \mathbf{2} & \mathbf{1}\end{array}\right)$ and...
and BA, which ever exists when ![A=\left[\begin{array}{lll} 1 & 2 & 3 & 4 \end{array}\right] \text { and } B=\left[\begin{array}{l} 1 \\ 2 \\ 3 \\ 4 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dd6c4ca0935c74e87d91cc74e4077943_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $\boldsymbol{A}=\left(\begin{array}{llll}\mathbf{1} & 2 & 3 & 4\end{array}\right)$ and $\boldsymbol{B}=\left(\begin{array}{l}\mathbf{1} \\ \mathbf{2} \\ \mathbf{3} \\...
![\mathrm{A}=\left[\begin{array}{rrr} 0 & 1 & -5 \\ 2 & 4 & 0 \end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{cc} 1 & 3 \\ -1 & 0 \\ 0 & 5 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-02b69ceb338bd244cfc287cc78e2993a_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ccc}\mathbf{0} & \mathbf{1} & -\mathbf{5} \\ \mathbf{2} & \mathbf{4} & \mathbf{0}\end{array}\right)$ and...
![\mathrm{A}=\left[\begin{array}{cc} -1 & 1 \\ -2 & 2 \\ -3 & 3 \end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{ccc} 3 & -2 & 1 \\ 0 & 1 & 2 \\ -3 & 4 & -5 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-499601ef9942d093f177dcd042e96b76_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}-\mathbf{1} & \mathbf{1} \\ -\mathbf{2} & \mathbf{2} \\ -\mathbf{3} & \mathbf{3}\end{array}\right)$ and...
![A=\left[\begin{array}{cc} 2 & -1 \\ 3 & 0 \\ -1 & 4 \end{array}\right] \text { and } B=\left[\begin{array}{cc} -2 & 3 \\ 0 & 4 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-21ad4759d8ac8e3299f333785850143c_l3.png "Rendered by QuickLaTeX.com")
Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}\mathbf{2} & \mathbf{- 1} \\ \mathbf{3} & \mathbf{0} \\ \mathbf{- 1} & \mathbf{4}\end{array}\right)$ and...
![\left[\begin{array}{cc}x-y & 2 y \\ 2 y+z & x+y\end{array}\right]=\left[\begin{array}{ll}1 & 4 \\ 9 & 5\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2ca4ce3421a2dc3e7599f50fd58c71fa_l3.png "Rendered by QuickLaTeX.com")
then write the value of 
Solution: $\text { If }\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{ll} e & f \\ g & h \end{array}\right]$ Therefore $a=e, b=f, c=g, d=h$ It is given...
from the following equation : ![2\left[\begin{array}{ll} 1 & 3 \\ 0 & \mathrm{x} \end{array}\right]+\left[\begin{array}{ll} \mathrm{y} & 0 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fae2ef579b0094ec795f1c62b9a4b3ba_l3.png "Rendered by QuickLaTeX.com")
Solution: It is given that $\begin{array}{l} 2\left[\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 1 & 2...
and , when ![2\left[\begin{array}{cc}x & 5 \\ 7 & y-3\end{array}\right]+\left[\begin{array}{cc}3 & -4 \\ 1 & 2\end{array}\right]=\left[\begin{array}{cc}7 & 6 \\ 15 & 14\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b289d871572f69a53ee3333c591b7fdd_l3.png "Rendered by QuickLaTeX.com")
Solution: (i) Given $2\left(\begin{array}{lc} x & 5 \\ 7 y-3 \end{array}\right)+\left(\begin{array}{c} 3-4 \\ 12 \end{array}\right)=\left(\begin{array}{cc} 7 & 6 \\ 1514 \end{array}\right)$...
