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Unit-3(Theorems)

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
    
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