19.4 - Multiplication of a Matrix by a scalar

In the previous section, we saw addition of matrices. In this section, we will see multiplication of matrix by a scalar.

This can be written in 3 steps:
1. In the previous section, we saw the industrialist and his two factories . Suppose that, the industrialist wants to triple the production in factory B.
2. The new quantities in factory B can be obtained simply by multiplying each element of matrix B by 3.
• We get:
3B = $3 \times\left[\begin{array}{r}        
15    &{25}    \\
30    &{37}    \\
60    &{42}    \\
\end{array}\right] =
\left[\begin{array}{r}        
3 × 15    &{3 × 25}    \\
3 × 30    &{3 × 37}    \\
3 × 60    &{3 × 42}    \\
\end{array}\right] =
\left[\begin{array}{r}        
45    &{75}    \\
90    &{111}    \\
180    &{126}    \\
\end{array}\right]        
$
3. Using symbols, we can write:
kA = k[a_ij]_m×n = [ka_ij]_m×n

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Negative of a matrix

This can be written in 3 steps:
1. We have seen the technique used for multiplication by a scalar. If the scalar is −1, we will get the negative of the matrix.
2. Let us see an example:
If A = $\left[\begin{array}{r}        
5    &{2}    \\
−3    &{7}    \\
x    &{4}    \\
\end{array}\right]$, then:
−A = $−1 \times\left[\begin{array}{r}        
5    &{2}    \\
−3    &{7}    \\
x    &{4}    \\
\end{array}\right] =
\left[\begin{array}{c}        
−1 × 5    &{−1 × 2}    \\
−1 × −3    &{−1 × 7}    \\
−1 × x    &{−1 × 4}    \\
\end{array}\right] =
\left[\begin{array}{r}        
−5    &{−2}    \\
3    &{−7}    \\
−x    &{−4}    \\
\end{array}\right]        
$
3. We can write:
If A is a given matrix, then negative of A = −A = (−1)A

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Difference of matrices

This can be written in 3 steps:
1. If A and B are two matrices, then the difference A−B is defined as:
A−B = A + (−B)
2. That means, to find the difference, we take the negative of B and then add it to A.
3. Let us see an example:
If A = $\left[\begin{array}{r}           
2    &{5}    &{−1}    \\
6    &{3}    &{9}    \\
\end{array}\right]           
$ and B = $\left[\begin{array}{r}           
11    &{6}    &{4}    \\
−3    &{12}    &{7}    \\
\end{array}\right]           
$, then find 2A−B
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2A−B}    & {~=~}    &{2\left[\begin{array}{r}    2 &{5} &{−1} \\ 6 &{3} &{9} \\ \end{array}\right]    − \left[\begin{array}{r}    11 &{6} &{4} \\ −3 &{12} &{7} \\ \end{array}\right]    }    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\begin{array}{r}


4
&{10}
&{−2}
\\ 12
&{6}
&{18}
\\ \end{array}\right]


+\left[\begin{array}{r}


−11
&{−6}
&{−4}
\\ 3
&{−12}
&{−7}
\\ \end{array}\right]


}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\left[\begin{array}{r}


4 – 11
&{10−6}
&{−2−4}
\\ 12+3
&{6−12}
&{18−7}
\\ \end{array}\right]


}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\left[\begin{array}{r}


−7
&{4}
&{−6}
\\ 15
&{−6}
&{11}
\\ \end{array}\right]


}    \\
\end{array}                           
$

---

Properties of scalar multiplication of a matrix

If A = [a_ij]_m×n and B = [b_ij]_m×n are two matrices of the same order m×n, then scalar multiplication will satisfy the following two properties:
(i) k(A+B) = kA + kB
(ii) (k+l)A = kA + lA

Part (i)
• Proof can be written in 10 steps:
1. A + B = [a_ij] + [b_ij] = [a_ij + b_ij]
We already saw this when we discussed matrix addition.
2. So k(A+B) = k[a_ij + b_ij]
• (a_ij + b_ij) is a number, written as the sum of two numbers.
• (a_ij + b_ij) represents each element in the sum matrix (A+B)
• We know that, for scalar multiplication, each element is to be multiplied by the scalar.
• So we get:
k[a_ij + b_ij] = [k(a_ij + b_ij)]
3. So the expression in (2) can be modified as:
k(A+B) = [k(a_ij + b_ij)]
4. a_ij and b_ij are numbers. So the expression in (3) can be modified as:
k(A+B) = [k a_ij + k b_ij]
5. Consider the R.H.S of the expression in (4).
It is the sum of two matrices: [k a_ij] and [k b_ij]
6. So the expression in (4) can be modified as:
k(A+B) = [k a_ij] + [k b_ij]
7. [k a_ij] is a matrix, which is obtained by multiplying the matrix [a_ij]  by the scalar k. That is: [k a_ij] = k[a_ij]
• Similarly, [k b_ij] is a matrix, which is obtained by multiplying the matrix [b_ij]  by the scalar k. That is: [k b_ij] = k[b_ij]
8. So the expression in (6) can be modified as:
k(A+B) = k [a_ij] + k [b_ij]
9. But k [a_ij] is kA. Similarly, k [b_ij] is kB.
10. So the expression in (8) can be modified as:
k(A+B) = kA + kB

