Scalar Multiplication of a Matrix

The operation of multiplying variables by a constant scalar factor may properly be called scalar multiplication and the rule of multiplication of matrix by a scalar is that
the product of an m × n matrix A = [aij] by a scalar quantity c is the m × n matrix [bij] where bij = caij.

It is denoted by cA or Ac

For example:

c $$\begin{bmatrix} a_{1 1}& a_{1 2} & a_{1 3}\\ a_{2 1}& a_{2 2} & a_{2 3}\\ a_{3 1}& a_{3 2} & a_{3 3} \end{bmatrix}$$

= $$\begin{bmatrix} ca_{1 1}& ca_{1 2} & ca_{1 3}\\ ca_{2 1}& ca_{2 2} & ca_{2 3}\\ ca_{3 1}& ca_{3 2} & ca_{3 3} \end{bmatrix}$$

= $$\begin{bmatrix} a_{1 1}c& a_{1 2}c & a_{1 3}c\\ a_{2 1}c& a_{2 2}c & a_{2 3}c\\ a_{3 1}c& a_{3 2}c & a_{3 3}c \end{bmatrix}$$

= $$\begin{bmatrix} a_{1 1}& a_{1 2} & a_{1 3}\\ a_{2 1}& a_{2 2} & a_{2 3}\\ a_{3 1}& a_{3 2} & a_{3 3} \end{bmatrix}$$ c.

The product of an m × n matrix A = (aij)m, n by a scalar k where k ∈ F, the field of scalars, is a matrix B = (bij)m, n defined by bij = kaij, i = 1, 2, 3, ....., m : j = 1, 2, 3, ....., n and is written as B = kA.

Let A be an m × n matrix and k, p are scalars. Then the following results are obvious.

(i) k(pA) = (kp)A,

(ii) 0A = Om, n,

(iii) kOm, n = Om, n,

(iv) kIn = $$\begin{bmatrix} k & 0 & ... & 0\\ 0 & k & ... & 0\\ ... & ... & ... & ...\\ 0 & 0 & ... & k \end{bmatrix}$$,

(v) 1A = A, where 1 is the identity element of F.

The scalar matrix of order n whose diagonal elements are all k can be expressed as kIn.

In general, if c is any number (scalar or any complex number) and a is a matrix of order m × n, then the matrix cA is obtained by multiplying each element of the matrix A by the scalar c.

In other words, A = [aij]m × n

then, cA = [kij]m × n, where kij = caij

Examples on scalar multiplication of a matrix:

1. If A = $$\begin{bmatrix} 3 & 1\\ 2 & 0 \end{bmatrix}$$ and c = 3, then

cA = 3$$\begin{bmatrix} 3 & 1\\ 2 & 0 \end{bmatrix}$$

= $$\begin{bmatrix} 3 × 3 & 3 × 1\\ 3 × 2 & 3 × 0 \end{bmatrix}$$

= $$\begin{bmatrix} 9 & 3 \\ 6 & 0 \end{bmatrix}$$

2. If A = $$\begin{bmatrix} 0 & -1 & 5\\ -3 & 2 & 1\\ 2 & 0 & -4 \end{bmatrix}$$ and c = -5, then

cA = -5$$\begin{bmatrix} 0 & -1 & 5\\ -3 & 2 & 1\\ 2 & 0 & -4 \end{bmatrix}$$

= $$\begin{bmatrix} (-5) × 0 & (-5) × (-1) & (-5) × 5\\ (-5) × (-3) & (-5) × 2 & (-5) × 1\\ (-5) × 2 & (-5) × 0 & (-5) × (-4) \end{bmatrix}$$

= $$\begin{bmatrix} 0 & 5 & -25 \\ 15 & -10 & -5 \\ -10 & 0 & 20 \end{bmatrix}$$