and , when ![\left[\begin{array}{l}\mathrm{x}+\mathrm{y} \\ \mathrm{x}-\mathrm{y}\end{array}\right]=\left[\begin{array}{l}8 \\ 4\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4f327f4e30d328717ac87c5768d0bd84_l3.png "Rendered by QuickLaTeX.com")
![\left[\begin{array}{cc}2 x+5 & 7 \\ 0 & 3 y-7\end{array}\right]=\left[\begin{array}{cc}x-3 & 7 \\ 0 & -5\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9bc3080cd3ca5d8961ae81f2e6b07c22_l3.png "Rendered by QuickLaTeX.com")
Solution: (i) Given $\left(\begin{array}{l}\boldsymbol{x}+\boldsymbol{y} \\ \boldsymbol{x}-\boldsymbol{y}\end{array}\right)=\left(\begin{array}{l}8 \\ \mathbf{4}\end{array}\right)$. By equality of...
![A=\operatorname{diag}[2,-5,9], B=\operatorname{diag}[-3,7,14]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e915667c4dc298d0e9ed08f771e99b84_l3.png "Rendered by QuickLaTeX.com")
and ![C=\operatorname{diag}[4,-6,3]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-274f356305cd7bdb13df4255156b7ef6_l3.png "Rendered by QuickLaTeX.com")
, find: 
Solution: If $Z=\operatorname{diag}[a, b, c]$, then it can be written as $Z=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right]$ Therefore, $A+2...
such that 
where ![A=\left[\begin{array}{ll}3 & 1 \\ 0 & 2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-687ea49b624f69078701c50bdfe70143_l3.png "Rendered by QuickLaTeX.com")
and ![B=\left[\begin{array}{cc}-2 & 1 \\ 0 & 3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-327689b71c17725ba5fadd5f6c04b7ea_l3.png "Rendered by QuickLaTeX.com")
Solution: It is given that $2 A-B+X=0$ $\begin{array}{l} 2\left(\left[\begin{array}{ll} 3 & 1 \\ 0 & 2 \end{array}\right]\right)-\left[\begin{array}{cc} -2 & 1 \\ 0 & 3...
![A=\left[\begin{array}{cc}-2 & 3 \\ 4 & 5 \\ 1 & -6\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c1a126b89e46fb513ab316d9ddf765f2_l3.png "Rendered by QuickLaTeX.com")
and ![B=\left[\begin{array}{cc}5 & 2 \\ -7 & 3 \\ 6 & 4\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ea823875b60d35239b7a90f86c04d7b6_l3.png "Rendered by QuickLaTeX.com")
, find a matrix such that 
Solution: It is given that $A+B-C=0$ $\begin{array}{c} {\left[\begin{array}{cc} -2 & 3 \\ 4 & 5 \\ 1 & -6 \end{array}\right]+\left[\begin{array}{cc} 5 & 2 \\ -7 & 3 \\ 6 & 4...
, if ![\left[\begin{array}{rrr}3 & 5 & -9 \\ -1 & 4 & -7\end{array}\right]+X=\left[\begin{array}{lll}6 & 2 & 3 \\ 4 & 8 & 6\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-45e666fb26d68da37727a27b2abe54f0_l3.png "Rendered by QuickLaTeX.com")
Solution: It is given that $\left[\begin{array}{ccc}3 & 5 & -9 \\ -1 & 4 & -7\end{array}\right]+x=\left[\begin{array}{lll}6 & 2 & 3 \\ 4 & 8 & 6\end{array}\right]$...
and , if ![2 A-B=\left[\begin{array}{rrr}6 & -6 & 0 \\ -4 & 2 & 1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7c7e70a825c505f4ed47602a40feb33a_l3.png "Rendered by QuickLaTeX.com")
and ![2 B+A=\left[\begin{array}{ccc}3 & 2 & 5 \\ -2 & 1 & -7\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-97bab0d2cf485a48b9cadb26dc60dbb6_l3.png "Rendered by QuickLaTeX.com")
Solution: $\operatorname{Add} 2(2 A-B)$ and $(2 B+A)$ $\begin{array}{l} 2(2 A-B)+(2 B+A)=2\left(\left[\begin{array}{ccc} 6 & -6 & 0 \\ -4 & 2 & 1...
