`star` Elementary Operation (Transformation) of a Matrix

`star` Invertible Matrices

`star `Inverse of a matrix by elementary operations

`star` Invertible Matrices

`star `Inverse of a matrix by elementary operations

`color{green} ✍️` There are six operations (transformations) on a matrix, three of which are due to rows and three due to columns, which are known as elementary operations or transformations.

(i) `color{green}{"The interchange of any two rows or two columns."`

`color{green} ✍️`Symbolically the interchange of `i^(th)` and `j^(th)` rows is denoted by `R_i ↔ R_j` and interchange of `i^(th)` and `j^(th)` column is denoted by `C_i ↔ C_j`.

For example, applying `R_1↔ R_2` to `A= [ (1,2,1),(-1,sqrt 3 , 1),(5,6,7)]` we get `[(-1, sqrt 3 , 1),(1,2,1),(5,6,7)]`

(ii) `color{green}{"The multiplication of the elements of any row or column by a non zero number."}`

`color{green} ✍️` Symbolically, the multiplication of each element of the `i^(th)` row by `k`, where `k ≠ 0` is denoted by `R_i → k R_i`.

The corresponding column operation is denoted by `C_i → kC_i`

For example, applying `C_3 -> 1/7 C_3 ` to `B= [ (1, 2,1),(-1, sqrt 3 , 1)]` we get `[(1, 2, 1/7),(-1, sqrt 3 , 1/7)]`

(iii) `color{green}{"The addition to the elements of any row or column,"}`

`color{green}{" the corresponding elements of any other row or column multiplied by any non zero number."}`

`color{green} ✍️` Symbolically, the addition to the elements of `i^(th)` row, the corresponding elements of `j^(th)` row multiplied by k is denoted by `R_i → R_i + kR_j`.

`color{green} ✍️` The corresponding column operation is denoted by `C_i → C_i + kC_j`.

For example, applying `R_2 → R_2 – 2R_1`, to `C= [(1,2),(2,-1)]` we get ` [(1,2),(0,-5)]`

(i) `color{green}{"The interchange of any two rows or two columns."`

`color{green} ✍️`Symbolically the interchange of `i^(th)` and `j^(th)` rows is denoted by `R_i ↔ R_j` and interchange of `i^(th)` and `j^(th)` column is denoted by `C_i ↔ C_j`.

For example, applying `R_1↔ R_2` to `A= [ (1,2,1),(-1,sqrt 3 , 1),(5,6,7)]` we get `[(-1, sqrt 3 , 1),(1,2,1),(5,6,7)]`

(ii) `color{green}{"The multiplication of the elements of any row or column by a non zero number."}`

`color{green} ✍️` Symbolically, the multiplication of each element of the `i^(th)` row by `k`, where `k ≠ 0` is denoted by `R_i → k R_i`.

The corresponding column operation is denoted by `C_i → kC_i`

For example, applying `C_3 -> 1/7 C_3 ` to `B= [ (1, 2,1),(-1, sqrt 3 , 1)]` we get `[(1, 2, 1/7),(-1, sqrt 3 , 1/7)]`

(iii) `color{green}{"The addition to the elements of any row or column,"}`

`color{green}{" the corresponding elements of any other row or column multiplied by any non zero number."}`

`color{green} ✍️` Symbolically, the addition to the elements of `i^(th)` row, the corresponding elements of `j^(th)` row multiplied by k is denoted by `R_i → R_i + kR_j`.

`color{green} ✍️` The corresponding column operation is denoted by `C_i → C_i + kC_j`.

For example, applying `R_2 → R_2 – 2R_1`, to `C= [(1,2),(2,-1)]` we get ` [(1,2),(0,-5)]`

`color{green} ✍️` If `A` is a square matrix of order `m`, and if there exists another square matrix `B` of the same order m,

such that `color{green}{AB = BA = I}`,

then `B` is called the inverse matrix of `A` and it is denoted by `A^(– 1)`. In that case A is said to be invertible.

For example, let `A= [(2,3),(1,2)]` and `B= [(2,-3),(-1,2)]` be two matrices.

`AB= [(2,3),(1,2)] [(2,-3),(-1,2)]`

`= [ (4-3, -6+6),(2-2 ,-3+4)]= [(1,0),(0,1)]=I`

`BA=[(1,0),(0,1)]=I` Thus `B` is the inverse of `A`, in other words `B = A^(– 1)` and `A` is inverse of `B`, i.e., `A = B^(–1)`

`color{red}{ul{"Note :"}}`

1. A rectangular matrix does not possess inverse matrix, since for products BA and AB to be defined and to be equal, it is necessary that matrices A and B should be square matrices of the same order.

