19.16 - Miscellaneous Examples

In the previous section, we completed a discussion on inverse matrix. In this section, we will see some miscellaneous examples.

Solved Example 19.19
If A = $\left[\begin{array}{r}               
\cos \theta     &{    \sin \theta    }    \\
-\sin \theta     &{    \cos \theta    }  \\
\end{array}\right]               
$, then prove that Aⁿ = $\left[\begin{array}{r}               
\cos n \theta     &{    \sin n \theta    }    \\
-\sin n \theta     &{    \cos n \theta    }  \\
\end{array}\right]               
$, where n is any natural number.
Solution:
We will prove this by using the principle of mathematical induction. Details can be seen here.

1. For any natural number, let P(n) denote the given statement. Then we can write:

P(n): If A = $\left[\begin{array}{r}               
\cos \theta     &{    \sin \theta    }    \\
-\sin \theta     &{    \cos \theta    }  \\
\end{array}\right]               
$, then Aⁿ = $\left[\begin{array}{r}               
\cos n \theta     &{    \sin n \theta    }    \\
-\sin n \theta     &{    \cos n \theta    }  \\
\end{array}\right]               
$

2. Basic step: (n=1)

• We see that, the statement is true when n = 1

3. Inductive step: (n=k) and (n= k+1)

(i) n = k

P(k): If A = $\left[\begin{array}{r}               
\cos \theta     &{    \sin \theta    }    \\
-\sin \theta     &{    \cos \theta    }  \\
\end{array}\right]               
$, then A^k = $\left[\begin{array}{r}               
\cos k \theta     &{    \sin k \theta    }    \\
-\sin k \theta     &{    \cos k \theta    }  \\
\end{array}\right]               
$

• We will assume that, this statement is true. 

(ii) n = k+1
• We need to prove the statement:
P(k+1): If A = $\left[\begin{array}{r}               
\cos \theta     &{    \sin \theta    }    \\
-\sin \theta     &{    \cos \theta    }  \\
\end{array}\right]               
$, then A^k+1 = $\left[\begin{array}{r}               
\cos (k+1) \theta     &{    \sin (k+1) \theta    }    \\
-\sin (k+1) \theta     &{    \cos (k+1) \theta    }  \\
\end{array}\right]               
$

• This can be proved as follows:

◼ Remarks:
In (2 magenta color), we are able to replace A^k. This is because, we assumed that, P(k) is true.

4. Thus P(k+1) is true whenever P(k) is true. Hence by the principle of mathematical induction, P(n) is true for any natural number n.

Solved Example 19.20
If A and B are symmetric matrices of the same order, then show that AB is symmetric if and only if A and B are commute, that is AB = BA
Solution:
1. If AB is to be symmetric, then it should be equal to it’s transpose. That is:
AB = (AB)'.
2. We know that, (AB)' = B'A'.
3. So the result in (1) becomes:
AB is symmetric if and only if AB = B'A'.
4. But given that, A and B are symmetric matrices. So we get:
A' = A and B' = B
5. So the result in (3) becomes:
AB is symmetric if and only if AB = BA.

---

• We can show the converse also:
AB = BA, if and only if AB is symmetric.
This can be shown in 3 steps:
1. If AB = BA, then we can write: AB = B'A'.
• This is because, A and B are said to be symmetric matrices. So A' = A and B' = B
2. We know that (AB)' = B'A'.
• So the result in (1) becomes:
If AB = BA, then AB = (AB)'.
3. But AB = (AB)' indicates that, AB is symmetric.
• So the result in (2) becomes:
If AB = BA, then AB is symmetric.

