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CBSE Assignments class 09 Mathematics

  Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX 

NCERT Sol Maths Class XII Ch-3 | Matrices

Solution of Important Question of NCERT Book
Class XII Chapter 3 Matrices
Get Free NCERT Solutions for Class 12 Maths Chapter 3 Matrices. Solution of important questions of NCERT book solved by Expert Teachers as per NCERT (CBSE) Book guidelines.

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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