Question

Properties of Scalar Multiplication of a Matrix

Answer 1

Property 1.

In the addition of matrices, scalar multiplication follows distributive law, i.e., if k is an arbitrary scalar, then

k(A. + B) = kA + kB

Proof: Let A = [a_ij]_{m × n} and B = [b_ij]_{m × n}

∵ A + B=[a_ij + b_ij]_{m × n}

k(A + B) = k[a_ij + b_ij]_{m × n}„ = [ka_ij + kb_ij]_{m × n}

= [ka_ij]_{m × n} + [kb_ij]_{m × n} = k[a_ij]_{m × n} + k[b_ij]_{m × n}

= kA + kB

Property 2.

It k and l are scalars and A = [a_ij]_{m × n} is matrix, then

(k + l)A =kA + IA

Proof:

A = [a_ij]_{m × n} and k and l are scalars, then

(k + l)A = (k + l) [a_ij] = [(k + l)a_ij]

= [ka_ij] + [la_ij]

= k[a_ij] + l[a_ij] = kA + IA

Thus, (k + l)A = kA + lA

Property 3.

If k and 1 are scalars and A = [a_ij]_{m × n} is - matrix, then

(i) (kl)A = l(lA) = l(kA)

(ii) (- k)A = - (kA) = k(- A)

(iii) IA = A

(iv) (- 1)A = - A .

Proof :

(i) Since, k and l are scalars, so kl is also scalar then (kl) A also a matrix of order m × n. So, k(lA) and l(kA) are martices of order rn x n.

Thus, (kl)A and k(lA) are two matrices of same order m × n such that

(i) [(kl)A]_ij = (kl) a_ij

(By the definition of scalar multiplication)

⇒ [(kl)A]_ij =k(la_ij) (By associative multiplication)

⇒ [(kl)A]_ij = k(lA)_ij

(By definition of scalar multiplication)

⇒ [(kl)A]_ij = [k(lA)]_ij

(By definition of scalar multiplication where i = 1, 2, ........, m and j = 1, 2,..n)

Thus, by definition of equality of matrices (kl)A = k(lA)

Similarly, we can prove that

(kl)A = k(lA) = l(kA)

(ii) Putting l = -1 in (kl) A = k(lA) = l(kA)

(- k)A = k(- A) = - (kA)

(iii) Putting k = -1 in (- k) A = k(- A) = - (kA)

-1(-1)A =(-1)(-A)

= - (- 1.A) = 1A = A

∴ 1A = A

(iv) Putting k = 1 in (- k) A = k(~ A) = - (kA)

(-1) A = 1(- A) = - (1A) = - A

∴ (- 1)A = - A