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]\), a^{2} + b^{2} + c^{2} + d^{2} = 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} = A^{3}

Solution:

∴ A^{4} = I

det A = 3(1) – 3(-2) + 4(-2) = 1

∵ A ≠ 0 ⇒ A^{-1} exists

∵ A^{4} = I

Multiply with A^{-1}

A^{4} (A^{-1}) = I (A^{-1})

⇒ A^{3} (AA^{-1}) = A^{-1}

⇒ A^{3} (I) = A^{-1}

∴ A^{-1} = A^{3}

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^{-1}AB = A^{-1}I

⇒ IB = A^{-1}

⇒ B = A^{-1}

∴ B = A^{-1}