Solved Example 19.21
Let $A = \left[\begin{array}{r}               
2     &{    -1    }    \\
3     &{    4    }  \\
\end{array}\right] ,~ B = \left[\begin{array}{r}               
5     &{    2    }    \\
7     &{    4    }  \\
\end{array}\right],~ C = \left[\begin{array}{r}               
2     &{    5    }    \\
3     &{    8    }  \\
\end{array}\right]$.
Find a matrix D such that CD – AB = O
Solution:
1. Finding the order of D:
• Both A and B are 2 × 2 matrices. So AB will be a 2 × 2 matrix.
• AB is being subtracted from CD. So CD will be a 2 × 2 matrix.
• In CD, the matrix C is a 2 × 2 matrix. So we have two points:
    ♦ CD is a 2 × 2 matrix.
    ♦ C is a 2 × 2 matrix.
• Let D be of the order m × n.
• Comparing the orders of C and D to form CD, we see that:
    ♦ m must be 2
    ♦ n must be 2
• So the order of D is 2 × 2

2. Let $D = \left[\begin{array}{r}               
a     &{    b    }    \\
c     &{    d    }  \\
\end{array}\right]$.

3. Given that, CD – AB = O
This can be rearranged as shown below:

◼ Remarks:
Magenta 2: We add AB on both sides.

4. By equality of matrices, we get four equations:
(i) 2a + 5c = 3
(ii) 2b + 5d = 0
(iii) 3a + 8c = 43
(iv) 3b + 8d = 22

5. Now we can solve the equations:
• Solving (i) and (iii), we get: a = -191 and c = 77
• Solving (ii) and (iv), we get: b = -110 and d = 44

6. So the required matrix can be written as:
$D = \left[\begin{array}{r}               
a     &{    b    }    \\
c     &{    d    }  \\
\end{array}\right] ~=~ \left[\begin{array}{r}               
-191     &{    -110    }    \\
77     &{    44    }  \\
\end{array}\right]$.

Alternate method:
Since all matrices involved are square matrices, we can apply inverse.

1. Given that, CD – AB = O
This can be rearranged as shown below:

◼ Remarks:
Magenta 2: We add AB on both sides.
Magenta 4: We pre multiply both sides by C^-1.

2. So our next task is to find C^-1.

3. Our next task is to find AB:

4. Our final task is to find D:

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The link below gives a few more examples:

Miscellaneous Exercise

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In the next chapter, we will see Determinants.

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19.10 - Transpose of a Matrix

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

Transpose of a Matrix

This can be written in 9 steps:
1. Suppose that, we are given a matrix A.
• Based on A, we can write a new matrix A' in such a way that:
    ♦ Rows in A, are the columns in A'.
    ♦ Columns in A, are the rows in A'.
2. In other words,
    ♦ First row of A, becomes the first column in A'.   
    ♦ Second row of A, becomes the second column in A'.   
    ♦ Third row of A, becomes the third column in A'.   
    ♦ so on . . .
3. This is same as:
    ♦ First column of A, becomes the first row in A'.   
    ♦ Second column of A, becomes the second row in A'.   
    ♦ Third column of A, becomes the third row in A'.
    ♦ so on . . .
4. Let us see an example:
If A = $\left[\begin{array}{r}               
-7    &{    0    }    \\
8    &{    \sqrt{2}    }    \\
-{\frac{1}{3}}    &{    3    }    \\
\end{array}\right]               
$, then A' = $\left[\begin{array}{r}                                       
-7    &{    8    }    &{    -{\frac{1}{3}}    }       \\
0    &{    \sqrt{2}    }    &{    3    }        \\
\end{array}\right]                                       
$
5. Consider any element, say $-{\frac{1}{3}}$ in A. It’s position is: row 3, column 1.
• But in A', the position of $-{\frac{1}{3}}$ is: row 1, column 3.
    ♦ That is., the row and column are interchanged.
6. This interchanging of rows and columns is applicable to all elements in general.
• We can write:
    ♦ In the original matrix, an element is at position i,j.
    ♦ Then in the new matrix, that element will be at the position j,i.
7. Also note that:
If the order of the original matrix is m×n, then the order of the new matrix will be n×m.
8. Based on the above steps, we can write the definition:
If A = [a_ij]_m×n, then A' = [a_ji]_n×m .
9. The new matrix A' is called the transpose of A. It can be denoted either as A' or A^T.

