Question 11. 
Solution:
![\begin{aligned} &\text { Let } A=\left[\begin{array}{cc} 2 & -6 \\ 1 & -2 \end{array}\right]\\ &\text { W.K.T. , } A=IA \Rightarrow\left[\begin{array}{cc} 2 & -6 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A }\\ &\Rightarrow\left[\begin{array}{ll} 1 & -2 \\ 2 & -6 \end{array}\right]=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] A\left[R_{1} \leftrightarrow R_{2}\right]\\ &\Rightarrow\left[\begin{array}{rr} 1 & -2 \\ 0 & -2 \end{array}\right]=\left[\begin{array}{rr} 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-f085b04f9c9448cfd7448d7370de6a8a_l3.png "Rendered by QuickLaTeX.com")
}![\begin{array}{l} \Rightarrow\left[\begin{array}{cc} 1 & -2 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 0 & 1 \\ -1 / 2 & 1 \end{array}\right] \mathrm{A}\left[\mathrm{R}{2} \rightarrow \frac{-1}{2} \mathrm{R}{2}\right] \\ \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} -1 & 3 \\ -1 / 2 & 1 \end{array}\right] \mathrm{A}\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}+2 \mathrm{R}_{2}\right] \\ \mathrm{Therefore, A}^{-1}=\left[\begin{array}{ll} -1 & 3 \\ -1 / 2 & 1 \end{array}\right] \end{array}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-745ed912355ddaf131a4d14dbb868e1c_l3.png "Rendered by QuickLaTeX.com")
Question 12. 
Solution:
![\begin{array}{l} \text { Let } A=\left[\begin{array}{cc} 6 & -3 \\ -2 & 1 \end{array}\right] \\ \text { W.K.T. , } A=IA \Rightarrow\left[\begin{array}{cc} 6 & -3 \\ -2 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \mathrm{A} \\ \Rightarrow\left[\begin{array}{cc} 1 & -1 / 2 \\ -2 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 / 6 & 0 \\ 0 & 1 \end{array}\right] \text { A }\left[R_{1} \rightarrow \frac{1}{6} R_{1}\right] \\ \Rightarrow\left[\begin{array}{cc} 1 & -1 / 2 \\ 0 & 0 \end{array}\right]=\left[\begin{array}{cc} 1 / 6 & 0 \\ 1 / 3 & 1 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}+2 R_{1}\right] \end{array}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-43263a12f1c71c331c41a7bcbd693cb9_l3.png "Rendered by QuickLaTeX.com")
Here, both the elements of R2 of L.H.S. are 0.
Therefore, A-1 does not exist.
Question 13. 
Solution:
![\begin{array}{l} \text { Let } A=\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right] \\ \text { W.K.T. , } A=IA \Rightarrow\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A } \\ \Rightarrow\left[\begin{array}{cc} 1 & -1 \\ -1 & 2 \end{array}\right]=\left[\begin{array}{cc} 1 & 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-ab635c4635ea43b2cc872a0158c0516f_l3.png "Rendered by QuickLaTeX.com")
![\begin{array}{l} \Rightarrow\left[\begin{array}{ll} 1 & -1 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}+R_{1}\right] \\ \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right] \text { A }\left[R_{1} \rightarrow R_{1}+R_{2}\right] \\ Therefore, A^{-1}=\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right] \end{array}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-c6effefe0431d49bced5ae5d4057cbda_l3.png "Rendered by QuickLaTeX.com")
Question 14. 
Solution:
![\begin{aligned} &\text { Let } A=\left[\begin{array}{ll} 2 & 1 \\ 4 & 2 \end{array}\right]\\ &\text { W.K.T. ,} A=IA \Rightarrow\left[\begin{array}{ll} 2 & 1 \\ 4 & 2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { A }\\ \end{aligned}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-e10dd037fb312be3e471739cf7489af2_l3.png "Rendered by QuickLaTeX.com")
![\begin{array}{l} \Rightarrow\left[\begin{array}{cc} 1 & 1 / 2 \\ 4 & 2 \end{array}\right]=\left[\begin{array}{cc} 1 / 2 & 0 \\ 0 & 1 \end{array}\right] \text { A }\left[R_{1} \rightarrow \frac{1}{2} R_{1}\right] \\ \Rightarrow\left[\begin{array}{ll} 1 & 1 / 2 \\ 0 & 0 \end{array}\right]=\left[\begin{array}{ll} 1 / 2 & 0 \\ -2 & 1 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}-4 R_{1}\right] \end{array}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-5be4fa9b5d5e43382948cad21b80cdc2_l3.png "Rendered by QuickLaTeX.com")
Here, both the elements of R2 of L.H.S. are 0.
