Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 4 Matrices Ex 4(b) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 4 Matrices Exercise 4(b)

Question 1.

State which of the following matrices are symmetric, skew-symmetric, both or not either:

Solution:

(i) Symmetric

(ii) Neither Symmetric nor skew-symmetric

(iii) Symmetric

(iv) Skew symmetric

(v) Both

(vi) Neither symmetric nor skew-symmetric

(vii) Skew symmetric

Question 2.

State ‘True’ or ‘False’:

(i) If A and B are symmetric matrices of the same order and AB – BA ≠ 0, then AB is not symmetric.

Solution:

True

(ii) For any square matrix A, AA’ is symmetric.

Solution:

True

(iii) If A is any skew-symmetric matrix, then A^{2} is also skew-symmetric.

Solution:

False

(iv) If A is symmetric, then A^{2}, A^{3}, …, A^{n} are all symmetric.

Solution:

True

(v) If A is symmetric then A – A^{1} is both symmetric and skew-symmetric.

Solution:

False

(vi) For any square matrix (A – A^{1})^{2} is skew-symmetric.

Solution:

True

(vii) A matrix which is not symmetric is skew-symmetric.

Solution:

False

Question 3.

(i) If A and B are symmetric matrices of the same order with AB ≠ BA, final whether AB – BA is symmetric or skew symmetric.

Solution:

A and B are symmetric matrices;

Thus A’ = A and B’ = B

Now (AB – BA)’ = (AB)’ – (BA)’

= B’A’ – A’B’

= BA – AB = – (AB – BA)

∴ AB – BA is skew symmetric.

(ii) If a symmetric/skew-symmetric matrix is expressed as a sum of a symmetric and a skew-symmetric matrix then prove that one of the matrices in the sum must be zero matrix.

Solution:

We know that zero matrix is both symmetric as well as skew-symmetric.

Let A is symmetric.

∴ A = A + O where A is symmetric and O is treated as skew-symmetric. If B is skew-symmetric then we can write B = O + B where O is symmetric and B is skew-symmetric.

Question 4.

A and B are square matrices of the same order, prove that

(i) If A, B and AB are all symmetric, then AB – BA = 0

Solution:

Let A, B and AB are all symmetric.

∴A’ = A, B’ = B and (AB)’ = AB

⇒ B’A’ = AB

⇒ BA = AB

⇒ AB – BA = 0

(ii) If A, B and AB are all skew symmetric then AB + BA = 0

Solution:

Let A, B and AB are all skew symmetric matrices

∴ A’ = -A, B’ = -B and (AB)’ = -AB

Now (AB)’ = -AB

⇒ B’A’ = -AB

⇒ (-B) (-A) = -AB

⇒ BA = -AB

⇒ AB + BA = 0

Question 5.

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

1 & 2 & 0 \\

0 & 1 & 3 \\

-2 & 5 & 3

\end{array}\right]\), then verify that A’ = \(\left[\begin{array}{ccc}

1 & 0 & -2 \\

2 & 1 & 5 \\

0 & 3 & 3

\end{array}\right]\)

(i) A+A’ is symmetric

(ii) A-A’ is skew-symmetric

Question 6.

Prove that a unit matrix is its own inverse. Is the converse true?

IfA = \(\left[\begin{array}{rrr}

0 & 1 & -1 \\

4 & -3 & 4 \\

3 & -3 & 4

\end{array}\right]\) show that A^{2} = I and hence A= A^{-1}.

Solution:

No the converse is not true for example:

Question 7.

Here A is an involuntary matrix, recall the definition given earlier.

Solution:

Question 8.

Show that \(\left[\begin{array}{ll}

\mathbf{0} & \mathbf{1} \\

\mathbf{1} & \mathbf{0}

\end{array}\right]\) is its own inverse.

Solution:

Question 9.

Express as a sum of a symmetric and a skew symmetric matrix.

Solutions:

Question 10.

What is the inverse of

Question 11.

Find inverse of the following matrices by elementary row/column operation (transformations):

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

1 & 2 \\

3 & 5

\end{array}\right]\)

Solution:

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

2 & 5 \\

1 & 3

\end{array}\right]\)

Solution:

(iii) \(\left[\begin{array}{cc}

4 & -2 \\

3 & 1

\end{array}\right]\)

Solution:

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

2 & 5 \\

1 & 3

\end{array}\right]\)

Solution:

(v) \(\left[\begin{array}{cc}

1 & 0 \\

2 & -3

\end{array}\right]\)

Solution:

(vi) \(\left[\begin{array}{cc}

1 & 0 \\

0 & -1

\end{array}\right]\)

Solution:

Question 12.

Find the inverse of the following matrices using elementary transformation:

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

\mathbf{0} & \mathbf{0} & 2 \\

\mathbf{0} & \mathbf{2} & \mathbf{0} \\

\mathbf{2} & \mathbf{0} & \mathbf{0}

\end{array}\right]\)

Solution:

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

0 & 1 & 2 \\

1 & 2 & 3 \\

3 & 1 & 1

\end{array}\right]\)

Solution:

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

3 & -2 & 3 \\

2 & 1 & -1 \\

4 & -3 & 2

\end{array}\right]\)

Solution:

(iv) \(\left[\begin{array}{lll}

1 & 1 & 2 \\

0 & 1 & 2 \\

1 & 2 & 1

\end{array}\right]\)

Solution:

(v) \(\left[\begin{array}{lll}

1 & 2 & 3 \\

2 & 1 & 4 \\

1 & 0 & 2

\end{array}\right]\)

Solution: