Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 2 Matrices Ex 2.4 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4

Question 1.

Find A^{T}, if

(i) A = \(\left[\begin{array}{cc}

1 & 3 \\

-4 & 5

\end{array}\right]\)

(ii) A = \(\left[\begin{array}{ccc}

2 & -6 & 1 \\

-4 & 0 & 5

\end{array}\right]\)

Solution:

Question 2.

If A = [a_{ij}]_{3×3} where a_{ij} = 2(i – j). Find A and A^{T}. State whether A and A^{T} both are symmetric or skew-symmetric matrices.

Solution:

Hence, A and A^{T} are both skew-symmetric matrices.

Question 3.

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

5 & -3 \\

4 & -3 \\

-2 & 1

\end{array}\right]\), prove that (A^{T})^{T} = A.

Solution:

Question 4.

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

1 & 2 & -5 \\

2 & -3 & 4 \\

-5 & 4 & 9

\end{array}\right]\), prove that A^{T} = A.

Solution:

Question 5.

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

2 & -3 \\

5 & -4 \\

-6 & 1

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

2 & 1 \\

4 & -1 \\

-3 & 3

\end{array}\right]\), C = \(\left[\begin{array}{cc}

1 & 2 \\

-1 & 4 \\

-2 & 3

\end{array}\right]\), then show that

(i) (A + B)^{T} = A^{T} + B^{T}

(ii) (A – C)^{T} = A^{T} – C^{T}

Solution:

Question 6.

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

5 & 4 \\

-2 & 3

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

-1 & 3 \\

4 & -1

\end{array}\right]\), then find C^{T}, such that 3A – 2B + C = I, where I is the unit matrix of order 2.

Solution:

Question 7.

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

7 & 3 & 0 \\

0 & 4 & -2

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

0 & -2 & 3 \\

2 & 1 & -4

\end{array}\right]\), then find

(i) A^{T} + 4B^{T}

(ii) 5A^{T} – 5B^{T}

Solution:

Question 8.

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

1 & 0 & 1 \\

3 & 1 & 2

\end{array}\right]\), B = \(\left[\begin{array}{lll}

2 & 1 & -4 \\

3 & 5 & -2

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

0 & 2 & 3 \\

-1 & -1 & 0

\end{array}\right]\), verify that (A + 2B + 3C)^{T} = A^{T} + 2B^{T} + 3C^{T}

Solution:

Question 9.

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

-1 & 2 & 1 \\

-3 & 2 & -3

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

2 & 1 \\

-3 & 2 \\

-1 & 3

\end{array}\right]\), prove that (A + B^{T})^{T} = A^{T} + B.

Solution:

From (1) and (2),

(A + B^{T})^{T} = A^{T} + B.

Question 10.

Prove that A + A^{T} is symmetric and A – A^{T} is a skew-symmetric matrix, where

(i) A = \(\left[\begin{array}{ccc}

1 & 2 & 4 \\

3 & 2 & 1 \\

-2 & -3 & 2

\end{array}\right]\)

(ii) A = \(\left[\begin{array}{ccc}

5 & 2 & -4 \\

3 & -7 & 2 \\

4 & -5 & -3

\end{array}\right]\)

Solution:

Question 11.

Express each of the following matrix as the sum of a symmetric and a skew-symmetric matrix:

(i) \(\left[\begin{array}{ll}

4 & -2 \\

3 & -5

\end{array}\right]\)

(ii) \(\left[\begin{array}{ccc}

3 & 3 & -1 \\

-2 & -2 & 1 \\

-4 & -5 & 2

\end{array}\right]\)

Solution:

Question 12.

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

2 & -1 \\

3 & -2 \\

4 & 1

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

0 & 3 & -4 \\

2 & -1 & 1

\end{array}\right]\), verify that

(i) (AB)^{T} = B^{T}A^{T}

(ii) (BA)^{T} = A^{T}B^{T}

Solution: