Class 12 NCERT Solutions- Mathematics Part I – Chapter 3 Matrices – Exercise 3.4 | Set 1
Using elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 6.
Question 1.![\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-46bbe8806903fee589d9a9aed174228b_l3.png "Rendered by QuickLaTeX.com")
Solution:
![\begin{aligned} &\text { Since we know that, } A= IA \Rightarrow\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A }\\ &\Rightarrow\left[\begin{array}{cc} 1 & -1 \\ 0 & 5 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ -2 & 1 \end{array}\right] \mathrm{A}\left[\mathrm{R}{2} \rightarrow \mathrm{R}{2}-2 \mathrm{R}_{1}\right]\\ &\Rightarrow\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ -2 / 5 & 1 / 5 \end{array}\right] \mathrm{A}\left[\mathrm{R}{2} \rightarrow \frac{1}{5} \mathrm{R}{2}\right]\\ &\left.\Rightarrow\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 3 / 5 & 1 / 5 \\ -2 / 5 & 1 / 5 \end{array}\right] \text { A }\left[R_{1} \rightarrow R_{1}+R_{2}\right.\\ \end{aligned}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-4698e322f86de68c2026ad0e92849a06_l3.png "Rendered by QuickLaTeX.com")
![Therefore,A^{-1}=\left[\begin{array}{cc} 3 / 5 & 1 / 5 \\ -2 / 5 & 1 / 5 \end{array}\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-00b5773f1bce928210152afc0e9d0bc6_l3.png "Rendered by QuickLaTeX.com")
Question 2.![\left[\begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array}\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-255215a0a33ee2718b33a6cd25592fe1_l3.png "Rendered by QuickLaTeX.com")
Solution:
Let A=
[Tex]\begin{array}{l} \left(R_{1} \rightarrow R_{1}-R_{2}\right) \\ \left(R_{2} \rightarrow R_{2}-R_{1}\right) \end{array}[/Tex]
Question 3.![\left[\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-7a77f9614e75582ca00758f5f23835c2_l3.png "Rendered by QuickLaTeX.com")
Solution:
![\begin{aligned} &\text { Let } A=\left[\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right]\\ &\text { W.K.T., } A=IA \Rightarrow\left[\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A }\\ &\Rightarrow\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{rc} 1 & 0 \\ -2 & 1 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}-2 R_{1}\right]\\ &\left.\left.\Rightarrow \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\begin{array}{cc} - & -3 \\ -2 & 1 \end{array}\right] \text { A }\left[R_{1} \rightarrow R_{1}-3 R_{2}\right]\\ &Therefore,A^{-1}⇒\left[\begin{array}{cc} 7 & -3 \\ -2 & 1 \end{array}\right] \end{aligned}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-916dec9bad71613d88d9ede48a7339c2_l3.png "Rendered by QuickLaTeX.com")
Question 4.![\left[\begin{array}{ll} 2 & 3 \\ 5 & 7 \end{array}\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-3c0cd97db5de6e748a67bb57ff6aa099_l3.png "Rendered by QuickLaTeX.com")
Solution:
![\begin{aligned} &\text { Let } A=\left[\begin{array}{ll} 2 & 3 \\ 5 & 7 \end{array}\right]\\ &\text { W.K.T. , } A=I A \Rightarrow\left[\begin{array}{ll} 2 & 3 \\ 5 & 7 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A }\\ &\Rightarrow\left[\begin{array}{ll} 2 & 3 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{rc} 1 & 0 \\ -2 & 1 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}-2 R_{1}\right]\\ &⇒\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]=\left[\begin{array}{cc} -2 & -1 \\ 1 & 0 \end{array}\right] A\left[R_{1} \leftrightarrow R_{i}\right] \end{aligned}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-d25bab57cb9f313acd5d991fa14799d2_l3.png "Rendered by QuickLaTeX.com")
![\begin{array}{l} \Rightarrow\left[\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} -2 & 1 \\ 5 & -2 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}-2 R_{1}\right] \\ \Rightarrow\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} -7 & 3 \\ 5 & -2 \end{array}\right] \mathrm{A}\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}-\mathrm{R}_{2}\right] \\ Therefore, \mathrm{A}^{-1}=\left[\begin{array}{cc} -7 & 3 \\ 5 & -2 \end{array}\right] \end{array}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-dd0d72d76eca3d12b111cbed89b004e9_l3.png "Rendered by QuickLaTeX.com")
Question 5.![\left[\begin{array}{ll} 2 & 1 \\ -7 & 4 \end{array}\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-ac30c6f535a0f4ae5bbf475bdff7d19a_l3.png "Rendered by QuickLaTeX.com")
Solution:
![\begin{aligned} &\text { Let } A=\left[\begin{array}{ll} 2 & 1 \\ 7 & 4 \end{array}\right]\\ &\text { W.K.T. , } A=IA \Rightarrow\left[\begin{array}{ll} 2 & 1 \\ 7 & 4 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A }\\ &\Rightarrow\left[\begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ -3 & 1 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}-3 R_{1}\right]\\ &⇒\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{cc} 4 & -1 \\ -3 & 1 \end{array}\right] A\left[R_{1} \rightarrow R_{1}-R_{2}\right]\\ &⇒\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 4 & -1 \\ -7 & 2 \end{array}\right] A \quad\left[R, \rightarrow R,-R_{1}\right]\\ &Therefore,A=\left[\begin{array}{cc} 4 & -1 \\ -7 & 2 \end{array}\right] \end{aligned}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-3a1d7385613420b8c58376c8cf517501_l3.png "Rendered by QuickLaTeX.com")
Question 6.![\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-4db40b8594963786ce1a662c1129da66_l3.png "Rendered by QuickLaTeX.com")
Solution:
![\begin{aligned} &\text { Let } A=\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]\\ &\text { W.K.T. , } A=IA \Rightarrow\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A }\\ &⇒\left[\begin{array}{ll} 1 & 3 \\ 2 & 5 \end{array}\right]=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] A \quad\left[\mathrm{R}{1} \leftrightarrow \mathrm{R}{2}\right]\\ &\Rightarrow\left[\begin{array}{cc} 1 & 3 \\ 0 & -1 \end{array}\right]=\left[\begin{array}{cc} 0 & 1 \\ 1 & -2 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}-2 R_{1}\right] \end{aligned}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-c0e2dde793fab6c5441bd3c8cd7d204b_l3.png "Rendered by QuickLaTeX.com")
![\begin{array}{l} \Rightarrow\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 0 & 1 \\ -1 & 2 \end{array}\right] \text { A }\left[\mathrm{R}{2} \rightarrow(-1) \mathrm{R}{2}\right] \\ \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 3 & -5 \\ -1 & 2 \end{array}\right] \mathrm{A} \quad\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}-3 \mathrm{R}_{2}\right] \\ Therefore,\mathrm{A}^{-1}=\left[\begin{array}{rr} 3 & -5 \\ -1 & 2 \end{array}\right] \end{array}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-2014f081340f782fb28590601ede4e61_l3.png "Rendered by QuickLaTeX.com")
Using elementary transformation, find the inverse of each of the matrices, if it exists in Exercises 7 to 14.
Question 7.![\left[\begin{array}{ll} 3 & 1 \\ 5 & 2 \end{array}\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-e6e5e02bb3979617a52bfd96532ab0d5_l3.png "Rendered by QuickLaTeX.com")
Solution:
![\begin{array}{l} \text { Ans. Let } A=\left[\begin{array}{ll} 3 & 1 \\ 5 & 2 \end{array}\right] \\ \text { W.K.T. , } A=IA \Rightarrow\left[\begin{array}{ll} 3 & 1 \\ 5 & 2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] A \\ \Rightarrow\left[\begin{array}{ll} 6 & 2 \\ 5 & 2 \end{array}\right]=\left[\begin{array}{ll} 2 & 0 \\ 0 & 1 \end{array}\right] \text { A }\left[R_{1} \rightarrow 2 R_{1}\right] \\ \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 5 & 2 \end{array}\right]=\left[\begin{array}{cc} 2 & -1 \\ 0 & 1 \end{array}\right] \text { A }\left[R_{1} \rightarrow R_{1}-R_{2}\right] \end{array}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-a5afbee4e5747024be35f604224f230d_l3.png "Rendered by QuickLaTeX.com")
![\begin{array}{l} \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right]=\left[\begin{array}{cc} 2 & -1 \\ -10 & 6 \end{array}\right] \mathrm{A}\left[R_{2} \rightarrow R_{2}-5 R_{1}\right] \\ \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 2 & -1 \\ -5 & 3 \end{array}\right] \mathrm{A} \quad\left[\mathrm{R}{2} \rightarrow \frac{1}{2} \mathrm{R}{2}\right] \\ Therefore, \quad \mathrm{A}^{-1}=\left[\begin{array}{rr} 2 & -1 \\ -5 & 3 \end{array}\right] \end{array}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-65f121f3e46886fe15b38656c1f6596f_l3.png "Rendered by QuickLaTeX.com")
Question 8.
