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_{n}B) = (I_{n}C) [since, A^{-1}A = 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^{-1}A^{-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^{-1}A^{-1}) = A(BB^{-1})A^{-1}

= (AI_{n}) A^{-1}

= AA^{-1} = I_{n}

Also,

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

= (BI_{n}) B^{-1}

= BB^{-1} = I_{n}

Therefore,

(AB) (B^{-1}A^{-1}) = (B^{-1}A^{-1})(AB) = I_{n}

Hence,

(**AB) ^{-1} = B^{-1}A^{-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})^{T}A^{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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