Referring to the AP Inter 2nd Year Maths Study Material Chapter 3 Matrices Exercise 3c Solutions makes it easier to understand complex problems.
AP Inter 2nd Year Maths Matrices Solutions Exercise 3c
I.
Question 1.
Find the transpose of matrix \(\left[\begin{array}{c}
5 \\
\frac{1}{2} \\
-1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{c}
5 \\
\frac{1}{2} \\
-1
\end{array}\right]\) ⇒ A’ = \(\left[\begin{array}{lll}
5 & \frac{1}{2} & -1
\end{array}\right]\)
Question 2.
Find the transpose of matrix \(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\) ⇒ A’ = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 3
\end{array}\right]\)
Question 3.
Find the transpose of matrix \(\left[\begin{array}{ccc}
-1 & 5 & 6 \\
\sqrt{3} & 5 & 6 \\
2 & 3 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
-1 & 5 & 6 \\
\sqrt{3} & 5 & 6 \\
2 & 3 & -1
\end{array}\right]\) ⇒ A’ = \(\left[\begin{array}{ccc}
-1 & \sqrt{3} & 2 \\
5 & 5 & 3 \\
-6 & 6 & -1
\end{array}\right]\)
Question 4.
If A’ = \(\left[\begin{array}{cc}
-2 & 3 \\
1 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
-1 & 0 \\
1 & 2
\end{array}\right]\), then find (A + 2B)’
Solution:
Let A’ = \(\left[\begin{array}{cc}
-2 & 3 \\
1 & 2
\end{array}\right]\) ⇒ A = (A’)’ = \(\left[\begin{array}{cc}
-2 & 1 \\
3 & 2
\end{array}\right]\)
Now A + 2B = \(\left[\begin{array}{cc}
-2 & 1 \\
3 & 2
\end{array}\right]+2\left[\begin{array}{cc}
-1 & 0 \\
1 & 2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
-2 & 1 \\
3 & 2
\end{array}\right]+\left[\begin{array}{cc}
-2 & 0 \\
2 & 4
\end{array}\right]\) = \(\left[\begin{array}{cc}
-4 & 1 \\
5 & 6
\end{array}\right]\)
⇒ (A + 2B)’ = \(\left[\begin{array}{cc}
-4 & 5 \\
1 & 6
\end{array}\right]\)
Question 5.
Show that the matrix A = \(\left[\begin{array}{ccc}
1 & -1 & 5 \\
-1 & 2 & 1 \\
5 & 1 & 3
\end{array}\right]\) is a symmetric matrix.
Solution:
Given that A = \(\left[\begin{array}{ccc}
1 & -1 & 5 \\
-1 & 2 & 1 \\
5 & 1 & 3
\end{array}\right]\)
Now A’ = \(\left[\begin{array}{ccc}
1 & -1 & 5 \\
-1 & 2 & 1 \\
5 & 1 & 3
\end{array}\right]\) = A
Hence, A is a symmetric matrix
Question 6.
Show that the matrix A = \(\left[\begin{array}{ccc}
0 & 1 & -1 \\
-1 & 0 & 1 \\
1 & -1 & 0
\end{array}\right]\) is a skew symmetric matrix.
Solution:
A = \(\left[\begin{array}{ccc}
0 & 1 & -1 \\
-1 & 0 & 1 \\
1 & -1 & 0
\end{array}\right]\)
Now A’ = \(\left[\begin{array}{ccc}
0 & -1 & 1 \\
1 & 0 & -1 \\
-1 & 1 & 0
\end{array}\right]\) = –\(\left[\begin{array}{ccc}
0 & 1 & -1 \\
-1 & 0 & 1 \\
1 & -1 & 0
\end{array}\right]\) = -A
Hence, A is a skew symmetric matrix.
Question 7.
For the matrix A = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]\), verify that (A + A’) is a symmetric matrix.
Solution:
Given that A = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]\) A’ = \(\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right]\)
A + A’ = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]+\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right]\) = \(\left[\begin{array}{ll}
2 & 11 \\
11 & 14
\end{array}\right]\)
Also [(A + A’)]’ = \(\left[\begin{array}{ll}
2 & 11 \\
11 & 14
\end{array}\right]\) = (A + A’)
∴ (A + A’) is a symmetric matrix
Question 8.
