Exemplar Solution

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

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

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

$\begin{array}{l}LetA={\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]}_{2\times 2}\\ {\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 2\hfill \end{array}\right]}_{2\times 2}{\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]}_{2\times 2}{\left[\begin{array}{cc}\hfill -3\hfill & \hfill 2\hfill \\ \hfill 5\hfill & \hfill -3\hfill \end{array}\right]}_{2\times 2}={\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]}_{2\times 2}\\ \Rightarrow {\left[\begin{array}{cc}\hfill 2a+c\hfill & \hfill 2b+d\hfill \\ \hfill 3a+2c\hfill & \hfill 3b+2d\hfill \end{array}\right]}_{2\times 2}{\left[\begin{array}{cc}\hfill -3\hfill & \hfill 2\hfill \\ \hfill 5\hfill & \hfill -3\hfill \end{array}\right]}_{2\times 2}={\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]}_{2\times 2}\\ \Rightarrow \left[\begin{array}{cc}\hfill -6a-3c+10b+5d\hfill & \hfill 4a+2c-6b-3d\hfill \\ \hfill -9a-6c+15b+10d\hfill & \hfill 6a+4c-9b-6d\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\\ Equatingthecorrespondingelements,weget,\\ -6a-3c+10b+5d=1\dots \left(1\right)\\ -9a-6c+15b+10d=0\dots \left(2\right)\\ 4a+2c-6b-3d=0\dots \left(3\right)\\ 6a+4c-9b-6d=1\dots \left(4\right)\\ Multiplyingeq.\left(1\right)by2andsubtractingeq.\left(2\right),weget,\\ -12a-6c+20b+10d=2\\ -9a-6c+15b+10d=0\\ \text{\&t}\end{array}$