and , if ![A+B=\left[\begin{array}{ccc}1 & 0 & 2 \\ 5 & 4 & -6 \\ 7 & 3 & 8\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3a87601979fb533a239d62337a5b2064_l3.png "Rendered by QuickLaTeX.com")
and ![A-B=\left[\begin{array}{ccc}-5 & -4 & 8 \\ 11 & 2 & 0 \\ -1 & 7 & 4\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-58cb7bbac0f7ccbeaeb8025170c9380e_l3.png "Rendered by QuickLaTeX.com")
Solution: Add $(A+B)$ and $(A-B)$ We obtain $(A+B)+(A-B)=\left[\begin{array}{ccc}1 & 0 & 2 \\ 5 & 4 & -6 \\ 7 & 3 & 8\end{array}\right]+\left[\begin{array}{ccc}-5 & -4...
![5 \mathrm{~A}=\left[\begin{array}{ccc}5 & 10 & -15 \\ 2 & 3 & 4 \\ 1 & 0 & -5\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-68173ac372a0be08a045f2efa000c776_l3.png "Rendered by QuickLaTeX.com")
, find Solution: $5 A=\left[\begin{array}{ccc}5 & 10 & -15 \\ 2 & 3 & 4 \\ 1 & 0 & -5\end{array}\right]$ $A=\left[\begin{array}{ccc}\frac{5}{5} & \frac{10}{5} &...
![A=\left[\begin{array}{ccc}0 & 1 & -2 \\ 5 & -1 & -4\end{array}\right], B=\left[\begin{array}{ccc}1 & -3 & -1 \\ 0 & -2 & 5\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-41e2731669cc626117fcfaab4c782d70_l3.png "Rendered by QuickLaTeX.com")
and ![C=\left[\begin{array}{ccc}2 & -5 & 1 \\ -4 & 0 & 6\end{array}\right] .](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ec6db1baabfa8122c2209e13ff3fac41_l3.png "Rendered by QuickLaTeX.com")
Compute 
Solution: $\begin{array}{l} \left.5 A-3 B+4 C=5\left(\left[\begin{array}{ccc} 0 & 1 & -2 \\ 5 & -1 & -4 \end{array}\right]\right)-3\left(\begin{array}{ccc} 1 & -3 & -1 \\ 0...
![A=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], B=\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a7bc1c4d1eeefbbc4f83e854a1687521_l3.png "Rendered by QuickLaTeX.com")
and ![C=\left[\begin{array}{cc}-2 & 5 \\ 3 & 4\end{array}\right] .](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8bd7890a3a96a4c97189a908536ad3df_l3.png "Rendered by QuickLaTeX.com")
Find: 
Solution: $\begin{array}{l} \text { i. } A-2 B+3 C \\ \text { A- } 2 B+3 C=\left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right]-2\left(\left[\begin{array}{cc} 1 & 3 \\ -2 & 5...
and Find: 
Solution: i. $\begin{array}{l} A+2 B=\left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right]+2\left(\left[\begin{array}{cc} 1 & 3 \\ -2 & 5 \end{array}\right]\right) \\...
![A=\left[\begin{array}{ccc}3 & 1 & 2 \\ 1 & 2 & -3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-31a9f285753e96665b1043f971261649_l3.png "Rendered by QuickLaTeX.com")
and ![B=\left[\begin{array}{ccc}-2 & 0 & 4 \\ 5 & -3 & 2\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2db0d1cb1cd62becc5703078ca3469a4_l3.png "Rendered by QuickLaTeX.com")
, find 
Solution: $\begin{array}{l} 2 A=2\left(\left[\begin{array}{ccc} 3 & 1 & 2 \\ 1 & 2 & -3 \end{array}\right]\right) \\ =\left[\begin{array}{ccc} 6 & 2 & 4 \\ 2 & 4 & -6...
