Exemplar Solution

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Here,A=\left[\begin{array}{cc}\hfill cos\alpha \hfill & \hfill sin\alpha \hfill \\ \hfill -sin\alpha \hfill & \hfill cos\alpha \hfill \end{array}\right]\\ Giventhat:A^{-1}=A^{\text{'}}\\ Pre-multiplyingbothsidesbyA\\ A{A}^{-1}=A{A}^{\text{'}}\\ \Rightarrow I=A{A}^{\text{'}}\left[?A{A}^{-1}=I\right]\\ \Rightarrow \left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill cos\alpha \hfill & \hfill sin\alpha \hfill \\ \hfill -sin\alpha \hfill & \hfill cos\alpha \hfill \end{array}\right]\left[\begin{array}{cc}\hfill cos\alpha \hfill & \hfill -sin\alpha \hfill \\ \hfill sin\alpha \hfill & \hfill cos\alpha \hfill \end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill co{s}^2\alpha +si{n}^2\alpha \hfill & \hfill -sin\alpha cos\alpha +sin\alpha cos\alpha \hfill \\ \hfill -sin\alpha cos\alpha +cos\alpha sin\alpha \hfill & \hfill si{n}^2\alpha +co{s}^2\alpha \hfill \end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\\ Hence,itistrueforallvaluesof\alpha .\end{array}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 1\hfill \end{array}\right]\\ A^2=A.A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1+8\hfill & \hfill 2+2\hfill \\ \hfill 4+4\hfill & \hfill 8+1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 9\hfill & \hfill 4\hfill \\ \hfill 8\hfill & \hfill 9\hfill \end{array}\right]\\ \therefore A^2+2A+7I=\left[\begin{array}{cc}\hfill 9\hfill & \hfill 4\hfill \\ \hfill 8\hfill & \hfill 9\hfill \end{array}\right]+2\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 1\hfill \end{array}\right]+7\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\\ =\left[\begin{array}{cc}\hfill 9\hfill & \hfill 4\hfill \\ \hfill 8\hfill & \hfill 9\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 2\hfill & \hfill 4\hfill \\ \hfill 8\hfill & \hfill 2\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 7\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 7\hfill \end{array}\right]\\ =\left[\begin{array}{cc}\hfill 9+2+7\hfill & \hfill 4+4+0\hfill \\ \hfill 8+8+0\hfill & \hfill 9+2+7\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 18\hfill & \hfill 8\hfill \\ \hfill 16\hfill & \hfill 18\hfill \end{array}\right]\\ Hence,A^2+2A+7I=\left[\begin{array}{cc}\hfill 18\hfill & \hfill 8\hfill \\ \hfill 16\hfill & \hfill 18\hfill \end{array}\right]\end{array}$

This is a short answer type question as classified in NCERT Exemplar

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:3\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill a\hfill & \hfill 6\hfill \\ \hfill -1\hfill & \hfill 2d\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 4\hfill & \hfill a+b\hfill \\ \hfill c+d\hfill & \hfill 3\hfill \end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}\hfill 3a\hfill & \hfill 3b\hfill \\ \hfill 3c\hfill & \hfill 3d\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill a+4\hfill & \hfill 6+a+b\hfill \\ \hfill -1+c+d\hfill & \hfill 2d+3\hfill \end{array}\right]\\ Equatingthecorrespondingelements,weget,\\ 3a=a+4\Rightarrow 3a-a=4\Rightarrow 2a=4\Rightarrow a=2\\ 3b=6+a+b\Rightarrow 3b-b-a=6\Rightarrow 2b-a=6\Rightarrow 2b-2=6\\ 2b=8\Rightarrow b=4\\ 3c=-1+c+d\Rightarrow 3c-c-d=-1\Rightarrow 2c-d=-1\\ 3d=2d+3\Rightarrow 3d-2d=3\Rightarrow d=3\\ Now,2c-d=-1\\ \Rightarrow 2c-3=-1\Rightarrow 2c=3-1\Rightarrow 2c=2\Rightarrow c=1\\ \therefore a=2,b=4,c=1,d=3.\end{array}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:A=\left[\begin{array}{cc}\hfill 3\hfill & \hfill -5\hfill \\ \hfill -4\hfill & \hfill 2\hfill \end{array}\right]\\ A^2=A.A=\left[\begin{array}{cc}\hfill 3\hfill & \hfill -5\hfill \\ \hfill -4\hfill & \hfill 2\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 3\hfill & \hfill -5\hfill \\ \hfill -4\hfill & \hfill 2\hfill \end{array}\right]\\ =\left[\begin{array}{cc}\hfill 9+20\hfill & \hfill -15-10\hfill \\ \hfill -12-8\hfill & \hfill 20+4\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 29\hfill & \hfill -25\hfill \\ \hfill -20\hfill & \hfill 24\hfill \end{array}\right]\\ \therefore A^2-5A-14I=\left[\begin{array}{cc}\hfill 29\hfill & \hfill -25\hfill \\ \hfill -20\hfill & \hfill 24\hfill \end{array}\right]-5\left[\begin{array}{cc}\hfill 3\hfill & \hfill -5\hfill \\ \hfill -4\hfill & \hfill 2\hfill \end{array}\right]-14\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\\ =\left[\begin{array}{cc}\hfill 29\hfill & \hfill -25\hfill \\ \hfill -20\hfill & \hfill 24\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 15\hfill & \hfill -25\hfill \\ \hfill -20\hfill & \hfill 10\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 14\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 14\hfill \end{array}\right]\\ =\left[\begin{array}{cc}\hfill 29\hfill & \hfill -25\hfill \\ \hfill -20\hfill & \hfill 24\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 29\hfill & \hfill -25\hfill \\ \hfill -20\hfill & \hfill 24\hfill \end{array}\right]\\ =\left[\begin{array}{cc}\hfill 29-29\hfill & \hfill -25+25\hfill \\ \hfill -20+20\hfill & \hfill 24-24\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\\ Hence,A^2-5A-14I=0\\ Now,multiplyingbothsidesbyA,weget,\\ A^2.A-5A.A-14IA=0A\\ \Rightarrow A^3-5A^2-14A=0\\ \Rightarrow A^3=5A^2+14A\\ \Rightarrow A^3=5\left[\begin{array}{cc}\hfill 29\hfill & \hfill -25\hfill \\ \hfill -20\hfill & \hfill 24\hfill \end{array}\right]+14\left[\begin{array}{cc}\hfill 3\hfill & \hfill -5\hfill \\ \hfill -4\hfill & \hfill 2\hfill \end{array}\right]\\ \Rightarrow =\left[\begin{array}{cc}\hfill 145\hfill & \hfill -125\hfill \\ \hfill -100\hfill & \hfill 120\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 42\hfill & \hfill -70\hfill \\ \hfill -56\hfill & \hfill 28\hfill \end{array}\right]\\ \Rightarrow =\left[\begin{array}{cc}\hfill 145+42\hfill & \hfill -125-70\hfill \\ \hfill -100-56\hfill & \hfill 120+28\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 187\hfill & \hfill -195\hfill \\ \hfill -156\hfill & \hfill 148\hfill \end{array}\right]\\ Hence,A^3=\left[\begin{array}{cc}\hfill 187\hfill & \hfill -195\hfill \\ \hfill -156\hfill & \hfill 148\hfill \end{array}\right]\end{array}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}OrderofmatricesAandBis2\times 2.\\ \therefore OrderofmatrixCmustbe2\times 2.\\ LetC=\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]\\ \therefore 3A+5B+2C=0\\ \Rightarrow 3\left[\begin{array}{cc}\hfill 1\hfill & \hfill 5\hfill \\ \hfill 7\hfill & \hfill 12\hfill \end{array}\right]+5\left[\begin{array}{cc}\hfill 9\hfill & \hfill 1\hfill \\ \hfill 7\hfill & \hfill 8\hfill \end{array}\right]+2\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}\hfill 3\hfill & \hfill 15\hfill \\ \hfill 21\hfill & \hfill 36\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 45\hfill & \hfill 5\hfill \\ \hfill 35\hfill & \hfill 40\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 2a\hfill & \hfill 2b\hfill \\ \hfill 2c\hfill & \hfill 2d\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}\hfill 3+45+2a\hfill & \hfill 15+5+2b\hfill \\ \hfill 21+35+2c\hfill & \hfill 36+40+2d\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}\hfill 48+2a\hfill & \hfill 20+2b\hfill \\ \hfill 56+2c\hfill & \hfill 76+2d\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\\ Equatingthecorrespondingelements,weget,\\ 48+2a=0\Rightarrow 2a=-48\Rightarrow a=-24\\ 20+2b=0\Rightarrow 2b=-20\Rightarrow b=-10\\ 56+2c=0\Rightarrow 2c=-56\Rightarrow c=-28\\ 76+2d=0\Rightarrow 2d=-76\Rightarrow d=-38\\ Hence,C=\left[\begin{array}{cc}\hfill -24\hfill & \hfill -10\hfill \\ \hfill -28\hfill & \hfill -38\hfill \end{array}\right]\end{array}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:\left[\begin{array}{cc}\hfill xy\hfill & \hfill 4\hfill \\ \hfill z+6\hfill & \hfill x+y\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 8\hfill & \hfill w\hfill \\ \hfill 0\hfill & \hfill 6\hfill \end{array}\right]\\ Equatingthecorrespondingelements,\\ xy=8,w=4,z+6=0\Rightarrow z=-6,x+y=6\\ Now,solvingx+y=6\dots \left(1\right)\\ andxy=8\dots \left(2\right)\\ Fromeqn.\left(1\right),y=6-x\dots \left(3\right)\\ Puttingthevalueofyineqn.