2. If B is the inverse of A, then A is also the inverse of B.

such that `color{green}{AB = BA = I}`,

then `B` is called the inverse matrix of `A` and it is denoted by `A^(– 1)`. In that case A is said to be invertible.

For example, let `A= [(2,3),(1,2)]` and `B= [(2,-3),(-1,2)]` be two matrices.

`AB= [(2,3),(1,2)] [(2,-3),(-1,2)]`

`= [ (4-3, -6+6),(2-2 ,-3+4)]= [(1,0),(0,1)]=I`

`BA=[(1,0),(0,1)]=I` Thus `B` is the inverse of `A`, in other words `B = A^(– 1)` and `A` is inverse of `B`, i.e., `A = B^(–1)`

`color{red}{ul{"Note :"}}`

1. A rectangular matrix does not possess inverse matrix, since for products BA and AB to be defined and to be equal, it is necessary that matrices A and B should be square matrices of the same order.

2. If B is the inverse of A, then A is also the inverse of B.

`color{red}{" Inverse of a square matrix, if it exists, is unique"}`

`color{green} ✍️` Let `A = [a_(ij)]` be a square matrix of order `m,` If possible, let B and C be two inverses of `A.` We shall show that `B = C.`

Since B is the inverse of A

`AB = BA = I`

Since C is also the inverse of A

`AC = CA = I`

`color{green} ✍️` Thus `B = BI = B (AC) = (BA) C = IC = C`

`color{green} ✍️` Let `A = [a_(ij)]` be a square matrix of order `m,` If possible, let B and C be two inverses of `A.` We shall show that `B = C.`

Since B is the inverse of A

`AB = BA = I`

Since C is also the inverse of A

`AC = CA = I`

`color{green} ✍️` Thus `B = BI = B (AC) = (BA) C = IC = C`

`color{green} ✍️` If `A` and `B` are invertible matrices of the same order, then `(AB)^(–1) = B^(–1) A^(–1)`

`color{blue}{"Proof" :}`

`(AB) (AB)^(–1) =1`

`=>` `A^(–1) (AB) (AB)^(–1) =A^(–1)I` (Pre multiplying both sides by `A^(–1)`

`=>` `(A^(–1)A) B (AB)^(–1) =A^(–1)` (Since `A^(–1) I = A^(–1))`

`=>` ` IB (AB)^(–1) =A^(–1)`

`=>` `B (AB)^(–1) =A^(–1)`

`=>` `B^(–1) B (AB)^(–1) =B^(–1) A^(–1)`

`=>` `I (AB)^(–1) =B^(–1) A^(–1)`

Hence `color{blue}{(AB)^(–1) =B^(–1) A^(–1)}`

`color{blue}{"Proof" :}`

`(AB) (AB)^(–1) =1`

`=>` `A^(–1) (AB) (AB)^(–1) =A^(–1)I` (Pre multiplying both sides by `A^(–1)`

`=>` `(A^(–1)A) B (AB)^(–1) =A^(–1)` (Since `A^(–1) I = A^(–1))`

`=>` ` IB (AB)^(–1) =A^(–1)`

`=>` `B (AB)^(–1) =A^(–1)`

`=>` `B^(–1) B (AB)^(–1) =B^(–1) A^(–1)`

`=>` `I (AB)^(–1) =B^(–1) A^(–1)`

Hence `color{blue}{(AB)^(–1) =B^(–1) A^(–1)}`

`color{green} ✍️` If `A` is a matrix such that `A^(–1)` exists.

● write `A = IA`

● Apply a sequence of row operation on `A = IA` till we get, `I = BA. `

The matrix B will be the inverse of A.

● Similarly, if we wish to find `A^(–1)` using column operations, then, write `A = AI` and apply a sequence of column operations on `A = AI` till we get, `I = AB`

`color{red}{"Remark :"}` In case, after applying one or more elementary row (column) operations on

`A = IA (A = AI),` if we obtain all zeros in one or more rows of the matrix A on L.H.S., then `A^(–1)` does not exist

● write `A = IA`

● Apply a sequence of row operation on `A = IA` till we get, `I = BA. `

The matrix B will be the inverse of A.

