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.