20.16 - Miscellaneous Examples

In the previous section, we completed a discussion on determinants. We saw a solved example also. In this section, we will see some miscellaneous examples.

Solved example 20.27
If a, b, c are positive and unequal, show that the value of the determinant $\Delta ~=~\left |\begin{array}{r}                           
a    &{    b    }    &{    c    }    \\
b    &{    c    }    &{    a    }    \\
c    &{    a    }    &{    b    }    \\
\end{array}\right |
$ is negative.
Solution:
1. First we will simplify the given determinant:

◼ Remarks:
• 2 (magenta color): Apply R₁ → R₁ + R₂.
• 3 (magenta color): Apply R₁ → R₁ + R₃.
• 4 (magenta color): Apply C₃ → C₃ − C₁.
• 5 (magenta color): Apply C₂ → C₂ − C₁.
• 6 (magenta color): Expand along R₁.

2. Consider the above result. There are three terms:
(i) −(1/2)
(ii) (a+b+c)
(iii) (a−b)² + (a−c)² + (b−c)².

• Given that: a, b, c are +ve and unequal.
• So (ii) and (iii) cannot become -ve.
• Therefore, due to the presence of −(1/2), the result as a whole will become -ve.

Solved example 20.28
If a, b, c are in A.P, find the value of
$\Delta ~=~\left |\begin{array}{r}                           
2y+4    &{    5y+7    }    &{    8y+a    }    \\
3y+5    &{    6y+8    }    &{    9y+b    }    \\
4y+6    &{    7y+9    }    &{    10y+c    }    \\
\end{array}\right |
$.
Solution:

◼ Remarks:
• 2 (magenta color): Apply R₃ → R₃ − R₂.
• 3 (magenta color): Apply R₂ → R₂ − R₁.
• 4 (magenta color): Apply R₃ → R₃ − R₂.
• 5 (magenta color): Since, a, b, c are in A.P, we can put 2b = a+c.
• 6 (magenta color): All elements of R₃ are zeroes. So the value of the determinant is zero.

Solved example 20.29
Show that
$\Delta ~=~\left |\begin{array}{r}                         
(y+z)^2    &{    xy    }    &{    zx    }    \\
xy    &{    (x+z)^2    }    &{    yz    }    \\
xz    &{    yz    }    &{    (x+y)^2    }    \\
\end{array}\right | ~=~2xyz (x+y+z)^3
$.
Solution:

◼ Remarks:
• 2 (magenta color):
    ♦ Multiply R₁ by x
    ♦ Multiply R₂ by y
    ♦ Multiply R₃ by z
To balance these multiplications, the whole determinant should be multiplied by (1/xyz)
• 3 (magenta color):
Take out the common factors:
    ♦ x from C₁
    ♦ y from C₂
    ♦ z from C₃
• 4 (magenta color):
Apply two operations:
    ♦ C₂ → C₂ − C₁
    ♦ C₃ → C₃ − C₁
• 5 (magenta color):
    ♦ Apply the identity: a² − b² = (a+b)(a-b).
    ♦ This is applied to C₂ and C₃.
• 6 (magenta color): Take out (x+y+z) from C₁ and C₂.
• 7 (magenta color):
Apply R₁ → R₁ − R₂
• 8 (magenta color):
Apply R₁ → R₁ − R₃
• 9 (magenta color):
Apply C₂ → C₂ + (1/y)C₁
• 10 (magenta color):
Apply C₃ → C₃ + (1/z)C₁
• 11 (magenta color):
Expand along R₁.