![A=\left[\begin{array}{cc}3 & 5 \\ -2 & 0 \\ 6 & -1\end{array}\right], B=\left[\begin{array}{cc}-1 & -3 \\ 4 & 2 \\ -2 & 3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-66f7d37c6136b09956a37ec4bd7b2766_l3.png "Rendered by QuickLaTeX.com")
and ![C=\left[\begin{array}{cc}0 & 2 \\ 3 & -4 \\ 1 & 6\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e8b8119fe41187456a97ee682923debd_l3.png "Rendered by QuickLaTeX.com")
, verify that 
Solution: $\begin{array}{l} (A+B)+C=\left(\left[\begin{array}{cc} 3 & 5 \\ -2 & 0 \\ 6 & -1 \end{array}\right]+\left[\begin{array}{cc} -1 & -3 \\ 4 & 2 \\ -2 & 3...
![A=\left[\begin{array}{ccc}2 & -3 & 5 \\ -1 & 0 & 3\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c18faf5289add2dd71e42331e8e0f3fb_l3.png "Rendered by QuickLaTeX.com")
and ![B=\left[\begin{array}{ccc}3 & 2 & -2 \\ 4 & -3 & 1\end{array}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ac37f82b6c044fe5ac67eaa50137a95c_l3.png "Rendered by QuickLaTeX.com")
, verify that 
Solution: $\begin{array}{l} A+B=\left[\begin{array}{ccc} 2 & -3 & 5 \\ -1 & 0 & 3 \end{array}\right]+\left[\begin{array}{ccc} 3 & 2 & -2 \\ 4 & -3 & 1...
is ![[-2,2]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a9b3b8b8ebc21c0ad1406661bbf93140_l3.png "Rendered by QuickLaTeX.com")
![(-\infty,-2]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c5486be18af34dbc37a82d81baec606e_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(D) is correct. $f(x)=x+\frac{1}{x}$ The range of the function can be given by putting values of $\mathrm{x}$ and find $\mathrm{y}$. $$\begin{tabular}{|l|l|} \hline $\mathrm{X}$ &...
. Then, range 
![(-\infty, 1]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2eb654b4016c2cc85fa136277a9d7f78_l3.png "Rendered by QuickLaTeX.com")
![[-1,1]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bd9ec9b7fdf72f6f954becf26cf7129a_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(B) is correct. $f(x)=\frac{1}{\left(1-x^{2}\right)}$ The range of $f(x)$ can be found out by putting $f(x)=y$ $\begin{array}{l} \mathrm{y}=\frac{1}{\left(1-x^{2}\right)} \\...
. Then, dom (f) and range (f) are respectively and 
and and Solution: Option(A) is correct. $f(x)=x^{3}$ $f(x)$ can assume any value, therefore domain of $f(x)$ is $R$ Range of the function can be positive or negative Real numbers, as the cube of any number...
. Then, 
![[1,2]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ab425e31cf7af5fb3d46646ff538fdcd_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(C) is correct. $f(x)=\sqrt{\log \left(2 x-x^{2}\right)}$ For $f(x)$ to be defined $2 x-x^2$ should be positive. Solving inequality, (Log taken to the opposite side of the equation...
. Then, 
? 
![\left[0, \frac{2}{3}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b52292076dc04c74fe44cded15544d8b_l3.png "Rendered by QuickLaTeX.com")
![\left[\frac{-2}{3}, \frac{2}{3}\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c6d1d6173855bcc8e2e22c311f052282_l3.png "Rendered by QuickLaTeX.com")
Solution: Option(B) is correct. $\begin{array}{l} \mathrm{f}(\mathrm{x})=\cos ^{-1}(3 \mathrm{x}-1) \end{array}$ Domain for function $\cos ^{-1} \mathrm{x}$ is $[-1,1]$ and range is $[0, \pi]$ When...