Class 12th

Candidates seeking admission to the certificate programme must have completed Class 12. The college offers certificate programmes to students in both short- and long-term modes. Gamics Institute, Indore admissions are merit-based. Furthermore, the selection criteria for all courses are based on merit. The courses are available in various streams such as Animation, Design, Mass Communication, and Media.

Candidates seeking admission to the BSc programme at the UG level must be Class 12 passed from a recognised board/ council or equivalent. Apart from this, candidates seeking admission to diploma-level programmes must be Class 10 passed by a recognised board. The college offers various programmes to students at UG and Diploma levels in full-time mode. IIMS School of Hotel Management admissions are merit-based. Further, the college is affiliated with NCVTE and offers BSc Catering and Hotel Administration and Diploma courses. 

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

$\begin{array}{l}GiventhatBisaskewsymmetricmatrix\therefore B^{\text{'}}=-B\\ LetP=A^{\text{'}}BA\\ \Rightarrow P^{\text{'}}={\left(A^{\text{'}}BA\right)}^{\text{'}}\\ \Rightarrow =A^{\text{'}}{B}^{\text{'}}{\left(A^{\text{'}}\right)}^{\text{'}}\left[{\left(A^{\text{'}}B\right)}^{\text{'}}=B^{\text{'}}{A}^{\text{'}}\right]\\ \Rightarrow =A^{\text{'}}\left(-B\right)A\\ \Rightarrow =A^{\text{'}}BA=-P\\ So{P}^{\text{'}}=-P\\ Hence,A^{\text{'}}BAisaskewsymmetricmatrix.\end{array}$

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

$\begin{array}{l}Toshowthat:{\left(I+A\right)}^3=7A+I\\ L.H.S.{\left(I+A\right)}^3=I^3+A^3+3I^{2}A+3I{A}^2\\ \Rightarrow =I+A^2.A+3IA+3I{A}^2\\ \Rightarrow =I+A.A+3IA+3IA\left[?A^2=A\right]\\ \Rightarrow =I+A^2+3IA+3IA\\ \Rightarrow =I+A+3IA+3IA\left[?A^2=A\right]\\ \Rightarrow =I+A+3A+3A\Rightarrow 7A+IR.H.S.\\ L.H.S.=R.H.S.Hence,proved.\end{array}$

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

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

$\begin{array}{l}LetA=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill b\hfill & \hfill -1\hfill \\ \hfill c\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]A^{\text{'}}=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 2\hfill & \hfill c\hfill \\ \hfill a\hfill & \hfill b\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right]\\ Forskewsymmetricmatrix,A^{\text{'}}=-A\\ \left[\begin{array}{ccc}\hfill 0\hfill & \hfill 2\hfill & \hfill c\hfill \\ \hfill a\hfill & \hfill b\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right]=-\left[\begin{array}{ccc}\hfill 0\hfill & \hfill a\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill b\hfill & \hfill -1\hfill \\ \hfill c\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]\\ \Rightarrow \left[\begin{array}{ccc}\hfill 0\hfill & \hfill 2\hfill & \hfill c\hfill \\ \hfill a\hfill & \hfill b\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill -a\hfill & \hfill -3\hfill \\ \hfill -2\hfill & \hfill -b\hfill & \hfill 1\hfill \\ \hfill -c\hfill & \hfill -1\hfill & \hfill 0\hfill \end{array}\right]\\ Equatingthecorrespondingelements,weget\\ a=-2,b=-b\Rightarrow 2b=0\Rightarrow b=0andc=-3\\ Hence,a=-2,b=0andc=-3.\end{array}$

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