Solved example 20.30
Use product
$\left [\begin{array}{r}                         
1    &{    -1    }    &{    2    }    \\
0    &{    2    }    &{    -3    }    \\
3    &{    -2    }    &{   4    }    \\
\end{array}\right ] \left [\begin{array}{r}                         
-2    &{    0    }    &{    1    }    \\
9    &{    2    }    &{    -3    }    \\
6    &{    1    }    &{   -2    }    \\
\end{array}\right ]$
to solve the system of equations
x - y + 2z = 1
2y − 3z = 1
3x − 2y + 4z = 2
Solution:
1. Use matrix multiplication to find the product.
• We get:
$\left [\begin{array}{r}                         
1    &{    -1    }    &{    2    }    \\
0    &{    2    }    &{    -3    }    \\
3    &{    -2    }    &{   4    }    \\
\end{array}\right ] \left [\begin{array}{r}                         
-2    &{    0    }    &{    1    }    \\
9    &{    2    }    &{    -3    }    \\
6    &{    1    }    &{   -2    }    \\
\end{array}\right ]~ = \left [\begin{array}{r}                         
1    &{    0    }    &{   0    }    \\
0    &{    1    }    &{   0    }    \\
0    &{    0    }    &{   1    }    \\
\end{array}\right ]
$

2. The product is an identity matrix. So it is clear that:
Inverse of $\left [\begin{array}{r}                         
1    &{    -1    }    &{    2    }    \\
0    &{    2    }    &{    -3    }    \\
3    &{    -2    }    &{   4    }    \\
\end{array}\right ]$ is $\left [\begin{array}{r}                         
-2    &{    0    }    &{    1    }    \\
9    &{    2    }    &{    -3    }    \\
6    &{    1    }    &{   -2    }    \\
\end{array}\right ]$

3. The given system can be written in the form AX = B.
$A = \left [\begin{array}{r}                         
1    &{    -1    }    &{    2    }    \\
0    &{    2    }    &{    -3    }    \\
3    &{    -2    }    &{   4    }    \\
\end{array}\right ],~X = \left[\begin{array}{r}       
x        \\
y        \\
z        \\
\end{array}\right]~~ \text{and}~~B = \left[\begin{array}{r}                           
1        \\
1        \\
2        \\
\end{array}\right]
$

4. So X = A⁻¹ B.
• Check whether A⁻¹ exists:
We have already obtained the inverse. Therefore, A⁻¹ exists.

5. Use matrix multiplication to find A⁻¹B.
• We get: X = A⁻¹ B =
$\left [\begin{array}{r}                         
-2    &{    0    }    &{    1    }    \\
9    &{    2    }    &{    -3    }    \\
6    &{    1    }    &{   -2    }    \\
\end{array}\right ]~\left[\begin{array}{r}                           
1        \\
1        \\
2        \\
\end{array}\right]~ = \left[\begin{array}{r}                        0        \\
5        \\
3        \\
\end{array}\right]
$

6. So the solution is: x = 0, y = 5 and z = 3

Solved example 20.31
Prove that
$ \Delta ~=~ \left |\begin{array}{r}                         
a+bx    &{    c+dx    }    &{    p+qx    }    \\
ax+b    &{    cx+d    }    &{    px+q    }    \\
u    &{    v    }    &{   w    }    \\
\end{array}\right | ~=~ (1 - x^2) \left |\begin{array}{r}                         
a    &{    c    }    &{    p    }    \\
b    &{    d    }    &{    q    }    \\
u    &{    v    }    &{   w    }    \\
\end{array}\right |$
Solution:
1. First we will split the given matrix by applying property V.

◼ Remarks:
• 2 (magenta color):
We split R₁ so that, a, c and p are obtained in the first row. These are the elements that we want in the R₁ of the final result.

2. Now we simplify |A|:

◼ Remarks:
• 2 (magenta color): We split R₂ of |A|.
• 3 (magenta color): Consider the first determinant in (2). Every element in R₂ is proportional to the corresponding elements in R₁, by the same ratio 'x'. So this determinant becomes zero.

3. Next we simplify |B|:

◼ Remarks:
• 2 (magenta color): We split R₂ of |B|.
• 3 (magenta color): Consider the second determinant in (2). Every element in R₁ is proportional to the corresponding elements in R₂, by the same ratio 'x'. So this determinant becomes zero.
• 4 (magenta color): We take out the common factor 'x' from R₁ and R₂.
• 5 (magenta color): We want the elements a, c and p in R₁. So we interchange R₁ and R₂. The sign of the determinant will change when the two rows are interchanged.

