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,

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

(I_nB) = (I_nC) [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

= (AI_n) A^-1

= AA^-1 = I_n

Also,

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

= (BI_n) B^-1

= BB^-1 = I_n

Therefore,

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

Hence,

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

Theorem5: If A is invertible square matrix, then show that A^T 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 A^T is also invertible.

Now,

Now,

AA^-1 = A^-1 A = I_n

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

(A^-1)^TA^T = A^T (A^-1)^T = I_n

Hence,

(A^T)^-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| I_n

As, AB is invertible, then by cancellation law,

(adjAB) = (adjA) (adjB)

Theorem7: If A is invertible square matrix, prove that

(adjA)^T = (adjA^T)

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

Then,

|A| ≠ 0

Therefore,

|A^T| = |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 . A^T = |A|. I_n^T

Or (adjA)^T.A^T = |A|. I_n(i)

Also,

(adjA^T).A^T = |A^T|. I_n

(adjA^T) .A^T = |A|. I_n (ii)

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

(adjA)^T.A^T = (adjA^T).A^T

But A^T being invertible, by cancellation law, we have

(adjA)^T = (adjA^T)

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