# Class 12 NCERT Solutions- Mathematics Part I – Chapter 3 Matrices – Exercise 3.4 | Set 1

Last Updated : 03 Apr, 2024

### Question 1. 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

## [Tex]\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right][/Tex]

Solution:

[Tex]\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}[/Tex]

[Tex]Therefore,A^{-1}=\left[\begin{array}{cc} 3 / 5 & 1 / 5 \\ -2 / 5 & 1 / 5 \end{array}\right][/Tex]

### [Tex]\left[\begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array}\right][/Tex]

Solution:

Let A=[Tex]\left[\begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array}\right][/Tex]

[Tex]\begin{array}{l} We Know That, A=IA â‡’\left[\begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] A \\ \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{lr} 1 & -1 \\ 0 & 1 \end{array}\right] A \\ \Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{lr} 1 & -1 \\ -1 & 2 \end{array}\right] A \\ Therefore, A^{-1}=\left[\begin{array}{lr} 1 & -1 \\ -1 & 2 \end{array}\right] \end{array}   [/Tex][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]

### [Tex]\left[\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right][/Tex]

Solution:

[Tex]\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}â‡’\left[\begin{array}{cc} 7 & -3 \\ -2 & 1 \end{array}\right] \end{aligned}[/Tex]

### [Tex]\left[\begin{array}{ll} 2 & 3 \\ 5 & 7 \end{array}\right][/Tex]

Solution:

[Tex]\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]\\ &â‡’\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}[/Tex]

[Tex]\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}[/Tex]

### [Tex]\left[\begin{array}{ll} 2 & 1 \\ -7 & 4 \end{array}\right][/Tex]

Solution:

[Tex]\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]\\ &â‡’\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]\\ &â‡’\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}[/Tex]

### [Tex]\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right][/Tex]

Solution:

[Tex]\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 }\\ &â‡’\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}[/Tex]

[Tex]\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}[/Tex]

## Using elementary transformation, find the inverse of each of the matrices, if it exists.

### [Tex]\left[\begin{array}{ll} 3 & 1 \\ 5 & 2 \end{array}\right][/Tex]

Solution:

[Tex]\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}[/Tex]

[Tex]\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}[/Tex]

### [Tex]\begin{bmatrix}4 & 5 \\3 & 4 \\\end{bmatrix}[/Tex]

Solution:

[Tex]\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}[/Tex]

### [Tex]\begin{bmatrix}3 & 10 \\2 & 7 \\\end{bmatrix}[/Tex]

Solution:

[Tex]\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}[/Tex]

[Tex]\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}[/Tex]

### [Tex]\begin{bmatrix}3& -1 \\-4 & 2 \\\end{bmatrix}[/Tex]

Solution:

[Tex]\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}[/Tex]

[Tex]\begin{array}{l} \Rightarrowâ‡’\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}[/Tex]

### [Tex]\begin{bmatrix}2&-6\\1 & -2\\\end{bmatrix}[/Tex]

Solution:

[Tex]\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}[/Tex]

}[Tex]\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}[/Tex]

### [Tex]\begin{bmatrix}6&-3\\-2 & 1\\\end{bmatrix}[/Tex]

Solution:

[Tex]\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}[/Tex]

Here, both the elements of R2 of L.H.S. are 0.

Therefore, A-1 does not exist.

### [Tex]\begin{bmatrix}2&-3\\-1 & 2\\\end{bmatrix}[/Tex]

Solution:

[Tex]\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}[/Tex]

[Tex]\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}[/Tex]

### [Tex]\begin{bmatrix}2&1\\4 & 2\\\end{bmatrix}[/Tex]

Solution:

[Tex]\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}[/Tex]

[Tex]\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}[/Tex]

Here, both the elements of R2 of L.H.S. are 0.

Therefore, A-1 does not exist.

### [Tex]\begin{bmatrix}2 & -3 & 3\\2 & 2 & 3\\3 & -2 & 2\end{bmatrix}[/Tex]

Solution:

Let A= [Tex]\begin{bmatrix}2 & -3 & 3\\2 & 2 & 3\\3 & -2 & 2\end{bmatrix}[/Tex]

W.K.T. , A=IA

[Tex]\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}   [/Tex][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]

[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}[/Tex]

[Tex]\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}[/Tex]

[Tex]\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}[/Tex]

Therefore, A-1 =[Tex]\left[\begin{array}{ccc} -2 / 5 & 0 & 3 / 5 \\ -1 / 5 & 1 / 5 & 0 \\ 2 / 5 & 1 / 5 & -2 / 5 \end{array}\right][/Tex]

### [Tex]\begin{bmatrix}1 & 3 & -2\\-3 & 0 & -5\\2 & 5 & 0\end{bmatrix}[/Tex]

Solution:

Let A=[Tex]\begin{bmatrix}1 & 3 & -2\\-3 & 0 & -5\\2 & 5 & 0\end{bmatrix}[/Tex]

W.K.T. , A=IA

[Tex]\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}   [/Tex][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]

### [Tex]\begin{bmatrix}2 & 0 & -1\\5 & 1 & 0\\0 & 1 & 3\end{bmatrix}[/Tex]

Solution:

Let A=[Tex]\begin{bmatrix}2 & 0 & -1\\5 & 1 & 0\\0 & 1 & 3\end{bmatrix}[/Tex]

W.K.T. , A=IA

[Tex]\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}[/Tex]

[Tex]\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}[/Tex]

[Tex]Therefore,A^{-1}=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right][/Tex]

Previous
Next