Math Assignment Help With Inverse Of A Matrix

3.7.2 Inverse of a matrix:

A non-zero square matrix A of order n is said to be invertible if a square matrix B of order n also exist, such that AB = BA = In.

Theorem 1: An invertible matrix processes a unique inverse.

Proof: Let A be an invertible square matrix of order n and matrix B and C be the inverse of A.

Then,

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AB = BA = In

And

AC = CA = In

Now,

BA = In

Therefore,

(BA)C = In.C = C

And,

AC = In

B (AC) = B.In = B

But we know that,

(BA) C = B (AC)

Therefore,

B = C

Theorem2: A square matrix A is invertible if A is non-singular, i.e.

|A| ≠ 0

Proof: Let A is an invertible square matrix of order n and another square matrix also exists of the same order, such that,

AB = BA =In

AB = In

Therefore,

|AB| = |In|

|A|.|B| = 1

|A| ≠ 0

Therefore A is a non-singular matrix. Thus A is invertible then A is non-singular.

Theorem3: (Cancellation Law) If A B and C are square matrices of order n, then

AB = AC.

If A is non-singular, then B = C

Proof: As A is a non-singular, A-1 exists.

Therefore,

AB = AC

A-1 (AB) = A-1 (AC)

(A-1 A) B = (A-1 A) C

(InB) = (InC) [since, A-1A = In]

B = C

Theorem4: (Reversal law) If A and B are invertible matrices of the same order, then show that AB is also invertible and (AB)-1 = B-1A-1

Proof: Let A and B be the two invertible matrices, each of order n.

Such that |A| ≠ 0 and |B| ≠ 0

|AB| = |A|. |B| ≠ 0

This shows that AB is non-singular

So,

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

= (AIn) A-1

= AA-1 = In

Also,

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

= (BIn) B-1

= BB-1 = In

Therefore,

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

Hence,

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

Theorem5: If A is invertible square matrix, then show that AT is also invertible and

(AT)-1 = (A-1)T

Proof: Let A is an invertible square matrix of order n.

Then,

|A| ≠ 0

Therefore,

|AT| = |A| ≠ 0

Therefore AT is also invertible.

Now,

Now,

AA-1 = A-1 A = In

(AA-1)T = (A-1 A )T= InT = In

(A-1)TAT = AT (A-1)T = In

Hence,

(AT)-1 = (A-1)T

Theorem6: If A and B are invertible square matrix of the same order, then prove that

(adjAB) = (adjA) (adjB)

Proof: let A and B be invertible square matrix of order n.

Then,

|A| ≠ 0 and |B| ≠ 0

|AB| = |A|.|B| ≠ 0

We know that,

(AB)(adjAB) = |AB| .In(i)

Also,

(AB).(adjA.adjB)

= A(B.adjB) .(adjA)

= A(|B| .In).(adjA)

= |B| . (A.adjA)

= |B|.|A|.In

= |A|.|B| .In

= |AB|.In

Thus,

(AB).(adjA.adjB) = |AB|. In(ii)

So from (i) and (ii) we get

(AB)(adjAB) = (AB).(adjA.adjB) = |AB| In

As, AB is invertible, then by cancellation law,

(adjAB) = (adjA) (adjB)

Theorem7: If A is invertible square matrix, prove that

(adjA)T = (adjAT)

Proof: let A be an invertible square matrix of order n,

Then,

|A| ≠ 0

Therefore,

|AT| = |A| ≠ 0

This shows that AT is non-singular and therefore invertible.

We know that,

A(adjA) = |A| .In

Therefore,

[A.(adjA)]T = (|A|.In)T

Or (adjA)T . AT = |A|. InT

Or (adjA)T.AT = |A|. In(i)

Also,

(adjAT).AT = |AT|. In

(adjAT) .AT = |A|. In (ii)

Thus, from (i) and (ii) we get:

(adjA)T.AT = (adjAT).AT

But AT being invertible, by cancellation law, we have

(adjA)T = (adjAT)

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