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:

(ii) $$\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$$
Solution:

(iii) $$\left[\begin{array}{lll} 1 & 0 & 2 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array}\right]$$
Solution:

(iv) $$\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 0 & 1 \\ 2 & 2 & 1 \end{array}\right]$$
Solution:

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:

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:

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:

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:

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:

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:

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:

∴ 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