We will discuss about the properties of scalar multiplication of a matrix.
If X and Y are two m × n matrices (matrices of the same order) and k, c and 1 are the numbers (scalars). Then the following results are obvious.
I. k(A + B) =
kA + kB
II. (k + c)A = kA + cA
III. k(cA) = (kc)A
IV. 1A = A
Proof: Let A = [a_{ij}] and B = [b_{ij}] are two m × n matrices.
I. k(A + B) = k([a_{ij}] + [b_{ij}])
= k[a_{ij} + b_{ij}], (by using the definition of addition of matrices)
= [k(a_{ij} + b_{ij})], (by using the definition of scalar multiplication of matrices)
= [ka_{ij} + kb_{ij}]
= [ka_{ij}] + [kb_{ij}]
= k[a_{ij}] + k[b_{ij}]
= kA + kB
Therefore, k(A + B) = kA + kB (proved).
II. (k + c)A = (k + c) [a_{ij}]
= [(k + c) (a_{ij})], (by using the definition of scalar multiplication of matrices)
= [ka_{ij} + ca_{ij}]
= [ka_{ij}] + [ca_{ij}]
= k[a_{ij}] + c[a_{ij}]
= kA + cA
Therefore, (k + c)A = kA + cA (proved).
III. k(cA) = k(c[a_{ij}])
= k[ca_{ij}], (by using the definition of scalar multiplication of matrices)
= [k(ca_{ij})]
= [(kc) a_{ij}], (by using the definition of scalar multiplication of matrices)
= (kc) [a_{ij}]
= (kc)A
Therefore, k(cA) = (kc)A (proved).
IV. 1A = 1[a_{ij}]
= [1 ∙ a_{ij}]
= [a_{ij}]
= A
Therefore, 1A = A (proved).
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