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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