---

Properties of transpose of matrices

• We have to be familiar with four properties:
For any matrices A and B of suitable orders, we have
I. (A')' = A                II. (kA)' = kA' (where k is any constant)
III. (A+B)' = A' + B'        IV. (AB)' = B' A'.

[We mention ‘suitable orders’ because, any two given matrices cannot be added or multiplied. If we want (A+B), then A and B must be of the same order. Similarly, if we want (AB), then number of columns in A must be equal to the number of rows in B]

• We will see the proofs of the four properties in higher classes. At present, we will verify them using examples.
Let A = $\left[\begin{array}{r}                       
9    &{    6    }    &{    -4    }\\
0    &{    \sqrt{2}    }    &{    1    }\\
\end{array}\right]                       
$ and B = $\left[\begin{array}{r}                       
-1    &{    2    }    &{    5    }\\
11    &{    7    }    &{    3    }\\
\end{array}\right] $.

1. Verifying property I:

2. Verifying property II:

• From (4) and (6), we get:
(kB)' = kB' (where k is any constant) 

3. Verifying property III:

• From (5) and (7), we get:
(A+B)' = A' + B'

4. Verifying property IV:
Let X = $\left[\begin{array}{r}                       
9    &{    6    }    &{    -4    }\\
0    &{    {2}    }    &{    1    }\\
\end{array}\right]                       
$ and Y = $\left[\begin{array}{r}                       
-1    &{    2    }    &{    5    }\\
11    &{    7    }    &{    3    }\\
7    &{    1    }    &{    4    }\\
\end{array}\right] $.

• From (3) and (7), we get:
(XY)' = Y' X'
(note that, positions of X and Y are interchanged)

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In the next section, we will see symmetric and skew symmetric matrices.

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19.9 - Multiplicative Identity

In the previous section, we saw two properties of multiplication of matrices. In this section, we will see the third property.

Property III: The existence of multiplicative identity
• Consider any square matrix A of the order (m×m).
• We can write an identity matrix I of the same order.
That I will satisfy the equation: AI = IA = A
• Let us see an example. It can be written in 4 steps:

1. Let A = $\left[\begin{array}{r}                           
-8    &{    2    }    &{    5    }    \\
0    &{    -7    }    &{    -3    }    \\
3    &{    2    }    &{    4    }    \\
\end{array}\right]$

• Then I = $\left[\begin{array}{r}                           
1    &{    0    }    &{    0    }    \\
0    &{    1    }    &{    0    }    \\
0    &{    0    }    &{    1    }    \\
\end{array}\right]                           
$               

2. First we find AI:

3. Next we find IA:

4. Based on (2) and (3), we can write:
AI = IA = A

◼ We will see the actual proof in higher classes.

---

Now we will see a solved example.

Solved example 19.14
If A = $\left[\begin{array}{r}                           
1    &{    3    }    &{    2    }    \\
2    &{    0    }    &{    -1    }    \\
1    &{    2    }    &{    3    }    \\
\end{array}\right]                           
$, then show that A³ - 4A² - 3A + 11I = O
Solution:
1. First we will write A²:

2. Next we will write A³:

3. Next we will write 4A²:

3. Next we will write 3A:

4. Finally we will write 11I:

5. Substituting the values, we get:

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The link below gives a few more solved examples:

Exercise 19.2

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In the next section, we will see transpose of a matrix.

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19.8 - Properties of Multiplication of Matrices

In the previous section, we saw that, multiplication of matrices is not commutative. In this section, we will see some properties of multiplication of matrices.

Multiplication of matrices possesses three properties.
Property I: The associative law
• If A, B and C are three matrices, then (AB)C = A(BC)
• For this property to be applicable, the following matrices should be defined:
AB, BC, A(BC) and (AB)C.
• Let us see an example. It can be written in 4 steps:

1. Let the three matrices be:
A = $\left[\begin{array}{r}                           
7    &{    -4    }    &{    3    }    \\
-2    &{    6    }    &{    8    }    \\
-3    &{    9    }    &{    -2    }    \\
\end{array}\right]                           
$, B = $\left[\begin{array}{r}               
2    &{    1    }    \\
-4    &{    6    }    \\
5    &{    3    }    \\
\end{array}\right]               
$ and C = $\left[\begin{array}{r}                                       
3    &{    5    }    &{    9    }    &{    0    }    \\
6    &{    7    }    &{    4    }    &{    2    }    \\
\end{array}\right]                                       
$

2. First we find (AB)C:

3. Next we find A(BC):

4. Based on (2) and (3), we can write:
(AB)C = A(BC)

◼ We will see the actual proof in higher classes.

Property II: The distributive law
• If A, B and C are three matrices, then:
(i) A(B+C) = AB + AC
(ii) (A+B)C = AC + BC
• For this property to be applicable, the following matrices should be defined:
AB, BC, AC, A(B+C) and (A+B)C.

• Let us see an example for part (i). It can be written in 4 steps:

1. Let the three matrices be:
A = $\left[\begin{array}{r}                           
10    &{    -7    }    &{    14    }    \\
7    &{    31    }    &{    3    }    \\
\end{array}\right]$,
B = $\left[\begin{array}{r}               
-7    &{    0    }    \\
8    &{    -2    }    \\
6    &{    3    }    \\
\end{array}\right]$
and C = $\left[\begin{array}{r}                
3    &{    -4    }    \\
12    &{    15    }    \\
9    &{    8    }    \\
\end{array}\right]$

2. First we find A(B+C):

3. Next we find AB + AC:

4. Based on (2) and (3), we can write:
A(B+C) = AB+AC

◼ We will see the actual proof in higher classes.

---

• Let us see an example for part (ii). It can be written in 4 steps:

1. Let the three matrices be:
A = $\left[\begin{array}{r}                           
-8    &{    2    }    &{    5    }    \\
0    &{    -7    }    &{    -3    }    \\
3    &{    2    }    &{    4    }    \\
\end{array}\right]$,
B = $\left[\begin{array}{r}                            
3    &{    9    }    &{    12    }    \\
7    &{    -4    }    &{    5    }    \\
8    &{    6    }    &{    15    }    \\
\end{array}\right]$
and C = $\left[\begin{array}{r}                                       
6     \\
4    \\
-7     \\
\end{array}\right]                                       
$

2. First we find (A+B)C:

3. Next we find AC+BC:

4. Based on (2) and (3), we can write:
(A+B)C = AC + BC

◼ We will see the actual proof in higher classes.

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In the next section, we will see the third property. 

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19.7 - Non-commutativity of Multiplication of Matrices

In the previous section, we saw the basics about multiplication of matrices. In this section, we will see that, multiplication of matrices is not commutative.

This can be explained in 12 steps:
1. If AB = BA for all A and B, we can say that, multiplication is commutative.
2. If we can show atleast one example where AB is not equal to BA, we will be able to say that, multiplication of matrices is not commutative.
3. Let A = $\left[\begin{array}{r}           
3    &{1}    &{-7}    \\
-5    &{10}    &{4}    \\
\end{array}\right]           
$ and B = $\left[\begin{array}{r}       
12    &{4}    \\
2    &{3}    \\
6    &{8}    \\
\end{array}\right]       
$.
• A is of the order 2×3
• B is of the order 3×2
• So both AB and BA are defined.
4. AB can be calculated as shown below:

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


3
&{1}
&{-7}
\\ -5
&{10}
&{4}
\\ \end{array}\right]


\left[\begin{array}{r}

12
&{4}
\\ 2
&{3}
\\ 6
&{8}
\\ \end{array}\right]

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

3(12)+1(2)+(-7)(6)
&{3(4)+1(3)+(-7)(8)}
\\ (-5)(12)+10(2)+4(6)
&{(-5)(4)+10(3)+4(8)}
\\ \end{array}\right]

