Solution of Important Question of NCERT Book
Class XII Chapter 3 Matrices
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Free NCERT Solutions for Class 12 Maths Chapter 3 Matrices. Solution of important questions of
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NCERT Exercise 3.2
Q.18: If I is an identity matrix of order 2 x 2 then show that
\[I+A=I-A\begin{bmatrix} cos\alpha &-sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\; \; where\]
\[A=\begin{bmatrix} 0 &-tan\frac{\alpha }{2} \\ tan\frac{\alpha }{2}& 0 \end{bmatrix}\]
Solution
\[Let\; \; tan\frac{\alpha }{2}=t\]\[Then\; \; A=\begin{bmatrix} 0 &-t \\ t&0 \end{bmatrix}\]
\[LHS= I+A=\begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}+\begin{bmatrix} 0 &-t \\ t&0 \end{bmatrix}=\begin{bmatrix} 1 &-t \\ t & 1 \end{bmatrix}\]
\[Now:\; \; cos\alpha =\frac{1-tan^{2}\frac{\alpha }{2}}{1+tan^{2}\frac{\alpha }{2}}=\frac{1-t^{2}}{1+t^{2}}\;\; and\]
\[sin\alpha =\frac{2tan^{2}\frac{\alpha }{2}}{1+tan^{2}\frac{\alpha }{2}}=\frac{2t^{2}}{1+t^{2}}\]
\[\begin{bmatrix} cos\alpha &-sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}=\frac{1}{1+t^{2}}\begin{bmatrix} 1-t^{2} & -2t\\ 2t& 1-t^{2} \end{bmatrix}\]
\[RHS=(I-A)\begin{bmatrix} cos\alpha &-sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\]
\[=\left (\begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix}-\begin{bmatrix} 0 &-t \\ t & 0 \end{bmatrix} \right )\frac{1}{1+t^{2}}\begin{bmatrix} 1-t^{2}&-2t \\ 2t & 1-t^{2} \end{bmatrix}\]
\[=\frac{1}{1+t^{2}}\begin{bmatrix} 1&t \\ -t & 1 \end{bmatrix}\begin{bmatrix} 1-t^{2} & -2t\\ 2t& 1-t^{2} \end{bmatrix}\]
\[=\frac{1}{1+t^{2}}\begin{bmatrix} 1-t^{2}+2t^{2}&-2t+t-t^{3} \\ -t+t^{3}+2t & 2t^{2}+1-t^{2} \end{bmatrix}\]
\[=\frac{1}{1+t^{2}}\begin{bmatrix} 1+t^{2}&-t-t^{3} \\ t+t^{3} & t^{2}+1 \end{bmatrix}\]
\[=\frac{1}{1+t^{2}}\begin{bmatrix} 1+t^{2}&-t(1+t^{2}) \\ t(1+t^{2}) & (1+t^{2}) \end{bmatrix}\]
\[=\begin{bmatrix} 1 &-t \\ t& 1 \end{bmatrix}=LHS\]
NCERT Exercise 3.4
Q 15. Find inverse of A by using elementary operations where \[A=\begin{bmatrix} 2 &-3 &3 \\ 2& 2 &3 \\ 3 &-2 & 2 \end{bmatrix}\]
Solution: We solve this question by using elementary row operations
For applying elementary row operation we have A = IA \[\begin{bmatrix} 2 &-3 &3 \\ 2& 2 &3 \\ 3 &-2 & 2 \end{bmatrix}=\begin{bmatrix} 1 &0 &0 \\ 0& 1 & 0\\ 0& 0 & 1 \end{bmatrix}A\]
\[Operating:\; \; R_{1}\rightarrow 2R_{1}-R_{3}\]
\[\begin{bmatrix} 1 &-4 &4 \\ 2& 2 &3 \\ 3 &-2 & 2 \end{bmatrix}=\begin{bmatrix} 2 &0 &-1 \\ 0& 1 & 0\\ 0& 0 & 1 \end{bmatrix}A\]
\[Operating:\; \; R_{2}\rightarrow R_{2}-2R_{1}\; \\ Operating:\; \; R_{3}\rightarrow R_{3}-3R_{1}\]
\[\begin{bmatrix} 1 &-4 &4 \\ 0& 10 &-5 \\ 0 &10 & -10 \end{bmatrix}=\begin{bmatrix} 2 &0 &-1 \\ -4& 1 & 2\\ -6& 0 & 4 \end{bmatrix}A\]
\[Operating:\; \; R_{3}\rightarrow R_{3}-R_{2}\]
\[\begin{bmatrix} 1 &-4 &4 \\ 0& 10 &-5 \\ 0 &0 & -5 \end{bmatrix}=\begin{bmatrix} 2 &0 &-1 \\ -4& 1 & 2\\ -2& -1 & 2 \end{bmatrix}A\]
\[Operating:\; \; R_{2}\rightarrow R_{2}-R_{3}\]
\[\begin{bmatrix} 1 &-4 &4 \\ 0& 10 &0 \\ 0 &0 & -5 \end{bmatrix}=\begin{bmatrix} 2 &0 &-1 \\ -2& 2 & 0\\ -2& -1 & 2 \end{bmatrix}A\]
\[Operating:\; \; R_{2}\rightarrow \frac{1}{10}R_{2}\\Operating:\; \; R_{3}\rightarrow \frac{-1}{5}R_{3}\]
\[\begin{bmatrix} 1 &-4 &4 \\ 0& 1 &0 \\ 0 &0 & 1 \end{bmatrix}=\begin{bmatrix} 2 &0 &-1 \\ -1/5& 1/5 & 0\\ 2/5& 1/5 & -2/5 \end{bmatrix}A\]
\[Operating:\; \; R_{1}\rightarrow R_{1}+4R_{2}\]
\[\begin{bmatrix} 1 &0 &4 \\ 0& 1 &0 \\ 0 &0 & 1 \end{bmatrix}=\begin{bmatrix} 6/5 &4/5 &-1 \\ -1/5& 1/5 & 0\\ 2/5& 1/5 & -2/5 \end{bmatrix}A\]
\[Operating:\; \; R_{1}\rightarrow R_{1}-4R_{3}\]
\[\begin{bmatrix} 1 &0 &0 \\ 0& 1 &0 \\ 0 &0 & 1 \end{bmatrix}=\begin{bmatrix} -2/5 &0 &3 /5\\ -1/5& 1/5 & 0\\ 2/5& 1/5 & -2/5 \end{bmatrix}A\]
\[A^{-1}I=A^{-1}=\begin{bmatrix} -2/5 &0 &3 /5\\ -1/5& 1/5 & 0\\ 2/5& 1/5 & -2/5 \end{bmatrix}\]
Q 16. Find inverse of A by using elementary operations where \[A=\begin{bmatrix} 1 &3 &-2 \\ -3& 0 &-5 \\ 2 &5 & 0 \end{bmatrix}\]
Solution: We solve this question by using elementary row operations
For applying elementary row operation we have A = IA \[\begin{bmatrix} 1 &3 &-2 \\ -3& 0 &-5 \\ 2 &5 & 0 \end{bmatrix}=\begin{bmatrix} 1 &0 &0 \\ 0& 1 & 0\\ 0& 0 & 1 \end{bmatrix}A\]
\[Operating:\; \; R_{2}\rightarrow R_{2}+3R_{1}\\Operating:\; \; R_{3}\rightarrow R_{3}-2R_{1}\]
\[\begin{bmatrix} 1 &3 &-2 \\ 0& 9 &-11 \\ 0 &-1 & 4 \end{bmatrix}=\begin{bmatrix} 1 &0 &0 \\ 3& 1 & 0\\ -2& 0 & 1 \end{bmatrix}A\]
\[Operating:\; \; R_{1}\rightarrow R_{1}+3R_{3}\\Operating:\; \; R_{2}\rightarrow R_{2}+8R_{3}\]
\[\begin{bmatrix} 1 &0 &10 \\ 0& 1 &21 \\ 0 &-1 & 4 \end{bmatrix}=\begin{bmatrix} -5 &0 &3 \\ -13& 1 & 8\\ -2& 0 & 1 \end{bmatrix}A\]
\[Operating:\; \; R_{3}\rightarrow R_{3}+R_{2}\]
\[\begin{bmatrix} 1 &0 &10 \\ 0& 1 &21 \\ 0 &0 & 25 \end{bmatrix}=\begin{bmatrix} -5 &0 &3 \\ -13& 1 & 8\\ -15& 1 & 9 \end{bmatrix}A\]
\[Operating:\; \; R_{3}\rightarrow \frac{1}{25}R_{3}\]
\[\begin{bmatrix} 1 &0 &10 \\ 0& 1 &21 \\ 0 &0 & 1 \end{bmatrix}=\begin{bmatrix} -5 &0 &3 \\ -13& 1 & 8\\ -3/5& 1/25 & 9/25 \end{bmatrix}A\]
\[Operating:\; \; R_{1}\rightarrow R_{1}-10R_{3}\\Operating:\; \; R_{2}\rightarrow R_{2}-21R_{3}\]
\[\begin{bmatrix} 1 &0 &0 \\ 0& 1 &0 \\ 0 &0 & 1 \end{bmatrix}=\begin{bmatrix} 1 &-2/5 &-3/5 \\ -2/5& 4/25 & 11/25\\ -3/5& 1/25 & 9/25 \end{bmatrix}A\]
\[A^{-1}I=A^{-1}=\begin{bmatrix} 1 &-2/5 &-3/5 \\ -2/5& 4/25 & 11/25\\ -3/5& 1/25 & 9/25 \end{bmatrix}\]
Q 17. Find inverse of matrix A by using elementary operations where \[A=\begin{bmatrix} 2 &0 &-1 \\ 5 & 1 & 0\\ 0 & 1 & 3 \end{bmatrix}\]
Solution: We solve this question by using elementary row operations
For applying elementary row operation we have A = IA \[\begin{bmatrix} 2 &0 &-1 \\ 5 & 1 & 0\\ 0 & 1 & 3 \end{bmatrix}=\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{bmatrix}A\]
\[R_{1}\rightarrow 3R_{1}-R_{2}\]
\[\begin{bmatrix} 1 &-1 &-3 \\ 5 & 1 & 0\\ 0 & 1 & 3 \end{bmatrix}=\begin{bmatrix} 3 &-1 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{bmatrix}A\]
\[R_{2}\rightarrow R_{2}-5R_{1}\]
\[\begin{bmatrix} 1 &-1 &-3 \\ 0 & 6 & 15\\ 0 & 1 & 3 \end{bmatrix}=\begin{bmatrix} 3 &-1 &0 \\ -15 &6 &0 \\ 0 &0 &1 \end{bmatrix}A\]
\[R_{1}\rightarrow R_{1}+R_{3}\]
\[\begin{bmatrix} 1 &0 &0 \\ 0 & 6 & 15\\ 0 & 1 & 3 \end{bmatrix}=\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &0 \\ 0 &0 &1 \end{bmatrix}A\]
\[R_{2}\rightarrow R_{2}-5R_{3}\]
\[\begin{bmatrix} 1 & 0 &0 \\ 0 &1 &0 \\ 0& 1 &3 \end{bmatrix}=\begin{bmatrix} 3 &-1 &1 \\ -15&6 &-5 \\ 0 & 0 & 1 \end{bmatrix}\]
\[R_{3}\rightarrow R_{3}-R_{2}\]
\[\begin{bmatrix} 1 &0 &0 \\ 0 & 1 & 0\\ 0 & 1 & 3 \end{bmatrix}=\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 15 &-6 &6 \end{bmatrix}A\]
\[R_{3}\rightarrow \frac{1}{3}R_{3}\]
\[\begin{bmatrix} 1 &0 &0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}A\]\[A^{-1}I = A^{-1}=\begin{bmatrix} 3 &-1 &1 \\ -15 &6 &-5 \\ 5 &-2 &2 \end{bmatrix}A\]
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