Theorem-1 :- For any square matrix A with real number entries, A + A′ is a symmetric matrix and A – A′ is a skew symmetric matrix.
Solution :-
Proof : Let B = A + A′, then
B′ = (A + A′)′
= A′ + (A′)′ [as (A + B)′ = A′ + B′]
= A′ + A [as (A′)′ = A]
= A + A′ [as A + B = B + A]
= B
Therefore B = A + A′ is a symmetric matrix
Now let C = A – A′
C′ = (A – A′)′
= A′ – (A′)′
= A′ – A
= – (A – A′) = – C
Therefore C = A – A′ is a skew symmetric matrix.
Theorem-2 :- Any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix.
Solution :-
Proof : Let A be a square matrix, then we can write
A = 1/2(A + A') + 1/2(A - A')
From the Theorem 1, we know that (A + A′) is a symmetric matrix and (A – A′) is
a skew symmetric matrix.
Since for any matrix A, (kA)′ = kA′, it follows that 1/2(A + A') is symmetric matrix
and 1/2(A - A') is skew symmetric matrix.
Thus, any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix.
Theorem-3 :- (Uniqueness of inverse) Inverse of a square matrix, if it exists, is unique.
Solution :-
Proof : Let A = [aᵢⱼ] be a square matrix of order m. If possible, let B and C be two
inverses of A. We shall show that B = C.
Since B is the inverse of A
AB = BA = I ... (1)
Since C is also the inverse of A
AC = CA = I ... (2)
Thus B = BI = B (AC) = (BA) C = IC = C
Theorem-4 :- If A and B are invertible matrices of the same order, then (AB)-1 = B-1 A-1.
Solution :-
Proof : From the definition of inverse of a matrix, we have
(AB) (AB)-1 = 1
A-1 (AB) (AB)-1 = A-1I [Pre multiplying both sides by A-1]
(A-1A) B (AB)-1 = A-1 [Since A-1 I = A-1]
IB (AB)-1 = A-1
B (AB)-1 = A-1
B-1 B (AB)-1 = B-1 A-1
I (AB)-1 = B-1 A-1
Hence, (AB)-1 = B-1 A-1