Part (ii)
• Proof can be written in 6 steps:
1. (k+l)A = (k+l)[a_ij]
This is because, we are multiplying each element of A by the scalar (k+l).
2. When we do the multiplication, we get the new matrix: [(k+l)a_ij].
• So the expression in (1) can be modified as:
(k+l)A = [(k+l)a_ij]
3. Consider the R.H.S of the expression in (2).
k and l are scalars. a_ij is a number.
• So the expression in (2) can be modified as:
(k+l)A = [(k a_ij) + (l a_ij)]
4. Consider the R.H.S of the expression in (3). It is the sum of two matrices.
• So the expression in (3) can be modified as:
(k+l)A = [k a_ij] + [l a_ij]
5. Consider the R.H.S of the expression in (4).
    ♦ The matrix [a_ij] is being multiplied by the scalar k.
    ♦ The matrix [a_ij] is being multiplied by the scalar l.
• So the expression in (4) can be modified as:
(k+l)A = k[a_ij] + l[a_ij]
6. The expression in (5) is same as:
(k+l)A = kA + lA

---

Now we will see a solved example:
Solved example 19.9
If A = $\left[\begin{array}{r}       
8    &{0}    \\
4    &{-2}    \\
3    &{6}    \\
\end{array}\right]       
$ and B = $\left[\begin{array}{r}       
2    &{-2}    \\
4    &{2}    \\
-5    &{1}    \\
\end{array}\right]       
$, then find the matrix X, such that 2A + 3X = 5B
Solution:
• The given expression is: 2A + 3X = 5B
• This expression can be modified as shown below:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2A + 3X}    & {~=~}    &{5B}    \\
{~\color{magenta}    2    }    &{\implies}    &{2A + 3X – 2A}    & {~=~}    &{5B – 2A}    \\
{~\color{magenta}    3    }    &{\implies}    &{2A – 2A + 3X}    & {~=~}    &{5B – 2A}    \\
{~\color{magenta}    4    }    &{\implies}    &{O + 3X}    & {~=~}    &{5B – 2A}    \\
{~\color{magenta}    5    }    &{\implies}    &{3X}    & {~=~}    &{5B – 2A}    \\
{~\color{magenta}    6    }    &{\implies}    &{X}    & {~=~}    &{\frac{1}{3}(5B – 2A)}    \\
\end{array}                           
$

◼ Remarks:
2. The matrix -2A is added on both sides.
3. Matrix addition is commutative. So the position of -2A in the L.H.S can be changed.
4. -2A is the additive inverse of 2A. Their sum will be O.

• Now we can input the values of A and B to find X. This is shown below:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{X}    & {~=~}    &{\frac{1}{3} \left(5 \left[\begin{array}{r} 2&{-2}\\ 4&{2}\\ -5&{1}\\ \end{array}\right] - 2\left[\begin{array}{r} 8&{0}\\ 4&{-2}\\ 3&{6}\\ \end{array}\right]   \right)}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{3} \left(\left[\begin{array}{r} 10&{-10}\\ 20&{10}\\ -25&{5}\\ \end{array}\right] + \left[\begin{array}{r} -16&{0}\\ -8&{4}\\ -6&{-12}\\ \end{array}\right]   \right)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{3} \left[\begin{array}{r}

10-16
&{-10+0}
\\ 20 – 8
&{10+4}
\\ -31
&{5-12}
\\ \end{array}\right]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{3} \left[\begin{array}{r}

-6
&{-10}
\\ 12
&{14}
\\ -31
&{-7}
\\ \end{array}\right]

}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\left[\begin{array}{c}

-2
&{\frac{-10}{3}}
\\ 4
&{\frac{14}{3}}
\\ \frac{-31}{3}
&{\frac{-7}{3}}
\\ \end{array}\right]

}    \\
\end{array}$                           

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In the next section, we will see a few more solved examples. 

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