Therefore, A-1 does not exist.
Question 15. 
Solution:
Let A=
W.K.T. , A=IA
![\begin{array}{l} \Rightarrow\left[\begin{array}{ccc} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathbf{A} \\ {\left[\begin{array}{rrr} -1 & -1 & 1 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}} \\ {\left[\begin{array}{rrr} 1 & 1 & -1 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{array}\right]=\left[\begin{array}{rrr} -1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}} \end{array}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-932191714532ca365ac7a111db3c8802_l3.png "Rendered by QuickLaTeX.com")
[Tex]\begin{array}{l} {\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}-\mathrm{R}_{3}\right]}\\\\ {\left[\mathrm{R}{1} \rightarrow(-1) \mathrm{R}{1}\right]} \end{array}[/Tex]
![\begin{array}{l} \Rightarrow\left[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & 0 & 5 \\ 0 & -5 & 5 \end{array}\right]=\left[\begin{array}{rrr} -1 & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 0 & -2 \end{array}\right] \mathrm{A}\left[\mathrm{R}{2} \rightarrow \mathrm{R}{2}-2 \mathrm{R}{1} \text { and } \mathrm{R}{3} \rightarrow \mathrm{R}{3}-3 \mathrm{R}{1}\right] \\ \left.=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & -5 & 5 \\ 0 & 0 & 5 \end{array}\right]=\begin{array}{rrr} -1 & 0 & 1 \\ 3 & 0 & -2 \\ 2 & 1 & -2 \end{array}\right] \mathrm{A}\left[\mathrm{R}{2} \leftrightarrow \mathrm{R}{3}\right] \end{array}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-91f67b2420edbfdfae4b8ada502ee5d4_l3.png "Rendered by QuickLaTeX.com")
![\begin{array}{l} {\left[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 5 \end{array}\right]=\left[\begin{array}{ccc} -1 & 0 & 1 \\ -3 / 5 & 0 & 2 / 5 \\ 2 & 1 & -2 \end{array}\right] \mathrm{A} \quad\left[\mathrm{R}{2} \rightarrow\left(\frac{-1}{5}\right) \mathrm{R}{2}\right]} \\ \left.\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 5 \end{array}\right]=\begin{array}{rrr} -2 / 5 & 0 & 3 / 5 \\ 2 & 0 & 2 / 5 \\ 2 & 1 & -2 \end{array}\right] \mathrm{A}\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}-\mathrm{R}_{2}\right] \end{array}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-b6e8066e18bd4d075b11f91f7d6d2841_l3.png "Rendered by QuickLaTeX.com")
![\begin{array}{l} {\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -2 / 5 & 0 & 3 / 5 \\ -3 / 5 & 0 & 2 / 5 \\ 2 / 5 & 1 / 5 & -2 / 5 \end{array}\right] \mathrm{A}\left[\mathrm{R}{3} \rightarrow \frac{1}{5} \mathrm{R}{3}\right]} \\ {\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -2 / 5 & 0 & 3 / 5 \\ -1 / 5 & 1 / 5 & 0 \\ 2 / 5 & 1 / 5 & -2 / 5 \end{array}\right] \mathrm{A}\left[R_{2} \rightarrow R_{2}+R_{3}\right]} \end{array}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-581bfa9be11cdb6df7e74ace30d43010_l3.png "Rendered by QuickLaTeX.com")
Therefore, A-1 =![\left[\begin{array}{ccc} -2 / 5 & 0 & 3 / 5 \\ -1 / 5 & 1 / 5 & 0 \\ 2 / 5 & 1 / 5 & -2 / 5 \end{array}\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-54b0164036fca5396e3ad3e54c1f3a76_l3.png "Rendered by QuickLaTeX.com")
Question 16. 
Solution:
Let A=
W.K.T. , A=IA
![\begin{aligned} &\left[\begin{array}{ccc} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \text { A }\\ &\left[\begin{array}{rrr} 1 & 3 & -2 \\ 0 & 9 & -11 \\ 0 & -1 & 4 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}+3 R_{1} \text { and } R_{3} \rightarrow R_{3}-2 R_{1}\right]\\ &\left[\begin{array}{ccc} 1 & 3 & -2 \\ 0 & -1 & 4 \\ 0 & 9 & -11 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ -2 & 0 & 1 \\ 3 & 1 & 0 \end{array}\right] \text { A }\left[R_{2} \leftrightarrow R_{3}\right]\\ &\left[\begin{array}{rrr} 1 & 3 & -2 \\ 0 & 1 & -4 \\ 0 & 9 & -11 \end{array}\right]=\left[\begin{array}{rrr} 1 & 0 & 0 \\ 2 & 0 & -1 \\ 3 & 1 & 0 \end{array}\right] \text { A }\left[R_{2} \rightarrow(-1) R_{2}\right.\\ \end{aligned}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-63fe42ad154e06a34f52e7aa06d8a820_l3.png "Rendered by QuickLaTeX.com")
[Tex]\begin{aligned} &\left[\begin{array}{lll} 1 & 0 & 10 \\ 0 & 1 & -4 \\ 0 & 0 & 25 \end{array}\right]=\left[\begin{array}{ccc} -5 & 0 & 3 \\ 2 & 0 & -1 \\ -15 & 1 & 9 \end{array}\right] \text { A }\left[R_{1} \rightarrow R_{1}-3 R_{2}\right. \text { and } R_{3} \rightarrow R_{3}-9 R_{2}\\ &\left[\begin{array}{ccc} 1 & 0 & 10 \\ 0 & 1 & -4 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -5 & 0 & 3 \\ 2 & 0 & -1 \\ -3 / 5 & 1 / 25 & 9 / 25 \end{array}\right] \text { A }\left[R_{1} \rightarrow \frac{1}{25} R_{3}\right]\\ &\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & -\frac{2}{5} & -\frac{3}{5} \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25} \end{array}\right] A\left[R_{1} \rightarrow R_{1}-10 R_{1} \text { and } R_{2} \rightarrow R_{2}+4 R_{3}\right]\\ &Therefore,A^{-1}=\left[\begin{array}{ccc} 1 & -2 / 25 & -3 / 25 \\ -2 / 25 & 4 / 25 & 11 / 25 \\ -3 / 25 & 1 / 25 & 9 / 25 \end{array}\right]\\ \end{aligned}[/Tex]
Question 17. 
Solution:
Let A=
W.K.T. , A=IA
![\begin{aligned} &\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\ &\left[\begin{array}{ccc} 2 & 0 & -1 \\ 1 & 1 & 2 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \text { A }\left[R_{2} \rightarrow R_{2}-2 R_{1}\right]\\ &\left[\begin{array}{ccc} 1 & 1 & 2 \\ 2 & 0 & -1 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{rrr} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right] \text { A }\left[R_{1} \leftrightarrow R_{2}\right]\\ &\left[\begin{array}{ccc} 1 & 1 & 2 \\ 0 & -2 & -5 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} -2 & 1 & 0 \\ 5 & -2 & 0 \\ 0 & 0 & 1 \end{array}\right] A\left[R_{2} \rightarrow R_{2}-2 R_{1}\right]\\ &\left[\begin{array}{ccc} 1 & 1 & 2 \\ 0 & 1 & 3 \\ 0 & -2 & -5 \end{array}\right]=\left[\begin{array}{ccc} -2 & 1 & 0 \\ 0 & 0 & 1 \\ 5 & -2 & 0 \end{array}\right] A \quad R_{2} \leftrightarrow R_{3} \end{aligned}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-eda89883bbc02503618bea4de055e782_l3.png "Rendered by QuickLaTeX.com")
![\begin{array}{l} {\left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -2 & 1 & -1 \\ 0 & 0 & 1 \\ 5 & -2 & 2 \end{array}\right] \text { A }\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}-\mathrm{R}{2} \text { and } \mathrm{R}{3} \rightarrow \mathrm{R}{3}+2 \mathrm{R}{2}\right]} \\ {\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right] \text { A }\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}+\mathrm{R}{3} \text { and } \mathrm{R}{2} \rightarrow \mathrm{R}{2}-3 \mathrm{R}{3}\right]} \\ \end{array}](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-5b8f97e869904e191f1ef71ce493227d_l3.png "Rendered by QuickLaTeX.com")
![Therefore,A^{-1}=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-303f45d4ec9e5ffb4143936bde9bcc59_l3.png "Rendered by QuickLaTeX.com")
Question 18. Matrices A and B will be inverse of each other only if:
(A) AB = BA
(B) AB = BA = 0
(C) AB = 0, BA = I
(D) AB = BA = I
Solution:
According to the definition of inverse of square matrix,
Option (D) is correct
i.e. AB=BA=I
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