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Definition, Theorem, Formulas, Solved Example Problems | Inverse of a Non-Singular Square Matrix - Properties of inverses of matrices | 12th Mathematics : UNIT 1 : Applications of Matrices and Determinants

Chapter: 12th Mathematics : UNIT 1 : Applications of Matrices and Determinants

Properties of inverses of matrices

We state and prove some theorems on non-singular matrices.

Properties of inverses of matrices

We state and prove some theorems on non-singular matrices.

 

Theorem 1.4

If A is non-singular, then


, where λ is a non-zero scalar.

Proof

Let A be non-singular. Then |A| ≠ 0 and A−1 exists. By definition,


 

Theorem 1.5 (Left Cancellation Law)

Let A, B, and C be square matrices of order n. If A is non-singular and AB = AC, then B = C.

Proof

Since A is non-singular, A1 exists and AA1 = A1 A = In . Taking AB = AC and pre-multiplying both sides by A1, we get A1 ( AB) = A1 ( AC). By using the associative property of matrix multiplication and property of inverse matrix, we get B = C.

 

Theorem1.6 (Right Cancellation Law)

Let A, B, and C be square matrices of order n. If A is non-singular and BA = CA, then B = C.

Proof

Since A is non-singular, A1 exists and AA1 = A1 A = In . Taking BA = CA and post-multiplying both sides by A1, we get (BA) A1 = (CA) A1. By using the associative property of matrix multiplication and property of inverse matrix, we get B = C.

Note

If A is singular and AB = AC or BA = CA, then B and C need not be equal. For instance, consider the following matrices:


We note that |A| = 0 and AB = AC ; . but B  ≠ C

 

Theorem 1.7 (Reversal Law for Inverses)

If A and B are non-singular matrices of the same order, then the product AB is also non-singular and ( AB)1 = B1 A1.

Proof

Assume that A and B are non-singular matrices of same order n. Then,| A | ≠ 0, | B | ≠ 0, both A1 and B1 exist and they are of order n. The products AB and B1 A1 can be found and they are also of order n. Using the product rule for determinants, we get |AB| =| A || B | 0. So, AB is non-singular and

( AB)(B-1 A-1 ) = ( A(BB-1 )) A-1 = ( AIn ) A-1 = AA-1 = In ;

(B-1 A-1 )( AB) = (B-1 ( A-1 A))B = (B-1In )B = B-1B = In .

Hence ( AB)1 = B1 A1.

 

Theorem 1.8 (Law of Double Inverse)

If A is non-singular, then A1 is also non-singular and ( A1 )1 = A.

Proof

Assume that A is non-singular. Then |A | 0, and A1 exists.

Now, |A-1| = 1/|A| ≠ 0

 A-1 is also non-singular, and AA-1 = A-1A = I

Now, A-1A = I  ⇒ (AA-1)-1 = I ⇒ (A-1)-1 A-1 = I.

Post-multiplying by A on both sides of equation (1), we get (A-1)-1 =A.

 

Theorem 1.9

If A is a non-singular square matrix of order n , then


Proof

Since A is a non-singular square matrix, we have |A|0 and so, we get


 

Note

If A is a non-singular matrix of order 3, then, |A | ≠ 0 . By theorem 1.9 (ii), we get |adjA| = | A|2 and so, | adj A | is positive. Then, we get |A| = ± √|adjA| .

So, we get 

Further, by property (iii), we get 

Hence, if A is a non-singular matrix of order 3, then we get 

 

Example 1.4

If A is a non-singular matrix of odd order, prove that |adj A| is positive.

Solution

Let A be a non-singular matrix of order 2m + 1 , where m = 0,1, 2, .. . . Then, we get |A| ≠ 0 and, by theorem 1.9 (ii), we have |adj A| = |A|(2m+1)-1 = |A|2m 

Since |A|2m is always positive, we get that |adj A| is positive.

 

Example 1.5

Find a matrix  A if adj( A)

Solution


 

Example 1.6

If adj A = find A1.

Solution


 

Example1.7

If A is symmetric, prove that adj A is also symmetric.

Solution

Suppose A is symmetric. Then, AT = A and so, by theorem 1.9 (vi), we get

adj (AT) = (adj A) T  ⇒ adj A = (adj A)T ⇒ adj A is symmetric

 

Theorem 1.10

If A and B are any two non-singular square matrices of order n , then

adj( AB) = (adj B)(adj A).

Proof

Replacing A by AB in adj(A) = |A|A−1 , we get

adj(AB) = |AB| (AB)-1 = (| B | B-1) (| A | A-1) = adj(B) adj(A)

 

Example 1.8

Verify the property  ( AT  )1  = ( A1 )T  with  A .

Solution

For the given A, we get |A |= (2) (7) -  (9)(1) = 14 − 9 = 5 .


From (1) and (2), we get (A-1) = (AT)-1. Thus, we have verified the given property.

 

Example 1.9

Verify ( AB)1 = B1 A1 with 

Solution


As the matrices in (1) and (2) are same, (AB) −1 = B-1 A-1 is verified.

 

Example 1.10

If  A  = , find x and y such that A2 + xA + yI2=O2, Hence, find A1.

Solution


So, we get 22 + 4x + y =0, 31+5x+y=0, 27+3x=0 and 18+2x=0

Hence x = −9 and y =14.Then, we get A2 - 9A + 14I2 = O2

Post-multiplying this equation by A−1 , we get A – 9I2 + 14A-1 = O2.  Hence, we get


 

Tags : Definition, Theorem, Formulas, Solved Example Problems | Inverse of a Non-Singular Square Matrix , 12th Mathematics : UNIT 1 : Applications of Matrices and Determinants
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12th Mathematics : UNIT 1 : Applications of Matrices and Determinants


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