Scalar Multiplication of a Matrix

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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})

`

10th Grade Math

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