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

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

I.

Question 1.

If A = \(\left[\begin{array}{ccc}

2 & 0 & 1 \\

-1 & 1 & 5

\end{array}\right]\) and B = \(\left[\begin{array}{ccc}

-1 & 1 & 0 \\

0 & 1 & -2

\end{array}\right]\), then find (AB’)’.

Solution:

Question 2.

If A = \(\left[\begin{array}{cc}

-2 & 1 \\

5 & 0 \\

-1 & 4

\end{array}\right]\) and B = \(\left[\begin{array}{ccc}

-2 & 3 & 1 \\

4 & 0 & 2

\end{array}\right]\) then find 2A + B’ and 3B’ – A.

Solution:

Question 3.

If A = \(\left[\begin{array}{cc}

2 & -4 \\

-5 & 3

\end{array}\right]\), then find A + A’ and A.A’

Solution:

Question 4.

If A = \(\left[\begin{array}{ccc}

-1 & 2 & 3 \\

2 & 5 & 6 \\

3 & x & 7

\end{array}\right]\) is a symmetric matrix, then find x.

Hint: ‘A’ is a symmetric matrix ⇒ A^{T} = A

Solution:

A is a symmetric matrix

⇒ A’ = A

Equating 2nd row, 3rd column elements we get x = 6.

Question 5.

If A = \(\left[\begin{array}{ccc}

0 & 2 & 1 \\

-2 & 0 & -2 \\

-1 & x & 0

\end{array}\right]\) is a skew-symmetric matrix, find x.

Solution:

∵ A is a skew-symmetric matrix

⇒ A^{T} = -A

Equating second-row third column elements we get x = 2.

Question 6.

Is \(\left[\begin{array}{ccc}

0 & 1 & 4 \\

-1 & 0 & 7 \\

-4 & -7 & 0

\end{array}\right]\) symmetric or skewsymmetric?

Solution:

∴ A is a skew-symmetric matrix.

II.

Question 1.

If A = \(\left[\begin{array}{cc}

\cos \alpha & \sin \alpha \\

-\sin \alpha & \cos \alpha

\end{array}\right]\), show that A.A’ = A’. A = I_{2}

Solution:

From (1), (2) we get A.A’ = A’. A = I_{2}

Question 2.

If A = \(\left[\begin{array}{ccc}

1 & 5 & 3 \\

2 & 4 & 0 \\

3 & -1 & -5

\end{array}\right]\) and B = \(\left[\begin{array}{ccc}

2 & -1 & 0 \\

0 & -2 & 5 \\

1 & 2 & 0

\end{array}\right]\) then find 3A – 4B’.

Solution:

Question 3.

If A = \(\left[\begin{array}{cc}

7 & -2 \\

-1 & 2 \\

5 & 3

\end{array}\right]\) and B = \(\left[\begin{array}{cc}

-2 & -1 \\

4 & 2 \\

-1 & 0

\end{array}\right]\) then find AB’ and BA’.

Solution:

Question 4.

For any square matrix A, Show that AA’ is symmetric.

Solution:

A is a square matrix

(AA’)’ = (A’)’A’ = A.A’

∵ (AA’)’ = AA’

⇒ AA’ is a symmetric matrix.