● Similarly, if we wish to find `A^(–1)` using column operations, then, write `A = AI` and apply a sequence of column operations on `A = AI` till we get, `I = AB`

`color{red}{"Remark :"}` In case, after applying one or more elementary row (column) operations on

`A = IA (A = AI),` if we obtain all zeros in one or more rows of the matrix A on L.H.S., then `A^(–1)` does not exist

Q 3154680554

By using elementary operations, find the inverse of the matrix

`A = [ ( 1,2 ), ( 2,-1 ) ]`

Class 12 Chapter 3 Example 23

`A = [ ( 1,2 ), ( 2,-1 ) ]`

Class 12 Chapter 3 Example 23

In order to use elementary row operations we may write A = IA.

or ` [ (1,2), ( 2,-1) ] = [ ( 1,0), (0,1) ] A` , then ` [ (1,2), (0,-5) ] = [ (1,0 ), (-2,1) ] A` (applying `R_2 → R_2 – 2R_1` )

or `[ (1,2), (0,1) ] = [ (1,0 ), ( 2/5 , -1/5) ] A` (applying `R_2 -> -1/5 R_2` )

or ` [ (1,0), (0,1 ) ] = [ ( 1/5, 2/5), ( 2/5 , -1/5) ] A` (applying `R_1 → R_1 – 2R_2`)

Thus `A^(-1) = [ (1/5, 2/5 ), ( 2/5 , -1/5) ]`

Q 3114780650

Obtain the inverse of the following matrix using elementary operations

`A = [ (0,1,2 ), ( 1,2,3), (3,1,1) ]`

Class 12 Chapter 3 Example 24

`A = [ (0,1,2 ), ( 1,2,3), (3,1,1) ]`

Class 12 Chapter 3 Example 24

Write A = I A, i.e., ` [ ( 0,1,2 ), ( 1,2,3 ), ( 3,1,1) ] = [ (1,0,0 ), ( 0,1,0 ), ( 0 , 0, 1 ) ] A`

or ` [ (1,2,3 ), ( 0,1,2), ( 3,1,1) ] = [ (0,1,0 ), (1,0,0), ( 0 ,0 ,1 ) ] A` (applying `R_1↔ R_2` )

or ` [ ( 1,2,3 ) ,( 0,1,2) ,( 0, -5 ,-8 ) ] =[ (0,1, 0 ), (1,0,0 ), ( 0, -3 ,1 ) ] A` (applying `R_3 → R_3 – 3R_1`)

or `[ (1,0,-1), ( 0,1,2), ( 0, -5 , -8 ) ] = [ ( -2 ,1, 0 ), ( 1,0 ,0 ), ( 0 , -3 ,1 ) ] A` (applying `R_1 → R_1 – 2R_2` )

or `[ ( 1,0 , -1 ), ( 0,1,2 ), ( 0,0,2 ) ] = [ ( -2,1,0), ( 1,0,0 ), ( 5,-3 , 1 ) ] A` (applying `R_3 → R_3 + 5R_2`)

or `[ (1,0 , -1 ), ( 0 ,1,2 ), ( 0,0,1) ] = [ (-2,1, 0 ), ( 1,0 , 0 ), ( 5/2 , -3/2 , 1/2 ) ]A` (applying `R_3 →1/2 R_3` )

or ` [ (1, 0, 0 ), ( 0,1,2 ), ( 0, 0,1 ) ] = [ ( 1/2 , -1/2 , 1/2 ), ( 1, 0 ,0 ), ( 5/2 , -3/2 , 1/2 ) ] A` (applying `R_1 → R_1 + R_3`)

or ` [ ( 1, 0 ,0 ), ( 0,1, 0 ), ( 0 , 0 , 1) ] = [ (1/2, -1/2 , 1/2 ), ( -4 , 3, -1 ), ( 5/2 , -3/2 , 1/2) ]A` (applying `R_2 → R_2 – 2R_3`)

Hence `A^(-1) = [ ( 1/2 , -1/2 , 1/2 ), ( -4 ,3, -1 ), ( 5/2 , -3/2 ,1/2 ) ]`

Q 3144780653

Find `P^(-1 )`, if it exists, given `P= [ (10 ,-2), ( -5 ,1 ) ]`

Class 12 Chapter 3 Example 25

Class 12 Chapter 3 Example 25

We have P = I P, i.e. ` [ ( 10, -2 ), ( -5,1 ) ] = [ (1,0 ), ( 0,1 ) ] P`

or ` [ ( 1, -1/5 ), ( -5 , 1) ] = [ (1/10 , 0 ), (0 ,1) ] P` (applying `R_1 → 1/10 R_1 ` )

or ` [ ( 1, -1/5) , ( 0 ,0 ) ] = [ (1/10 , 0 ), ( 1/2 ,1 ) ]P ` (applying `R_2 → R_2 + 5 R_1`)

We have all zeros in the second row of the left hand side matrix of the above

equation. Therefore, `P^(-1)` does not exist.