Mathematics Elementary Operation (Transformation) of a Matrix Invertible Matrices Inverse of a matrix by elementary operations For CBSE-NCERT
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### Topic covered

star Elementary Operation (Transformation) of a Matrix
star Invertible Matrices
star Inverse of a matrix by elementary operations

### Elementary Operation (Transformation) of a Matrix

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)]

### Invertible Matrices

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.

### Theorem (Uniqueness of inverse)

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

### Theorem : (AB)^(–1) = B^(–1) A^(–1)

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)}

### Inverse of a matrix by elementary operations

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
Q 3154680554

By using elementary operations, find the inverse of the matrix

A = [ ( 1,2 ), ( 2,-1 ) ]
Class 12 Chapter 3 Example 23
Solution:

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
Solution:

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
Solution:

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.