4. Finally we add |A| and |B|. We get:

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The link below gives a few more examples:

Miscellaneous Examples

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In the next section, we will see Continuity and Differentiability.

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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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19.9 - 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.

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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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19.1 - Types of Matrices

In the previous section, we saw some basic details about matrices. In this section, we will see some solved examples. Later in this section, we will see types of matrices.

Solved example 19.1
The following table gives the number of men and women workers in four factories I, II, III and IV.

Represent the above information in the form of a 4 × 2 matrix. Interpret the element in the third row and second column.
Solution:
1. A 4 × 2 matrix has four rows and two columns.
2. In the given problem, there are four factories. So each factory will have a unique row
3. There are two sets of workers. Men and women. So each set will have a unique column.
4. Thus we get the following matrix:
$\left[\begin{array}{r}       
{30}    &{25}    \\
{25}    &{31}    \\
{27}    &{26}    \\
{19}    &{21}    \\
\end{array}\right]$
5. Interpretation of element:
• Third row belongs to the third factory.
• Second column belongs to the women workers.
• So this element is the number of women workers in the third factory.

Solved example 19.2
If a matrix has 8 elements, what are the possible orders it can have?
Solution:
1. The general form of the order is m × n
2. The product of m and n = Number of elements in the matrix = 8
• Thus the possible values of m and n can be written as follows:
8 = (1 × 8) =  (2 × 4) =  (4 × 2) =  (8 × 1)
3. So the possible orders are:
(1 × 8) ⇒ 1 row and 8 columns.
(2 × 4) ⇒ 2 rows and 4 columns.
(4 × 2) ⇒ 4 rows and 2 columns.
(8 × 1) ⇒ 8 rows and 1 column.

Solved example 19.3
Construct a 3 × 2 matrix whose elements are given by $a_{ij} = \frac{1}{2}|i - 3j|$
Solution:
1. In the 3 × 2 matrix, there will be 3 rows and two columns.
2. Elements in the first row are:
   ♦ $a_{11} = \frac{1}{2}|1 - 3 × 1| = \frac{1}{2}| -2| = \frac{1}{2}× 2 = 1$

   ♦ $a_{12} = \frac{1}{2}|1 - 3 × 2| = \frac{1}{2}| -5| = \frac{1}{2}× 5 = \frac{5}{2}$

3. Elements in the second row are:
   ♦ $a_{21} = \frac{1}{2}|2 - 3 × 1| = \frac{1}{2}| -1| = \frac{1}{2}× 1 = \frac{1}{2}$

   ♦ $a_{22} = \frac{1}{2}|2 - 3 × 2| = \frac{1}{2}| -4| = \frac{1}{2}× 4 = 2$

4. Elements in the third row are:
   ♦ $a_{31} = \frac{1}{2}|3 - 3 × 1| = \frac{1}{2}| 0| = \frac{1}{2}× 0 = 0$

   ♦ $a_{32} = \frac{1}{2}|3 - 3 × 2| = \frac{1}{2}| -3| = \frac{1}{2}× 3 = \frac{3}{2}$

5. Thus, the required matrix is:
$\left[\begin{array}{r}       
{1}    &{\frac{5}{2}}    \\
{\frac{1}{2}}    &{2}    \\
{0}    &{\frac{3}{2}}    \\
\end{array}\right]       
$

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Types of Matrices

• We have to learn about 7 types of matrices.
I. Column matrix
• This can be written in 3 steps:
1. If a matrix has only one column, then it is called a column matrix.
• The matrix A in fig.19.9 below is an example.

