# Ncert Solutions For Class 12 Maths Ex 3.3

## Ncert Solutions For Class 12 Maths Chapter 3 Ex 3.3

Q-1: What is the transpose of each of the following matrices?

(i) $$\begin{bmatrix} 6 \\ \frac{ 1 }{ 2 }\\ -2 \end{bmatrix}$$

(ii) $$\begin{bmatrix} 2 & -2 \\ 3 & 4 \end{bmatrix}$$

Solution:

(i) Let M = $$\begin{bmatrix} 6 \\ \frac{ 1 }{ 2 }\\ -2 \end{bmatrix}$$

Then,

MT = $$\begin{bmatrix} 6 & \frac{ 1 }{ 2 } & -2 \end{bmatrix}$$

(ii) Let M = $$\begin{bmatrix} 2 & -2 \\ 3 & 4 \end{bmatrix}$$

Then,

MT = $$\begin{bmatrix} 2 & 3 \\ -2 & 4 \end{bmatrix}$$

Q-2: If A = $$\begin{bmatrix} -2 & 3 & 4 \\ 6 & 7 & 10 \\ -3 & 2 & 2 \end{bmatrix}$$ and B = $$\begin{bmatrix} -5 & 2 & -6 \\ 2 & 3 & 1 \\ 2 & 4 & 2 \end{bmatrix}$$, then verify that

(A + B) = A + B

Solution:

A = $$\begin{bmatrix} -2 & 3 & 4 \\ 6 & 7 & 10 \\ -3 & 2 & 2 \end{bmatrix}$$, B = $$\begin{bmatrix} -5 & 2 & -6 \\ 2 & 3 & 1 \\ 2 & 4 & 2 \end{bmatrix}$$

A + B = $$\begin{bmatrix} -2 & 3 & 4 \\ 6 & 7 & 10 \\ -3 & 2 & 2 \end{bmatrix}$$ + $$\begin{bmatrix} -5 & 2 & -6 \\ 2 & 3 & 1 \\ 2 & 4 & 2 \end{bmatrix}$$

= $$\begin{bmatrix} -2 – 5 & 3 + 2 & 4 – 6 \\ 6 + 2 & 7 + 3 & 10 + 1 \\ -3 + 2 & 2 + 4 & 2 + 2 \end{bmatrix}$$

= $$\begin{bmatrix} -7 & 5 & -2 \\ 8 & 10 & 10 \\ -1 & 6 & 4 \end{bmatrix}$$

Thus,

(A + B) = $$\begin{bmatrix} -7 & 8 & -1 \\ 5 & 10 & 6 \\ -2 & 10 & 4 \end{bmatrix}$$

A + B = $$\begin{bmatrix} -2 & 6 & -3 \\ 3 & 8 & 2 \\ 4 & 10 & 2 \end{bmatrix}$$ + $$\begin{bmatrix} -5 & 2 & 2 \\ 2 & 3 & 4 \\ -6 & 1 & 2 \end{bmatrix}$$

= $$\begin{bmatrix} -2 – 5 & 6 + 2 & -3 + 2 \\ 3 + 2 & 8 + 3 & 2 + 4 \\ 4 – 6 & 9 + 1 & 2 + 2 \end{bmatrix}$$

= $$\begin{bmatrix} -7 & 8 & -1 \\ 5 & 11 & 6 \\ -2 & 10 & 4 \end{bmatrix}$$

Thus, (A + B) = A + B

Hence, proved.

Q-3: If A= $$\begin{bmatrix} -3 & 4 \\ 2 & 3 \end{bmatrix}$$ and B = $$\begin{bmatrix} -2 & 1 \\ 2 & 3 \end{bmatrix}$$, now find (A + 2B)’.

Solution:

We know that:

A = $${ \left ( {A}’ \right ) }’$$

Thus,

$${ A }’$$ = $$\begin{bmatrix} -3 & 4 \\ 2 & 3 \end{bmatrix}$$

A = $$\begin{bmatrix} -3 & 2 \\ 4 & 3 \end{bmatrix}$$

Now,

A + 2B = $$\begin{bmatrix} -3 & 2 \\ 4 & 3 \end{bmatrix}$$ + 2$$\begin{bmatrix} -2 & 1 \\ 2 & 3 \end{bmatrix}$$

= $$\begin{bmatrix} -3 & 2 \\ 4 & 3 \end{bmatrix}$$ + $$\begin{bmatrix} -4 & 2 \\ 4 & 6 \end{bmatrix}$$

= $$\begin{bmatrix} -3 – 4 & 2 + 2 \\ 4 + 4 & 3 + 6 \end{bmatrix}$$

= $$\begin{bmatrix} -7 & 4 \\ 8 & 9 \end{bmatrix}$$

Therefore,$${ \left ( A + 2B \right ) }’$$ = $$\begin{bmatrix} -7 & 8 \\ 4 & 9 \end{bmatrix}$$

Q-4: Prove that $${ \left ( AB \right ) }’ = { B }'{ A }’$$ for the matrices A and B where,

A = $$\begin{bmatrix} 2 \\ -5 \\ 4 \end{bmatrix}$$, B = $$\begin{bmatrix} -2 & 3 & 2 \end{bmatrix}$$

Solution:

AB = $$\begin{bmatrix} 2 \\ -5 \\ 4 \end{bmatrix}$$$$\begin{bmatrix} -2 & 3 & 2 \end{bmatrix}$$

= $$\begin{bmatrix} -4 & 6 & 4 \\ 10 & -15 & -10 \\ -8 & 12 & 8 \end{bmatrix}$$

Thus,

$${ \left ( AB \right ) }’$$ = $$\begin{bmatrix} -4 & 10 & -8 \\ 6 & -15 & 12 \\ 4 & -10 & 8 \end{bmatrix}$$

Now,

$${ A }’$$ = $$\begin{bmatrix} 2 & -5 & 4 \end{bmatrix}$$

$${ B }’$$ = $$\begin{bmatrix} -2 \\ 3 \\ 2 \end{bmatrix}$$

Thus,

{ B }’ { A }’= $$\begin{bmatrix} -4 & 10 & -8 \\ 6 & -15 & 12 \\ 4 & -10 & 8 \end{bmatrix}$$

= $${ \left ( AB \right ) }’$$

Hence, proved.

Q-5: If M = $$\begin{bmatrix} cos \beta & sin \beta \\ -sin \beta & cos \beta \end{bmatrix}$$, then prove that M’M = I.

Solution:

M = $$\begin{bmatrix} cos \beta & sin \beta \\ -sin \beta & cos \beta \end{bmatrix}$$

M’ = $$\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}$$

M’M = $$\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}$$ $$\begin{bmatrix} cos \beta & sin \beta \\ -sin \beta & cos \beta \end{bmatrix}$$

= $$\begin{bmatrix} \left ( cos \beta \right )\left ( cos \beta \right ) + \left ( -sin \beta \right )\left ( -sin \beta \right ) & \left ( cos \beta \right )\left ( sin \beta \right ) + \left ( -sin \beta \right )\left ( cos \beta \right ) \\ \left ( sin \beta \right )\left ( cos \beta \right ) + \left ( cos \beta \right )\left ( -sin \beta \right ) & \left ( sin \beta \right )\left ( sin \beta \right ) + \left ( cos \beta \right )\left ( cos \beta \right ) \end{bmatrix}$$

= $$\begin{bmatrix} cos^{ 2 } \beta + sin^{ 2 } & cos \beta . sin \beta – sin \beta . cos \beta \\ sin \beta cos \beta – cos \beta . sin \beta & sin^{ 2 } \beta + cos^{ 2 } \beta \end{bmatrix}$$

= $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

= I

Hence, proved.