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

$\begin{array}{l}GiventhatA=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 3\hfill \\ \hfill -1\hfill & \hfill -2\hfill \end{array}\right]\\ A^2=A.A\\ =\left[\begin{array}{cc}\hfill 5\hfill & \hfill 3\hfill \\ \hfill -1\hfill & \hfill -2\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 5\hfill & \hfill 3\hfill \\ \hfill -1\hfill & \hfill -2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 25-3\hfill & \hfill 15-6\hfill \\ \hfill -5+2\hfill & \hfill -3+4\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 22\hfill & \hfill 9\hfill \\ \hfill -3\hfill & \hfill 1\hfill \end{array}\right]\\ A^2-3A-7I=0\\ L.H.S.\left[\begin{array}{cc}\hfill 22\hfill & \hfill 9\hfill \\ \hfill -3\hfill & \hfill 1\hfill \end{array}\right]-3\left[\begin{array}{cc}\hfill 5\hfill & \hfill 3\hfill \\ \hfill -1\hfill & \hfill -2\hfill \end{array}\right]-7\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}\hfill 22\hfill & \hfill 9\hfill \\ \hfill -3\hfill & \hfill 1\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 15\hfill & \hfill 9\hfill \\ \hfill -3\hfill & \hfill -6\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 7\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 7\hfill \end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}\hfill 22-15-7\hfill & \hfill 9-9-0\hfill \\ \hfill -3+3-0\hfill & \hfill 1+6-7\hfill \end{array}\right]\Rightarrow \left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]R.H.S.\\ Wearegiven{A}^2-3A-7I=0\\ \Rightarrow A^{-1}\left[A^2-3A-7I\right]=A^{-1}0\left[Pre-multiplyingbothsidesby{A}^{-1}\right]\\ \Rightarrow A^{-1}A.A-3A^{-1}.A-7A^{-1}I=0\left[A^{-1}0=0\right]\\ \Rightarrow I.A-3I-7A^{-1}I=0\\ \Rightarrow A-3I-7A^{-1}=0\\ \Rightarrow -7A^{-1}=3I-A\\ \Rightarrow A^{-1}=\frac{1}{-7}\left[3I-A\right]\\ \Rightarrow A^{-1}=\frac{1}{-7}\left[3\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)-\left[\begin{array}{cc}\hfill 5\hfill & \hfill 3\hfill \\ \hfill -1\hfill & \hfill -2\hfill \end{array}\right]\right]\\ =\frac{1}{-7}\left[\left(\begin{array}{cc}\hfill 3\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 3\hfill \end{array}\right)-\left[\begin{array}{cc}\hfill 5\hfill & \hfill 3\hfill \\ \hfill -1\hfill & \hfill -2\hfill \end{array}\right]\right]\\ =\frac{1}{-7}\left[\begin{array}{cc}\hfill 3-5\hfill & \hfill 0-3\hfill \\ \hfill 0+1\hfill & \hfill 3+2\hfill \end{array}\right]=\frac{1}{-7}\left[\begin{array}{cc}\hfill -2\hfill & \hfill -3\hfill \\ \hfill 1\hfill & \hfill 5\hfill \end{array}\right]\\ Hence,A^{-1}=\frac{1}{-7}\left[\begin{array}{cc}\hfill -2\hfill & \hfill -3\hfill \\ \hfill 1\hfill & \hfill 5\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}GiventhatA=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]andB=\left[\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]\\ A+B=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]\\ \Rightarrow A+B=\left[\begin{array}{cc}\hfill 0+0\hfill & \hfill 1-1\hfill \\ \hfill 1+1\hfill & \hfill 1+0\hfill \end{array}\right]\Rightarrow A+B=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right]\\ A-B=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]\\ \Rightarrow A-B=\left[\begin{array}{cc}\hfill 0-0\hfill & \hfill 1+1\hfill \\ \hfill 1-1\hfill & \hfill 1-0\hfill \end{array}\right]\Rightarrow A-B=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\\ \therefore \left(A+B\right).\left(A-B\right)=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0+0\hfill & \hfill 0+0\hfill \\ \hfill 0+0\hfill & \hfill 4+1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 5\hfill \end{array}\right]\\ Now,R.H.S.=A^2-B^2\\ =A.A-B.B\\ =\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 0\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right]\\ =\left[\begin{array}{cc}\hfill 0+1\hfill & \hfill 0+1\hfill \\ \hfill 0+1\hfill & \hfill 1+1\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 0-1\hfill & \hfill 0+0\hfill \\ \hfill 0+0\hfill & \hfill -1+0\hfill \end{array}\right]\\ =\left[\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right]-\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+1\hfill & \hfill 1-0\hfill \\ \hfill 1-0\hfill & \hfill 2+1\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 3\hfill \end{array}\right]\\ Hence,\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 5\hfill \end{array}\right]\ne \left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 3\hfill \end{array}\right]\\ Hence,\left(A+B\right).\left(A-B\right)\ne A^2-B^2\end{array}$

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

$\begin{array}{l}Thegivenequationcanbewrittenas\\ \left[\begin{array}{cc}\hfill 2x^2\hfill & \hfill 2x\hfill \\ \hfill 3x\hfill & \hfill x^2\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 16\hfill & \hfill 10x\hfill \\ \hfill 8\hfill & \hfill 8x\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2x^2+16\hfill & \hfill 48\hfill \\ \hfill 20\hfill & \hfill 12x\hfill \end{array}\right]\\ \Rightarrow \left[\begin{array}{cc}\hfill 2x^2+16\hfill & \hfill a_{12}\hfill \\ \hfill 3x+8\hfill & \hfill x^2+8x\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2x^2+16\hfill & \hfill 48\hfill \\ \hfill 20\hfill & \hfill 12x\hfill \end{array}\right]\\ Equatingthecorrespondingelements,weget\\ 12x=48,3x+8=20,x^2+8x=12x\\ \therefore x=\frac{48}{12}=4,3x=20-8=12,\Rightarrow x^2=12x-8x=4x\\ \therefore x=4,\Rightarrow x^2-4x=0\\ x=0,x=4\\ Hence,thenon-zerovaluesofxis4.\end{array}$