For the matrix A = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]\), verify that (A – A’) is a skew symmetric matrix
Solution:
A – A’ = \(\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]-\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right]\) = \(\left[\begin{array}{ll}
0 & -1 \\
1 & 0
\end{array}\right]\)
∴ (A – A’)’ = \(\left[\begin{array}{ll}
0 & 1 \\
-1 & 0
\end{array}\right]\) = –\(\left[\begin{array}{ll}
0 & -1 \\
1 & 0
\end{array}\right]\) = -(A – A’)
∴ (A – A’) is a skew symmetric matrix
Question 9.
If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\) then verify that A’A = I
Solution:
Question 10.
If A = \(\left[\begin{array}{cc}
\sin \alpha & \cos \alpha \\
-\cos \alpha & \sin \alpha
\end{array}\right]\), then verify that A’A = I
Solution:
II.
Question 1.
Find \(\frac{1}{2}\)(A + A’) and \(\frac{1}{2}\)(A – A’) when A = \(\left[\begin{array}{ccc}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right]\)
Solution:
Question 2.
Express the matrix \(\left[\begin{array}{cc}
3 & 5 \\
1 & -1
\end{array}\right]\) as the sum of a symmetric and a skew symmetric matrix.
Solution:
Hence Q’ = \(\left[\begin{array}{cc}
0 & -2 \\
2 & 0
\end{array}\right]\) = -Q
Thus, Q = \(\frac{1}{2}\)(A – A’) is a skew symmetric matrix.
Expressing A as the sum of P and Q:
P + Q = \(\left[\begin{array}{cc}
3 & 3 \\
3 & -1
\end{array}\right]+\left[\begin{array}{cc}
0 & 2 \\
-2 & 0
\end{array}\right]\) = \(\left[\begin{array}{cc}
3 & 5 \\
1 & -1
\end{array}\right]\) = A
Question 3.
Express the matrix \(\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]\) as the sum of a symmetric and a skew symmetric matrix.
Solution:
Expressing A as the sum of P and Q
P + Q = \(\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]+\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]=\left[\begin{array}{ccc}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]\) = A
Question 4.
Express the matrix \(\left[\begin{array}{ccc}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right]\) as the sum of a symmetric and a skew symmetricmatrix
Solution:
Question 5.
Express the matrix \(\left[\begin{array}{cc}
1 & 5 \\
-1 & 2
\end{array}\right]\) as the sum of a symmetric and a skew symmetric matrix.
Solution:
Question 6.
If A = \(\left[\begin{array}{ccc}
-1 & 2 & 3 \\
5 & 7 & 9 \\
-2 & 1 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-4 & 1 & -5 \\
1 & 2 & 0 \\
1 & 3 & 1
\end{array}\right]\), then verify that (A + B)’ = A’ + B’
Solution:
Now, A’ + B’ = \(\left[\begin{array}{ccc}
-1 & 5 & -2 \\
2 & 7 & 1 \\
3 & 9 & 1
\end{array}\right]+\left[\begin{array}{ccc}
-4 & 1 & 1 \\
1 & 2 & 3 \\
-5 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
-5 & 6 & -1 \\
3 & 9 & 4 \\
-2 & 9 & 2
\end{array}\right] .\) …………… (2)
∴ From (1) & (2), (A + B)’ = A’ + B’
Question 7.
If A = \(\left[\begin{array}{ccc}
-1 & 2 & 3 \\
5 & 7 & 9 \\
-2 & 1 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-4 & 1 & -5 \\
1 & 2 & 0 \\
1 & 3 & 1
\end{array}\right]\) then verify that (A – B)’ = A’ – B’
Solution:
∴ From (1) & (2), (A – B)’ = A’ – B’
Question 8.
If A’ = \(\left[\begin{array}{cc}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]\), then verify that (A + B)’ = A’ + B’
Solution:
∴ From (1) & (2), (A + B)’ = A’ + B’
Question 9.
If A = \(\left[\begin{array}{cc}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right] .\) then verify that (A – B)’ = A’ – B’
Solution:
∴ From (1) & (2), (A – B)’ = A’ – B’
Question 10.
For the matrices A and B, verify that (AB)’ = B’A’, where A = \(\left[\begin{array}{c}
1 \\
-4 \\
3
\end{array}\right]\), B = \(\left[\begin{array}{lll}
-1 & 2 & 1
\end{array}\right]\)
Solution:
∴ From (1) & (2), (AB)’ = B’A’
Question 11.
For the matrices A and B, verify that (AB)’ = B’A’, where A = \(\left[\begin{array}{c}
0 \\
1 \\
2
\end{array}\right]\), B = \(\left[\begin{array}{lll}
1 & 5 & 7
\end{array}\right]\)
Solution:
∴ From (1) & (2), (AB)’ = B’A’