Solution:
![\begin{aligned} &\text { Ans. Let } A=\left[\begin{array}{ll} 4 & 5 \\ 3 & 4 \end{array}\right]\\ &\text { W.K.T. , } A=IA \Rightarrow\left[\begin{array}{ll} 4 & 5 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A }\\ &\Rightarrow\left[\begin{array}{ll} 1 & 1 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right] \text { A } \quad\left[R_{1} \rightarrow R_{1}-R_{2}\right]\\ &\Rightarrow\left[\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{lr} 1 & -1 \\ -3 & 4 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}-3 R_{1}\right]\\ &\Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 4 & -5 \\ -3 & 4 \end{array}\right] \text { A }\left[R_{1} \rightarrow R_{1}-R_{2}\right]\\ &Therefore,A^{-1}=\begin{bmatrix}4 & -5 \\-3 & 4 \\\end{bmatrix} \end{aligned}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-55e4ccaca3dfa67ce556b9d33f64f0bc_l3.png "Rendered by QuickLaTeX.com")
Question 9.
Solution:
![\begin{aligned} &\text { Let } A=\left[\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}\right]\\ &\text { W.K.T. , } A=IA \Rightarrow\left[\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A }\\ &\Rightarrow\left[\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right] \mathrm{A} \quad\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}-\mathrm{R}_{2}\right]\\ &\Rightarrow\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ -2 & 3 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}-2 R_{1}\right]\\ & \end{aligned}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-bd99459ec59450282e61f989bb4d65f7_l3.png "Rendered by QuickLaTeX.com")
![\begin{array}{l} \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 7 & -10 \\ -2 & 3 \end{array}\right] \mathrm{A} \quad\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}-3 \mathrm{R}_{2}\right] \\ Therefore, \quad \mathrm{A}^{-1}=\left[\begin{array}{rr} 7 & -10 \\ -2 & 3 \end{array}\right] \end{array}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-96a99384f3b53cb233a140fc2359da8d_l3.png "Rendered by QuickLaTeX.com")
Question 10.
Solution:
![\begin{array}{l} \text { W.K.T. , } A=I A \Rightarrow\left[\begin{array}{rr} 3 & -1 \\ -4 & 2 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \mathrm{A} \\ \Rightarrow\left[\begin{array}{cc} -1 & 1 \\ -4 & 2 \end{array}\right]=\left[\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right] \mathrm{A} \quad\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}+\mathrm{R}_{2}\right] \\ \Rightarrow\left[\begin{array}{rr} 1 & -1 \\ -4 & 2 \end{array}\right]=\left[\begin{array}{cc} -1 & -1 \\ 0 & 1 \end{array}\right] \mathrm{A}\left[R_{1} \rightarrow(-1) R_{1}\right] \end{array}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-03610d4b7ce0b3c077ad20021c43c524_l3.png "Rendered by QuickLaTeX.com")
![\begin{array}{l} \Rightarrow⇒\left[\begin{array}{cc} 1 & -1 \\ 0 & -2 \end{array}\right]=\left[\begin{array}{cc} -1 & -1 \\ -4 & 3 \end{array}\right] A\left[\mathrm{R}{2} \rightarrow \mathrm{R}{2}+4 \mathrm{R}_{1}\right] \\ \Rightarrow\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} -1 & -1 \\ 2 & 3 / 2 \end{array}\right] \mathrm{A}\left[\mathrm{R}{2} \rightarrow \frac{-1}{2} \mathrm{R}{2}\right] \\ \Rightarrow \quad\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & 1 / 2 \\ 2 & 3 / 2 \end{array}\right] \mathrm{A}\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}+\mathrm{R}_{2}\right] \\ \mathrm{Therefore,A}^{-1}= {\left[\begin{array}{cc} 1 & 1 / 2 \\ 2 & 3 / 2 \end{array}\right]} \end{array}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-9d37a3d1a151dba97b67a4dde2cdd7ff_l3.png "Rendered by QuickLaTeX.com")
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