19.6 - Multiplication of Matrices

In the previous section, we completed a discussion on multiplication of matrix by a scalar. In this section, we will see multiplication of matrices.

Multiplication of two matrices can be explained using an example. It can be written in 5 steps:
1. Consider three students: Student X, Student Y and Student Z.
• They want to buy some notebooks, pens and pencils.
    ♦ X requires 8 notebooks, 3 pens and 2 pencils.  
    ♦ Y requires 7 notebooks, 5 pens and 3 pencils.
    ♦ Z requires 11 notebooks, 4 pens and 2 pencils.
• This is shown as the matrix A in fig.19.20 below:

Fig.19.20

2. The students go to a nearby store. there:
    ♦ Notebooks cost Rs. 25 each.
    ♦ Pens cost Rs. 12 each.
    ♦ Pencils cost Rs. 5 each.
• This is shown as the matrix B in fig.19.20 above.
3. Using the data in the matrices A and B, we can calculate the payments to be made.
• For student X,
   ♦ Cost of notebooks = 8 × 25
   ♦ Cost of pens = 3 × 12
   ♦ Cost of pencils = 2 × 5
• So total payment to be made by student X =  (8 × 25) + (3 × 12) + (2 × 5) = 246
• This is same as: (a₁₁ × b₁₁) + (a₁₂ × b₂₁) + (a₁₃ × b₃₁) = c₁₁.
• The sum 246, is the element c₁₁ of matrix C. This is shown in fig.19.20 above.
4. In this way, we can find the payment to be made by each student. The method is shown below:

$\begin{array}{ll} {~\color{magenta}    1    }    &{\text{Student X}}    &{(8 × 25) + (3 × 12) + (2 × 5)}    & {~=~}    &{246}    \\
{~\color{magenta}    2    }    &{\implies}    &{(a_{11} × b_{11})\,+\,(a_{12} × b_{21})\,+\,(a_{13} × b_{31})}    & {~=~}    &{c_{11}}    \\
{~\color{magenta}    3    }    &{\text{Student Y}}    &{(7 × 25) + (5 × 12) + (3 × 5)}    & {~=~}    &{250}    \\
{~\color{magenta}    4    }    &{\implies}    &{(a_{21} × b_{11})\,+\,(a_{22} × b_{21})\,+\,(a_{23} × b_{31})}    & {~=~}    &{c_{21}}    \\
{~\color{magenta}    5    }    &{\text{Student Z}}    &{(11 × 25) + (4 × 12) + (2 × 5)}    & {~=~}    &{333}    \\
{~\color{magenta}    6    }    &{\implies}    &{(a_{31} × b_{11})\,+\,(a_{32} × b_{21})\,+\,(a_{33} × b_{31})}    & {~=~}    &{c_{31}}    \\
\end{array}$

• The reader is advised to check the above steps and recognize the neat pattern that exists between a_ij, b_ij and c_ij.

5. It is clear that,
    ♦ Each row of matrix A is multiplied by the column of matrix B.
    ♦ Such a multiplication and subsequent summation, gives the matrix C.
We can write: AB = C

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Let us see another example. It can be written in 6 steps:

1. Consider the same three students in the previous example: Student X, Student Y and Student Z.
Their requirements are also the same. So matrix A is the same. It is shown in fig.19.21 below:

Fig.19.21

2. This time the students decide to check another store. So we will name the first store as I and the second store as II.
• At store II:
    ♦ Notebooks cost Rs. 24 each.
    ♦ Pens cost Rs. 14 each.
    ♦ Pencils cost Rs. 6 each.
• The prices at the two stores can be written together as a matrix. This is shown as the matrix D in fig.19.21 above.
3. Using the data in the matrices A and D, we can calculate the payments to be made.
• For student X,
   ♦ Cost of notebooks = 8 × 24
   ♦ Cost of pens = 3 × 14
   ♦ Cost of pencils = 2 × 6
• So total payment to be made by student X =  (8 × 24) + (3 × 14) + (2 × 6) = 246
• This is same as: (a₁₁ × d₁₂) + (a₁₂ × d₂₂) + (a₁₃ × d₃₂) = e₁₂.
• The sum 246, is the element e₁₂ of matrix E. This is shown in fig.19.21 above.
4. In this way, we can find the payment to be made by each student. The method is shown below:

$\begin{array}{ll} {~\color{magenta}    1    }    &{\text{Student X}}    &{(8 × 24) + (3 × 14) + (2 × 6)}    & {~=~}    &{246}    \\
{~\color{magenta}    2    }    &{\implies}    &{(a_{11} × d_{12})\,+\,(a_{12} × d_{22})\,+\,(a_{13} × d_{32})}    & {~=~}    &{e_{11}}    \\
{~\color{magenta}    3    }    &{\text{Student Y}}    &{(7 × 24) + (5 × 14) + (3 × 6)}    & {~=~}    &{256}    \\
{~\color{magenta}    4    }    &{\implies}    &{(a_{21} × d_{21})\,+\,(a_{22} × d_{22})\,+\,(a_{23} × d_{32})}    & {~=~}    &{e_{22}}    \\
{~\color{magenta}    5    }    &{\text{Student Z}}    &{(11 × 24) + (4 × 14) + (2 × 6)}    & {~=~}    &{332}    \\
{~\color{magenta}    6    }    &{\implies}    &{(a_{31} × d_{12})\,+\,(a_{32} × d_{22})\,+\,(a_{33} × d_{32})}    & {~=~}    &{e_{32}}    \\
\end{array}$

• In this way the second column of matrix E is calculated.
• The reader is advised to check the above steps and recognize the neat pattern that exists between a_ij, d_ij and e_ij.

