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