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

$\begin{array}{l}GiventhatX=\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1-1\hfill \\ \hfill 5\hfill & \hfill -2-3\hfill \end{array}\right]andY=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1-1\hfill \\ \hfill 7\hfill & \hfill 24\hfill \end{array}\right]\\ (i)X+Y=\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1-1\hfill \\ \hfill 5\hfill & \hfill -2-3\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1-1\hfill \\ \hfill 7\hfill & \hfill 24\hfill \end{array}\right]\\ =\left[\begin{array}{cc}\hfill 3+2\hfill & \hfill 1+1-1-1\hfill \\ \hfill 5+7\hfill & \hfill -2+2-3+4\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 2-2\hfill \\ \hfill 12\hfill & \hfill 01\hfill \end{array}\right]\\ (ii)2X-3Y=2\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1-1\hfill \\ \hfill 5\hfill & \hfill -2-3\hfill \end{array}\right]-3\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1-1\hfill \\ \hfill 7\hfill & \hfill 24\hfill \end{array}\right]\\ =\left[\begin{array}{cc}\hfill 2\times 3\hfill & \hfill 2\times 1-2\times 1\hfill \\ \hfill 2\times 5\hfill & \hfill -2\times 2-2\times 3\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 3\times 2\hfill & \hfill 1\times 3-1\times 3\hfill \\ \hfill 3\times 7\hfill & \hfill 3\times 23\times 4\hfill \end{array}\right]\\ =\left[\begin{array}{cc}\hfill 6\hfill & \hfill 2-2\hfill \\ \hfill 10\hfill & \hfill -4-6\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 6\hfill & \hfill 3-3\hfill \\ \hfill 21\hfill & \hfill 612\hfill \end{array}\right]\\ =\left[\begin{array}{cc}\hfill 6-6\hfill & \hfill 2-3-2+3\hfill \\ \hfill 10-21\hfill & \hfill -4-6-6-12\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0\hfill & \hfill -11\hfill \\ \hfill -11\hfill & \hfill -10-18\hfill \end{array}\right]\\ (iii)X+Y+Z=0\\ \Rightarrow \left[\begin{array}{cc}\hfill 3\hfill & \hfill 1-1\hfill \\ \hfill 5\hfill & \hfill -2-3\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1-1\hfill \\ \hfill 7\hfill & \hfill 24\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill a\hfill & \hfill bc\hfill \\ \hfill d\hfill & \hfill ef\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 00\hfill \\ \hfill 0\hfill & \hfill 00\hfill \end{array}\right]\\ whereZ=\left[\begin{array}{cc}\hfill a\hfill & \hfill bc\hfill \\ \hfill d\hfill & \hfill ef\hfill \end{array}\right]\\ \Rightarrow =\left[\begin{array}{cc}\hfill 3+2+a\hfill & \hfill 1+1+b-1-1+c\hfill \\ \hfill 5+7+d\hfill & \hfill -2+2+e-3+4+f\hfill \end{array}\right]=[\begin{array}{cc}\hfill 0\hfill & \hfill 00\hfill \\ \hfill 0\hfill & \hfill 0\text{\&}\hfill \end{array}\end{array}$

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

$\begin{array}{l}TheorderofmatrixA=2\times 2andtheorderofmatrixB=2\times 3.Additionofmatricesisonly\\ possiblewhentheyhavesameorder.So,A+Bisnotpossible.\end{array}$

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

$\begin{array}{l}GiventhatA=B\\ \Rightarrow \left[\begin{array}{cc}\hfill a+4\hfill & \hfill 3b\hfill \\ \hfill 8\hfill & \hfill -6\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2a+2\hfill & \hfill b^2+2\hfill \\ \hfill 8\hfill & \hfill b^2-5b\hfill \end{array}\right]\\ Equatingthecorrespondingelements,weget\\ a+4=2a+2,3b=b^2+2and{b}^2-5b=-6\\ \Rightarrow 2a-a=2,b^2-3b+2=0,b^2-5b+6=0\\ \therefore a=2\\ \therefore b^2-3b+2=0\therefore b^2-5b+6=0\\ \Rightarrow b^2-2b-b+2=0,\Rightarrow b^2-3b-2b+6=0\\ \Rightarrow b\left(b-2\right)-1\left(b-2\right)=0,\Rightarrow b\left(b-3\right)-2\left(b-3\right)=0\\ \Rightarrow \left(b-1\right)\left(b-2\right)=0,\Rightarrow \left(b-2\right)\left(b-3\right)=0\\ \therefore b=1,2\therefore b=2,3\\ buthere2iscommon.\\ Hence,thevalueofa=2andb=2.\end{array}$