Question

Inverse of a Matrix

Answer 1

A square matrix of order n is invertible if there exists a square matrix B of the same order such that AB = In = BA.

In such a case we can say that the inverse of A is B and we write A^-1 = B.

Theorem 1.

The inverse of a square matrix exists if |A| ≠ 0.

Proof :

Let A be a matrix of order n and its inverse matrix B will of order n, then

AB = |1|

But |AB| = |A| . |B| = |I| = I

Thus |A| ≠0

Again if |A| ≠ 0 then

\(A\frac{(adjA)}{|A|} = \frac{(adjA)A}{|A|} = I\)

Thus B = \(\frac{(adjA)}{|A|} \) or A^-1 (Here B = A^-1)

i. e. AB = BA = I

Thus, B inverse matrix of A exists whereas |A| ≠ 0iff |A| ≠ 0.

Theorem 2.

The inverse of non-singular matrix is unique.

Proof :

Let matrix A is a non-singular matrix of order n. B and C are its invertible matrices.

Now AB = BA = I and AC = CA = f

AC = I ⇒ B(AC) = BI = B ...(i)

and BA = I ⇒ (BA)C = CI = C ...(ii)

but B(AQ = (BA)C

(By associativity of multiplication of matrices)

From (i) and (ii), B = C

Thus, invertible matrix has unique inverse.

Theorem 3.

Reversal law of inverse : If A and B are two square matrices of order n and are invertibe then

(AB)^-1 = B^-1A^-1

where A^-1 and B^-1 are inverse of A and B respectively.

Proof :

A^-1 and B^-1 are inverse matrices of A and B respectively, then

(AB) (B^-1A^-1) = A(BB^-1)A^-1

= AIA^-1 (∵ BB^-1 = 1)

By associativity of multiplication of matrices

= AA^-1 (∵ AI = A)

= I ...(1)

Similarly

(B^-1A^-1) (AB) = B^-1(A^-1A)B

= B^-1IB (∵A^-1A = 1)

By associativity of multiplication of matrices = B-1B (-.IB = B)

= 1 ...(2)

From (1) and (2), we get

(AB)^-1 = B^-1A^-1

Corollary:

(A₁A₂ ............ A_n)^-1 = A_n^-1A_n^-1.......... A₂^-1A₁^-1

Theorem 4.

Cancellation law: If A,B and C are square matrices of order n and A is a non-singular matrix, then

(i) AB = AC => B = C (Left cancellation law)

(ii) BA = CA =>B = C (Right cancellation law)

Proof: Since, A is a non-singular matrix, so A-1 exists.

AB = AC

⇒ A^-1(AB) = A^-1(AC)

⇒ (A^-1aA)B = (A^-1A)C [∵ A^-1(AB) = (A^-1A)B

A^-1(AC) = (A^-1A)C

By associativity of multiplication of matrices

⇒ IB=IC (∵ A^-1A = I)

⇒ B = C (∵ IB = B and IC = C)

Again BA = CA

⇒ (BA)A^-1 = (CA)A^-1 [∵ (BA)A^-1 = B(AA^-1)

(CA)A^-1 = C(AA^-1)

By associativity of multiplication of matrices]

⇒ B(AA^-1) = C(AA^-1)

⇒ BI = CI (∵ AA^-1 = 1)

⇒ B = C (∵ BI = B and Cl = C)

Theorem 5.

If matrix A is invertible then show that transpose of A is also invertible and (A')^-1 = (A^-1)'.

Proof:

Let A is an invertible matrices of order n, then

∵ |A| ≠ 0

∵ |A'| = |A| ≠ 0

i.e., A' is also invertible,

Now AA^-1 = A^-1A = I

⇒ (AA^-1)' = (A^-1A)' = r

⇒ (A^-1)'A' =A'(A^-1A) = I [∵ (AB)' = B'A' and I' = I

AB = BA = I => B^-1 = A]

Thus (A')^-1 =(A^-1)'

Theorem 6.

If matrix A is invertible and symmetric, then A-1 is also symmetric.

Proof:

A is an invertible symmetric matrix, then A' = A .

Again, we know that

(A^-1)' = (A')^-1

∴ (A^-1)' = A^-1 [∵ A' = A]

Thus, A^-1 is also symmetric.

Theorem 7.