Fig.19.9

2. The order of a column matrix will be in the form m × 1.
• So A is a 4 × 1 matrix.
3. The general form of a column matrix can be written as:
$A = \left[a_{ij} \right]_{m × 1}$

II. Row matrix
• This can be written in 3 steps:
1. If a matrix has only one row, then it is called a row matrix.
• The matrix B in fig.19.9 above is an example.
2. The order of a row matrix will be in the form 1 × n.
• So B is a 1 × 4 matrix.
3. The general form of a row matrix can be written as:
$B = \left[b_{ij} \right]_{1 × n}$

III. Square matrix
• This can be written in 3 steps:
1. If, in a matrix, the number of rows is equal to the number of columns, then it is called a square matrix.
• The matrix C in fig.19.9 above is an example.
2. The order of a row matrix will be in the form m × m.
• So C is a 4 × 4 matrix.
3. The general form of a square matrix can be written as:
$C = \left[c_{ij} \right]_{m × m}$
• We say that, C is a square matrix of order m.

IV. Diagonal matrix
• This can be written in 5 steps:
1. Consider any square matrix.
• We can think about a 'diagonal' from the top left element to the bottom right element.
2. All elements which lie along this diagonal, are called diagonal elements.
• All the remaining elements are called non-diagonal elements.
3. In a square matrix, if all the non-diagonal elements are zero, then it is called a diagonal matrix.
4. We can write the general form of a diagonal matrix as follows:
$D = \left[d_{ij} \right]_{m × m}$ is a diagonal matrix if d_ij = 0 when i ≠ j.
5. Fig.19.10 below shows some diagonal matrices.

Fig.19.10

• In the above fig.,
    ♦ A is a diagonal matrix of order 1.
    ♦ B is a diagonal matrix of order 2.
    ♦ C is a diagonal matrix of order 3.

V. Scalar matrix
• This can be written in 3 steps:
1. Consider any diagonal matrix.
• In that diagonal matrix, if all the diagonal elements are equal, then it is called a scalar matrix.
2. We can write the general form of a scalar matrix as follows:
• A square matrix $A = \left[a_{ij} \right]_{m × m}$ is a scalar matrix if two conditions are satisfied:
    ♦ a_ij = 0 when i ≠ j.
    ♦ a_ij = k when i = j, where k is a constant. 
3. Fig.19.11 below shows some scalar matrices.

Fig.19.11

• In the above fig.,
    ♦ A is a scalar matrix of order 1.
    ♦ B is a scalar matrix of order 2.
    ♦ C is a scalar matrix of order 3.

VI. Identity matrix
• This can be written in 5 steps:
1. Consider any scalar matrix.
• In that scalar matrix, if k = 1, then it is called an identity matrix.
2. We can write the general form of an identity matrix as follows:
• A square matrix $A = \left[a_{ij} \right]_{m × m}$ is an identity matrix if two conditions are satisfied:
    ♦ a_ij = 0 when i ≠ j.
    ♦ a_ij = 1 when i = j. 
3. Fig.19.12 below shows some identity matrices.

Fig.19.12

• In the above fig.,
    ♦ A is an identity matrix of order 1.
    ♦ B is an identity matrix of order 2.
    ♦ C is an identity matrix of order 3.
4. Denoting an identity matrix:
• If the order of an identity matrix is n, then we denote that matrix as I_n.
• So in the fig.19.12 above,
    ♦ A = I₁.
    ♦ B = I₂.
    ♦ C = I₃.
• If the order is clear from the context, we simply denote it as I.
5. Note that:
• Every identity matrix, is a scalar matrix.
• But every scalar matrix need not be an identity matrix.

VII. Zero matrix
• This can be written in 3 steps:
1. Consider any matrix (it need not be a square matrix).
• In that matrix, if all the elements are zero, then it is called a zero matrix.
2. Fig.19.13 below shows some zero matrices.

Fig.19.13

3. Denoting an identity matrix:
• We denote a zero matrix by O. It's order will be clear from the context.

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In the next section, we will see equality of matrices. 

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