Q-6: Prove that the matrix A = $$\begin{bmatrix} 2 & -2 & 6 \\ -2 & 3 & 2 \\ 6 & 2 & 4 \end{bmatrix}$$ is a symmetric matrix.

Solution:

Here, we can see that

$${ A }’$$ = $$\begin{bmatrix} 2 & -2 & 6 \\ -2 & 3 & 2 \\ 6 & 2 & 4 \end{bmatrix}$$ which is equal to A.

Hence, $${ A }’$$ = A

Therefore, the given matrix A = $$\begin{bmatrix} 2 & -2 & 6 \\ -2 & 3 & 2 \\ 6 & 2 & 4 \end{bmatrix}$$ is a symmetric matrix.

Q-7: Prove that ( A + $${ A }’$$ ) is a symmetric matrix for the matrix A = $$\begin{bmatrix} 2 & 6 \\ 7 & 8 \end{bmatrix}$$.

Solution:

$${ A }’$$ = $$\begin{bmatrix} 2 & 7 \\ 6 & 8 \end{bmatrix}$$

A + $${ A }’$$ = $$\begin{bmatrix} 2 & 6 \\ 7 & 8 \end{bmatrix}$$ + $$\begin{bmatrix} 2 & 7 \\ 6 & 8 \end{bmatrix}$$

= $$\begin{bmatrix} 4 & 13 \\ 13 & 16 \end{bmatrix}$$

$${ \left ( A + { A }’ \right ) }’$$ = $$\begin{bmatrix} 4 & 13 \\ 13 & 16 \end{bmatrix}$$ = A + $${ A }’$$

Hence, (A + $${ A }’$$ ) is a symmetric matrix.

Q-8: If M and N are the symmetric matrices with having same order, then MN – NM will be a

(a) Symmetric matrix

(b) Skew symmetric matrix.

(c) Identity matrix

(d) Zero matrix

Solution:

As per the data given in the question, M and N are two symmetric matrices.

Thus, we have

$${ A }’ = A$$ and $${ B }’ = B$$ ……..(i)

Let us consider,

$${ \left( MN – NM \right ) }’ = { \left ( MN \right ) }’ – { \left ( NM \right ) }’ \;\;\;\;\;\;\;\; \left [ {\left ( M – N \right )}’ = { M }’ – { N }’ \right ]$$

= $${ N }'{ M }’ – { M }'{ N }’ \;\;\;\;\;\;\;\; \left [ {\left ( MN \right )}’ = { N }'{ M }’ \right ]$$

= $$BA – MN \;\;\;\;\;\;\;\;\;\; \left [ by \; (i) \right ]$$

= – (MN – NM)

Thus, (MN – NM )’ = – (MN – NM)

Hence, (MN – NM) is a skew- symmetric matrix.

Therefore, the correct answer is (b).

Q-9: Consider M = $$\begin{bmatrix} cos \beta & – sin \beta \\ sin \beta & cos \beta \end{bmatrix}$$.

Then M + M’ = I, if β is:

(a) $$\frac{ \pi }{ 3 }$$

(b) $$\frac{ \pi }{ 6 }$$

(c) $$\frac{ 3 \pi }{ 2 }$$

(d) $$\frac{ \pi }{ 2 }$$

Solution:

M = $$\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}$$

M’ = $$\begin{bmatrix} cos \beta & sin \beta \\ -sin \beta & cos \beta \end{bmatrix}$$

M + M’ = I

⟹ $$\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}$$ + $$\begin{bmatrix} cos \beta & sin \beta \\ -sin \beta & cos \beta \end{bmatrix}$$ = $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

⟹ $$\begin{bmatrix} 2 cos \beta & 0 \\ 0 & 2 cos \beta \end{bmatrix}$$ = $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Thus,

$$2 cos \beta = 1$$

⟹ cos β = $$\frac{ 1 }{ 2 }$$

⟹ cos β = $$cos \frac{ \pi }{ 3 }$$

β = $$\frac{ \pi }{ 3 }$$

Therefore, (a) is the correct answer.