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

-4
&{-41}
\\ -16
&{42}
\\ \end{array}\right]

}    \\
\end{array}$                           

5. Also, BA can be calculated as shown below:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{BA}    & {~=~}    &{\left[\begin{array}{r} 12&{4}\\ 2&{3}\\ 6&{8}\\ \end{array}\right]\left[\begin{array}{r} 3&{1}&{-7}\\ -5&{10}&{4}\\ \end{array}\right]}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\begin{array}{r}


12(3)+4(-5)
&{12(1)+4(10)}
&{12(-7)+4(4)}
\\ 2(3)+3(-5)
&{2(1)+3(10)}
&{2(-7)+3(4)}
\\ 6(3)+8(-5)
&{6(1)+8(10)}
&{6(-7)+8(4)}
\\ \end{array}\right]


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


16
&{52}
&{-68}
\\ -9
&{32}
&{-2}
\\ -22
&{86}
&{-10}
\\ \end{array}\right]


}    \\
\end{array}$                           

6. We see that AB ≠ BA.
• Here, AB is of the order 2×2 and BA is of the order 3×3. So AB will never be equal to BA.
7. There may be some cases where AB and BA are of the same order. Let us see such an example:
8. Let A = $\left[\begin{array}{r}           
1    &{0}      \\
0    &{-1}   \\
\end{array}\right]           
$ and B =  $\left[\begin{array}{r}       
0    &{1}    \\
1    &{0}    \\
\end{array}\right]       
$.
9. Then AB can be calculated as shown below:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{AB}    & {~=~}    &{\left[\begin{array}{r}           
1    &{0}      \\
0    &{-1}   \\
\end{array}\right]
\left[\begin{array}{r}       
0    &{1}    \\
1    &{0}    \\
\end{array}\right] }    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\begin{array}{r}

1(0)+0(1)
&{1(1)+0(0)}
\\ (0)(0)+(-1)(1)
&{0(1)+(-1)(0)}
\\ \end{array}\right]

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

0
&{1}
\\ -1
&{0}
\\ \end{array}\right]

}    \\
\end{array}$   
10. Also, BA can be calculated as shown below:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{BA}    & {~=~}    &{
\left[\begin{array}{r}       
0    &{1}    \\
1    &{0}    \\
\end{array}\right] \left[\begin{array}{r}           
1    &{0}      \\
0    &{-1}   \\
\end{array}\right]}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\left[\begin{array}{r}

0(1)+1(0)
&{0(0)+1(-1)}
\\ 1(1)+0(0)
&{1(0)+0(-1)}
\\ \end{array}\right]

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

0
&{-1}
\\ 1
&{0}
\\ \end{array}\right]

}    \\
\end{array}$
11. Here, AB is of the order 2×2 and BA is also of the order 2×2. We see that AB ≠ BA
12. We can write:
Even if AB and BA are of the same order, we can show at least one example where AB ≠ BA. So multiplication of matrices is not commutative.

---

If A and B are both diagonal matrices, and of the same order, then AB will be equal to BA. Two examples are shown below:

Fig.19.23

• The reader may write the calculation steps of the above examples in his/her notebooks.

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Zero matrix as a product of two non zero matrices

• If a and b are two real numbers and ab=0, then a or b has to be zero.
• But this is not applicable for matrices. An example is shown below:

$\left[\begin{array}{r}       
0    &{-1}    \\
0    &{2}    \\
\end{array}\right]\left[\begin{array}{r}       
3    &{5}    \\
0    &{0}    \\
\end{array}\right] = \left[\begin{array}{r}       
0    &{0}    \\
0    &{0}    \\
\end{array}\right]   
$                           

• So we can write:
If the product of two matrices is a zero matrix, it is not necessary that, one of the matrices is a zero matrix.

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In the next section, we will see properties of multiplication of matrices. 

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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

---

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.

---

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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