Inter 1st Year Maths 1A Matrices Solutions Ex 3(e)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(e) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(e)

I.

Question 1.
Find the adjoint and inverse of the following matrices.
(i) \(\left[\begin{array}{cc}
2 & -3 \\
4 & 6
\end{array}\right]\)
Solution:
Inter 1st Year Maths 1A Matrices Solutions Ex 3(e) I Q1(i)

(ii) \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
Solution:
Inter 1st Year Maths 1A Matrices Solutions Ex 3(e) I Q1(ii)

Inter 1st Year Maths 1A Matrices Solutions Ex 3(e)

(iii) \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right]\)
Solution:
Inter 1st Year Maths 1A Matrices Solutions Ex 3(e) I Q1(iii)

(iv) \(\left[\begin{array}{lll}
2 & 1 & 2 \\
1 & 0 & 1 \\
2 & 2 & 1
\end{array}\right]\)
Solution:
Inter 1st Year Maths 1A Matrices Solutions Ex 3(e) I Q1(iv)

Question 2.
If A = \(\left[\begin{array}{cc}
a+i b & c+i d \\
-c+i d & a-i b
\end{array}\right]\), a2 + b2 + c2 + d2 = 1, then find the inverse of A.
Solution:
Inter 1st Year Maths 1A Matrices Solutions Ex 3(e) I Q2

Question 3.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 3 \\
0 & -1 & 4 \\
-2 & 2 & 1
\end{array}\right]\), then find A-1
Solution:
Inter 1st Year Maths 1A Matrices Solutions Ex 3(e) I Q3

Question 4.
If A = \(\left|\begin{array}{ccc}
-1 & -2 & -2 \\
2 & 1 & -2 \\
2 & -2 & 1
\end{array}\right|\), then show that the adjoint of A = 3A’ find A-1.
Solution:
Inter 1st Year Maths 1A Matrices Solutions Ex 3(e) I Q4
Inter 1st Year Maths 1A Matrices Solutions Ex 3(e) I Q4.1
Inter 1st Year Maths 1A Matrices Solutions Ex 3(e) I Q4.2

Inter 1st Year Maths 1A Matrices Solutions Ex 3(e)

Question 5.
If abc ≠ 0, find the inverse of \(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right]\)
Solution:
Inter 1st Year Maths 1A Matrices Solutions Ex 3(e) I Q5
Inter 1st Year Maths 1A Matrices Solutions Ex 3(e) I Q5.1

II.

Question 1.
If A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\) and B = \(\frac{1}{2}\left[\begin{array}{lll}
b+c & c-a & b-a \\
c-b & c+a & a-b \\
b-c & a-c & a+b
\end{array}\right]\) then show that ABA-1 is a diagonal matrix.
Solution:
Inter 1st Year Maths 1A Matrices Solutions Ex 3(e) II Q1
Inter 1st Year Maths 1A Matrices Solutions Ex 3(e) II Q1.1
Inter 1st Year Maths 1A Matrices Solutions Ex 3(e) II Q1.2

Question 2.
If 3A = \(\left[\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
-2 & 2 & -1
\end{array}\right]\) then show that A-1 = A’
Solution:
Inter 1st Year Maths 1A Matrices Solutions Ex 3(e) II Q2

Inter 1st Year Maths 1A Matrices Solutions Ex 3(e)

Question 3.
If A = \(\left[\begin{array}{rrr}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array}\right]\), then show that A-1 = A3
Solution:
Inter 1st Year Maths 1A Matrices Solutions Ex 3(e) II Q3
∴ A4 = I
det A = 3(1) – 3(-2) + 4(-2) = 1
∵ A ≠ 0 ⇒ A-1 exists
∵ A4 = I
Multiply with A-1
A4 (A-1) = I (A-1)
⇒ A3 (AA-1) = A-1
⇒ A3 (I) = A-1
∴ A-1 = A3

Question 4.
If AB = I or BA = I, then prove that A is invertible and B = A-1
Solution:
Given AB = I
⇒ AB| = |1|
⇒ |A| |B| = 1
⇒ |A| ≠ 0
∴ A is a non-singular matrix and BA = I
⇒ |BA| = |I|
⇒ |B| |A| = 1
⇒ |A| ≠ 0
∴ A is a non-singular matrix.
AB = I or BA = I, A is invertible.
∴ A-1 exists.
AB = I
⇒ A-1AB = A-1I
⇒ IB = A-1
⇒ B = A-1
∴ B = A-1