Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.7 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.7

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:

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

1 & 3 \\

-4 & 5

\end{array}\right]\)

∴ A^{T} = \(\left[\begin{array}{rr}

1 & -4 \\

3 & 5

\end{array}\right]\)

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

-4 & 0 & 5

2 & -6 & 1 \\

\end{array}\right]\)

∴ A^{T} = \(\left[\begin{array}{cc}

2 & -4 \\

-6 & 0 \\

1 & 5

\end{array}\right]\)

[Note: Answer given in the textbook is A^{T} = \(\left[\begin{array}{cc}

2 & -4 \\

6 & 0 \\

1 & 5

\end{array}\right]\). However, as per our calculation it is A^{T} = \(\left[\begin{array}{cc}

2 & -4 \\

-6 & 0 \\

1 & 5

\end{array}\right]\). ]

Question 2.

If [a_{ij}]_{3×3} where a_{ij} = 2(i – j), find A and

A^{T}. State whether A and A^{T} are symmetric or skew-symmetric matrices?

Solution:

A = [a_{ij}]_{3×3} = \(\left[\begin{array}{lll}

a_{11} & a_{12} & a_{13} \\

a_{21} & a_{22} & a_{23} \\

a_{31} & a_{32} & a_{33}

\end{array}\right]\)

Given a_{ij} = 2 (i — j)

∴ a_{11} = 2(1-1) = 0,

a_{12} = 2(1-2) = -2,

a_{13} = 2(1-3) = -4,

a_{21} = 2(2-1) = 2,

a_{22} = 2(2-2) = 0,

a_{23}=2(2-3) = -2,

a_{31} = 2(3-1) = 4,

a_{32} = 2(3-2) = 2,

a_{33}=2(3-3) = 0

∴ A^{T} = -A and A = -A^{T}

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

Questionn 3.

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

5 & -3 \\

4 & -3 \\

-2 & 1

\end{array}\right]\), prove that (2A)^{T} = 2A^{T}.

Solution:

From (i) and (ii), we get

(2A)^{T} = 2A^{T}

Question 4.

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

1 & 2 & -5 \\

2 & -3 & 4 \\

-5 & 4 & 9

\end{array}\right]\), prove that (3A)^{T} = 3A^{T}.

Solution:

From (i) and (ii), we get

(3A)^{T} = 3A^{T}

Question 5.

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

0 & 1+2 i & 1-2 \\

-1-2 i & 0 & -7 \\

2-i & 7 & 0

\end{array}\right]\),

where i = \(\sqrt{-1}\), prove that A^{T} = – A.

Solution:

Question 6.

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]\) and C = \(\left[\begin{array}{cc}

1 & 2 \\

-1 & 4 \\

-2 & 3

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

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

ii. (A – C)^{T} = A^{T} – C^{T}

Solution:

From (i) and (ii), we get

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

[Note: The question has been modified.]

From (i) and (ii), we get

(A – C)^{T} = A^{T} – C^{T</sup }

Question 7.

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

5 & 4 \\

-2 & 3

\end{array}\right]\) and \(\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:

3A – 2B + C = I

∴ C = I + 2B – 3A

Question 8.

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:

ii. ii. 5A^{T} – 5B^{T }= 5(A^{T} – B^{T})

Question 9.

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

1 & 0 & 1 \\

3 & 1 & 2

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

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:

A + 2B + 3C

∴ A^{T} + 2B^{T} + 3C^{T} = \(\left[\begin{array}{cc}

5 & 6 \\

8 & 8 \\

2 & -2

\end{array}\right]\)

From (i) and (ii), we get

(A + 2B + 3C)^{T} = A^{T} + 2B^{T} + 3C^{T}

Question 10.

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.

prove that (A + B^{T})^{T} = A^{T} + B

Solution:

From (i) and (ii), we get

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

Question 11.

Prove that A + A^{T} is a 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:

∴ (A + A^{T})^{T} = A + A^{T}, i.e., A + A^{T} = (A + A^{T})^{T}

∴ A + A^{T} is a symmetric matrix.

∴ (A – A^{T})^{T} = – (A – A^{T}),

i.e., A – A^{T} = -(A – A^{T})^{T}

∴ A – A^{T} is skew symmetric matrix.

∴ (A + A^{T})^{T} = A + A^{T}, i.e., A + A^{T} = (A + A^{T})^{T}

∴ A + A^{T} is a symmetric matrix.

∴ (A – A^{T})^{T} = – (A – A^{T}),

i.e., A – A^{T} = -(A – A^{T})^{T}

∴ A – A^{T} is skew symmetric matrix.

Question 12.

Express the following matrices as the sum of a symmetric and a skew symmetric matrix.

i. \(\left[\begin{array}{cc}

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:

A square matrix A can be expressed as the sum of a symmetric and a skew symmetric matrix as

A = \(\frac{1}{2}\) (A + A^{T}) + \(\frac{1}{2}\) (A – A^{T})

P is symmetric matrix …[∵ a_{ij} = a_{ji}]

and Q is a skew symmetric matrix [∵ -a_{ij} = -a_{ji}]

A = P + Q

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

4 & \frac{1}{2} \\

\frac{1}{2} & -5

\end{array}\right]+\left[\begin{array}{ll}

0 & \frac{-5}{2} \\

\frac{5}{2} & 0

\end{array}\right]\)

∴ P is symmetric matrix …[∵ a_{ij} = a_{ji}]

and Q is a skew symmetric matrix [∵ -a_{ij} = -a_{ji}]

∴ A = P + Q

∴ A = \(\left[\begin{array}{cc}

4 & \frac{1}{2} \\

\frac{1}{2} & -5

\end{array}\right]+\left[\begin{array}{ll}

0 & \frac{-5}{2} \\

\frac{5}{2} & 0

\end{array}\right]\)

Question 13.

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:

From (i) and (ii), we get

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

From (i) and (ii) we get

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

Question 14.

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

\cos \alpha & \sin \alpha \\

-\sin \alpha & \cos \alpha

\end{array}\right]\), show that A^{T}A = I, where I is the unit matrix of order 2.

Solution:

∴ A^{T}A = I, where I is the unit matrix of order 2.