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Inverse of a Matrix

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Inverse of a Matrix - Lesson Summary

The inverse of a matrix by using the elementary operations.

If A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is called the inverse matrix of A, and it is denoted by A-1.

we have a square matrix, A, B of the order, n by n.

A =  a 11 a 12 ⋯ a 1n a 21 a 22 ⋯ a 2n ⋮ ⋮ ⋮ ⋮ a n1 a n2 ⋯ a nn   B =  x 11 x 12 ⋯ x 1n x 21 x 22 ⋯ x 2n ⋮ ⋮ ⋮ ⋮ x n1 x n2 ⋯ x nn  


AB =  a 11 a 12 ⋯ a 1n a 21 a 22 ⋯ a 2n ⋮ ⋮ ⋮ ⋮ a n1 a n2 ⋯ a nn   ×  x 11 x 12 ⋯ x 1n x 21 x 22 ⋯ x 2n ⋮ ⋮ ⋮ ⋮ x n1 x n2 ⋯ x nn  = I     

Here, we can say that matrix A is invertible.

The inverse of a matrix A does not mean that it is one divided by matrix A. i.e. A-1 ≠ 1/A

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

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


Theorem: If the inverse of a matrix exists, it is unique.

Proof : Let A be a square matrix of the order, n by n.

A = [aij]nxn

Let X and Y be inverse matrices of A.

AX = XA = I

AY = YA = I

Now X = XI

    = X(AY) ['.' AY = I]

    = (XA)Y  ['.' Matrix multiplication is associative]

    = IY  ['.' XA = I]

 ⇒ X = Y

Theorem: If A and B are invertible matrices of the same order, then (AB)-1 = B-1A-1.

Proof :

LHS: (AB)-1

 (AB)(AB)-1 = I (The product of the matrices, AB and AB inverse, is equal to the identity matrix.)

⇒ A-1(AB)(AB)-1 = A-1I (Multiply with A-1 on both the sides)

⇒ (A-1A)B(AB)-1 = A-1  (Since matrix multiplication is associative)

⇒ IB(AB)-1 = A-1 ['.' A-1A = I]

⇒ B(AB)-1 = A-1

⇒ B-1B(AB)-1 = B-1A-1

⇒ I(AB)-1 = B-1A-1 ['.' B-1B = I]

⇒ (AB)-1 = B-1A-1

Inverse of a matrix by elementary operations

Conder a marix A

A = IA

AA-1 = I

The transformation is done by using either row operations or column operations.

A = IA, A = IA

Ex:

X = 1 3 -4 6

We can express this matrix as X = IX

1 3 -4 6  =  1 00 1  X       

R2 → R2 + 4R1     1 30 18   =   1 04 1  X   

R2 → 1/18 R2      1 30 1   =   1 02/9 1/18  X 

R1 → R1 - 3R2     1 00 1=  1/3 -1/6 2/9 1/18 X   

the inverse of matrix X is X-1 =  1/3 -1/6 2/9 1/18     

Now we will obtain the inverse of the same matrix, X, using column operations.

X = XI

1 3 -4 6    = X  1 0 0 1             

Now we will perform column operations until the matrix on the left hand side transforms into an identity matrix.

C2 → C2 - 3C1   1 0 -4 18    = X  1 -3 0 1    

C2 → 1/18 C2    1 0 -4 1    = X  1 -1/6 0 1/18

C1 → C1 + 4C2   1 0 0 1    = X  1/3 -1/6 2/9 1/18

X-1 = 1/3 -1/6 2/9 1/18

Note: While performing the row and column operations, if all the elements in a row or column are obtained as zero, then the matrix has no inverse.

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