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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