5. It is clear that,
    ♦ Each row of matrix A is multiplied by the second column of matrix D.
    ♦ Such a multiplication and subsequent summation, gives the second column of matrix E. (the first column is same as in the previous example)
We can write: AD = E

6. Now we can compare the matrix C from the first example and matrix E from the second example.
• If the students go to the store II,
    ♦ They will suffer a loss of Rs.6/- in the case of student Y.
    ♦ They will get a gain of Rs.1/- in the case of student Z.
    ♦ So they will suffer a net loss of Rs.5/- 

---

Let us see one more example.

If A = $\left[\begin{array}{r}           
3    &{-4}    &{1}    \\
0    &{7}    &{4}    \\
\end{array}\right]           
$ and B = $\left[\begin{array}{r}       
4    &{9}    \\
-8    &{2}    \\
5    &{-2}    \\
\end{array}\right]       
$, then find AB

This time, we will avoid the detailed steps:
• Multiply green rectangle by the yellow rectangle. The summation will give c₁₁. See fig.19.22 below:

Fig.19.22

• Multiply green rectangle by the magenta rectangle. The summation will give c₁₂.
• Multiply red rectangle by the yellow rectangle. The summation will give c₂₁.
• Multiply red rectangle by the magenta rectangle. The summation will give c₂₂.

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Based on the three examples, we can write two important points:
1. We are given two matrices A and B.
• We will be able to calculate AB only if:
Number of columns in A = Number of rows in B
• In other words:
    ♦ If A is of the order (m×n),
    ♦ Then B must be of the order (n×p).
2. When we multiply A of order (m×n) and B of order (n×p), the order of the resulting AB will be (m×p).

---

Now we will see a solved example:

Solved example 19.13
Find AB if A = $\left[\begin{array}{r}       
14    &{12}    \\
7    &{8}    \\
\end{array}\right]       
$ and B = $\left[\begin{array}{r}           
5    &{4}    &{8}    \\
9    &{7}    &{3}    \\
\end{array}\right]           
$.
Solution:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{AB}    & {~=~}    &{\left[\begin{array}{r}

14
&{12}
\\ 7
&{8}
\\ \end{array}\right]

\left[\begin{array}{r}


5
&{4}
&{8}
\\ 9
&{7}
&{3}
\\ \end{array}\right]


}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\begin{array}{r}


14(5)+12(9)
&{14(4)+12(7)}
&{14(8)+12(3)}
\\ 7(5)+8(9)
&{7(4)+8(7)}
&{7(8)+8(3)}
\\ \end{array}\right]


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


70+108
&{56+84}
&{112+36}
\\ 35+72
&{28+56}
&{56+24}
\\ \end{array}\right]


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


178
&{140}
&{148}
\\ 107
&{84}
&{80}
\\ \end{array}\right]


}    \\
\end{array}$

---

Let us see some interesting facts. They can be written in 4 steps:
1. In the above solved example, we saw that, AB is possible.
• This is because:
Number of columns in A = Number of rows in B = 3.
2. Is BA possible?
    ♦ Number of columns in B = 3
    ♦ Number of rows in A = 2
• We see that:
Number of columns in B ≠ Number of rows in A.
• So it is not possible to find BA. In other words, BA is not defined.
3. So we can write:
"AB being defined", gives no guarantee that, BA is also defined.
4. What is the condition for both AB and BA to be defined?
• Answer can be written in 8 steps:
(i) Given two matrices A and B
    ♦ A is of the order (m×n)
    ♦ B is of the order (k×l)
(ii) Suppose that, AB is defined. Then n = k = u
(iii) For checking BA, we consider the orders (k×l) and (m×n).
(iv) For BA to be defined, l and m must be equal. That is., l = m = v
(v) Based of (ii) and (iv), we can write:
    ♦ order of A = (m×n) = (v×u) 
    ♦ order of B = (k×l) = (u×v)
(vi) So we can write:
• If both AB and BA is to be defined, then:
Order of A must be (v×u) and that of B must be (u×v)
(that is., u and v are interchanged)
(vii) For example, A and B are of the orders (3×4) and (4×3) respectively, then both AB and BA are defined.
(viii) A particular case arises when A and B are square matrices of the same order.
• In such a situation, u will be equal to v. Then, interchanging u and v will give the same order.
• So we can write:
If A and B are two square matrices of the same order, then both AB and BA are defined.

---

In the next section, we will see that multiplication of matrices is not commutative. 

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