\left(2\right)weget,\\ x\left(6-x\right)=8\Rightarrow 6x-x^2=8\\ \Rightarrow x^2-6x+8=0\Rightarrow x^2-4x-2x+8=0\\ \Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\Rightarrow \left(x-4\right)\left(x-2\right)=0\\ \therefore x=4,2\\ Fromeqn.\left(3\right),y=2,4.\\ Hence,x=4or2,y=2or4,z=-6andw=4\end{array}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}(i)LetA=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill -5\hfill & \hfill 7\hfill \end{array}\right]\\ \left|A\right|=1\times 7-\left(-5\right)\times 3=7+15=22\ne 0\\ So,Aisinvertible.\\ LetA=IA\\ \left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill -5\hfill & \hfill 7\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]A\\ R_2\to R_2+5R_1\\ \Rightarrow \left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 22\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 5\hfill & \hfill 1\hfill \end{array}\right]A\\ R_2\to \frac{1}{22}{R}_2\\ \Rightarrow \left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill \frac{5}{22}\hfill & \hfill \frac{1}{22}\hfill \end{array}\right]A\\ R_1\to R_1-3R_2\\ \Rightarrow \left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill \frac{7}{22}\hfill & \hfill \frac{-3}{22}\hfill \\ \hfill \frac{5}{22}\hfill & \hfill \frac{1}{22}\hfill \end{array}\right]A\\ So,A^{-1}=\left[\begin{array}{cc}\hfill \frac{7}{22}\hfill & \hfill \frac{-3}{22}\hfill \\ \hfill \frac{5}{22}\hfill & \hfill \frac{1}{22}\hfill \end{array}\right]\Rightarrow \frac{1}{22}\left[\begin{array}{cc}\hfill 7\hfill & \hfill -3\hfill \\ \hfill 5\hfill & \hfill 1\hfill \end{array}\right]\\ Hence,inverseof\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill -5\hfill & \hfill 7\hfill \end{array}\right]is\frac{1}{22}\left[\begin{array}{cc}\hfill 7\hfill & \hfill -3\hfill \\ \hfill 5\hfill & \hfill 1\hfill \end{array}\right]\\ (ii)LetA=\left[\begin{array}{cc}\hfill 1\hfill & \hfill -3\hfill \\ \hfill -2\hfill & \hfill 6\hfill \end{array}\right]\\ \left|A\right|=1\times 6-\left(-3\right)\times \left(-2\right)=6-6=0\Rightarrow \left|A\right|=0\\ So,Aisnotinvertible.\\ Hence,inverseof\left[\begin{array}{cc}\hfill 1\hfill & \hfill -3\hfill \\ \hfill -2\hfill & \hfill 6\hfill \end{array}\right]isnotpossible.\end{array}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Toprovethat{\left(A^{\text{'}}\right)}^n={\left(A^n\right)}^{\text{'}}\\ LetP\left(n\right):{\left(A^{\text{'}}\right)}^n={\left(A^n\right)}^{\text{'}}\\ Step1:Putn=1,P\left(1\right):A^{\text{'}}=A^{\text{'}}whichistrueforn=1\\ Step2:Putn=K,P\left(K\right):{\left(A^{\text{'}}\right)}^K={\left(A^K\right)}^{\text{'}}Letitbetrueforn=K\\ Step3:Putn=K+1,P\left(K+1\right):{\left(A^{\text{'}}\right)}^{K+1}={\left(A^{K+1}\right)}^{\text{'}}\\ L.H.S.{\left(A^{\text{'}}\right)}^{K+1}={\left(A^{\text{'}}\right)}^K.\left(A^{\text{'}}\right)\\ ={\left(A^K\right)}^{\text{'}}.\left(A^{\text{'}}\right)\left[Fromstep2\right]\\ ={\left(A^K.A\right)}^{\text{'}}={\left(A^{K+1}\right)}^{\text{'}}R.H.S.\\ ThegivenstatementistrueforP\left(K+1\right)wheneveritistrueforP\left(K\right),whereK\in N.\end{array}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}GiventhatA=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 4\hfill & \hfill -3\hfill & \hfill 4\hfill \\ \hfill 3\hfill & \hfill -3\hfill & \hfill 4\hfill \end{array}\right]\\ L.H.S.A^2=A.A=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 4\hfill & \hfill -3\hfill & \hfill 4\hfill \\ \hfill 3\hfill & \hfill -3\hfill & \hfill 4\hfill \end{array}\right].\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 4\hfill & \hfill -3\hfill & \hfill 4\hfill \\ \hfill 3\hfill & \hfill -3\hfill & \hfill 4\hfill \end{array}\right]\\ =\left[\begin{array}{ccc}\hfill 0+4-3\hfill & \hfill 0-3+3\hfill & \hfill 0+4-4\hfill \\ \hfill 0-12+12\hfill & \hfill 4+9-12\hfill & \hfill -4-12+16\hfill \\ \hfill 0-12+12\hfill & \hfill 3+9-12\hfill & \hfill -3-12+16\hfill \end{array}\right]\\ =\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=1R.H.S.\\ Hence,A^2=Iisverified.\end{array}$