If A and B are two invertible matrices of same order then show that

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

Proof:

A and B are invertible matrices of same order, then

|A| ≠ 0 and |B| ≠ 0

∴ |AB| = |A||B| ≠ 0

Thus, (AB)^-1 exists.

Thus, adj (AB) = (adj B) (adj A)

Theorem 8.

For any sqaure matrix A, prove that (adj A)' = adj A'

Proof :

Let A be a sqaure matrix of order n. Then (adj A)' and adj A' will be matrices of order n and (i, j)th element of (adj A)'

= (i, j)th element of (adj A)

= Cofactor of (i, j)th element of A

= Cofactor of (j, j)th element of A'

= (i, j)th element of (adj A')

Thus, (adj A)' = (adj A').

Theorem 9.

If A be a non-singular matrix of order tt then show that

|adjA| = |A|^n-1.

Proof:

∵ A (adj A) = |A| I

⇒ |A (adj A) | = |A| |I| (∵ A = B ⇒ |A| = |B|)

⇒ |A| |adjA| = |A|ⁿ |I| [∵ |kI| = kⁿI]

⇒ |A| |adj A| = |A|ⁿ

⇒ |adjA| = |A|^n-1 [∵ |I| = I]

Thus, |adjA| = |A|^n-1.

Theorem 10.

If matrix A is non-singular matrix of order n, then show that

adj (adjA) = |A|^n-2A .

Proof:

We know that for any non-singular matrix B.

B(adj B) = |B|I

Let B = adj A

Then (adj A) [adj (adj A)] = |adj A| I

⇒ (adj A) [adj (adj A)] = | A |^n-1I [∵ (adjA) = |A|^n-1]

⇒ A(adj A) [adj (adj A)] = |A|^n-1A [∵ AI_n = A]

⇒ |A| [adj (adj A)] = |A|^n-1A [∵ A (adj A) = |A| I]

⇒ [adj (adj A)] = | A |^n-2A

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

| adj (adj A) | = |A|^(n-1)2

Proof:

∵ adj (adj A) |A|^n-2 (By Theorem 10)

⇒ |adj (adj A)| = || A |^n-2A|

⇒ |adj (adj A)| = | A |^n(n-2) |A| (∵ |kA| = kⁿ|A|)

⇒ |adj (adjA)| =A^{n2-2K + 1} (|A|^n(n-2) A| = | A |^{n2-2K + 1}]

⇒ |adj (adj A)| =A^(n-1)2

Answer 2

Theorem 11.

If product of two non-singular matrix is zero matrix then show that both are singular matrices.

Proof :

Let A and B are two non-singular matrices of order n and AB = O (zero matrix)

If possible, let B be a non-singular matrix. Thus, B-1 exists. .

Now AB = O

⇒ (AB)B^-1 = OB^-1

[Post multiplying both sides by B^-1]

⇒ A(BB^-1) = OB^-1 = O

[By associativity of multiplication of matrices

and OB^-1 = B^-1O = O]

⇒ AI_n = O

⇒ A = O [∵ AI_n = OA]

But, A is non-singular matrix so, B is singular matrix. Again, if possible A is a matrix so A-1 exists.

Now AB = O => A^-1(AB) = A^-1O

[Pre-multiplying both sides by A^-1]

⇒ (A^-1A)B = O

[By associtivity of multiplication of matrices]

⇒ I_nB = O [∵ A^-1A = I_n]

⇒ B = O [∵ I_nB = BI_n = B]

But A is non-singular matrix so B is singular matrix.

Theorem 12.

If A is non-singular matrix then show that

|A^-1| = |A^-1|, i.e., A^-1 = \(\frac{1}{|A|}\)

Proof:

| A | ≠ 0

Thus, A^-1 exists and

AA^-1 = I = A^-1A

⇒ |AA^-1| = |I |

⇒ |A| |A^-1| = 1 [∵ |AB| = |A| |B| and |I| = 1]

⇒ |A^-1| = \(\frac{1}{|A|}\) [∵ |A| ≠ 1]

Steps for finding the inverse of a square Matrix

To find inverse of matrix, following steps are used :

Step 1: Find determinant value of given matrix.

Step 2 : Replace each element of matrix by their cofactors and get cofactors matrix.

Step 3 : Find transpose of matrix obtained in step 2. Now, we get adj A.

Step 4 : Matrix obtained in step 3 is divided by determinant value of matrix. Thus, the matrix so obtained is the inverse of the given matrix.