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 = [a_{ij}] by a scalar quantity c is
the m × n matrix [b_{ij}] where b_{ij} = ca_{ij}.

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 = (a_{ij})_{m, n} by a scalar k where k ∈ F, the field of scalars, is a matrix B =
(b_{ij})_{m,
n} defined by b_{ij} = ka_{ij}, 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 = O_{m, n},

(iii) kO_{m, n} = O_{m, n},

(iv) k*I _{n}* = \(\begin{bmatrix} k & 0 & ... & 0\\ 0 &
k & ... & 0\\ ... & ... & ... & ...\\ 0 & 0 & ...
& k \end{bmatrix}\),

(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 k*I _{n}*.

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 = [a_{ij}]_{m × n}

then, cA =
[k_{ij}]_{m × n}, where k_{ij} = ca_{ij}

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

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

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

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