Higher Secondary Mathematics: 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: