Class 12 RD Sharma Solutions- Chapter 5 Algebra of Matrices - Exercise 5.4

Chapter 5 of RD Sharma's Class 12 Mathematics textbook focuses on the Algebra of Matrices. Exercise 5.4 specifically deals with elementary operations on matrices and their properties. This exercise is crucial for understanding how matrices can be manipulated and transformed, which is fundamental in various fields of mathematics and its applications.

Question 1: Let A = \begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix}    and B = \begin{vmatrix} 1 & 0 \\ 2 & -4 \end{vmatrix}    verify that

(i) (2A)^T = 2A^T

(ii) (A + B)^T = A^T + B^T

(iii) (A − B)^T = A^T − B^T

(iv) (AB)^T = B^T A^T

Solution:

(i) Given: A = \begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix}    and B = \begin{bmatrix} 1 & 0 \\ 2 & -4 \end{bmatrix}  

Assume,

(2A)^T = 2A^T

Substitute the value of A

\left (2 \begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix}  \right )^T = 2\begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix}^T\\ \begin{bmatrix} 4 & -6 \\ -14 & 10 \end{bmatrix}^T=2\begin{bmatrix} 2 & -7 \\ -3 & 5 \end{bmatrix}\\ \begin{bmatrix} 4 & -14 \\ -6 & 10 \end{bmatrix}=\begin{bmatrix} 4 & -14 \\ -6 & 10 \end{bmatrix}  

L.H.S = R.H.S

Hence, proved.

(ii) Given: A = \begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix}    and B = \begin{bmatrix} 1 & 0 \\ 2 & -4 \end{bmatrix}

Assume,

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

\left(\begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix} +\begin{bmatrix} 1 & 0 \\ 2 & -4 \end{bmatrix} \right )^T=\begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix} ^T+\begin{bmatrix} 1 & 0 \\ 2 & -4 \end{bmatrix}^T\\ \begin{bmatrix} 2+1 & -3+0 \\ -7+2 & 5-4 \end{bmatrix}=\begin{bmatrix} 2 & -7 \\ -3 & 5 \end{bmatrix}+\begin{bmatrix} 1 & 2 \\ 0 & -4 \end{bmatrix}\\ \begin{bmatrix} 3 & -3 \\ -5 & 1 \end{bmatrix}^T=\begin{bmatrix} 3 & -5 \\ -3 & 1 \end{bmatrix}\\ \begin{bmatrix} 3 & -5 \\ -3 & 1 \end{bmatrix}=\begin{bmatrix} 3 & -5 \\ -3 & 1 \end{bmatrix}  

L.H.S = R.H.S

Hence, proved.

(iii) Given: A= \begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix}    and B= \begin{bmatrix} 1 & 0 \\ 2 & -4 \end{bmatrix}

Assume,

(A − B)^T = A^T − B^T

\left(\begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix}-\begin{bmatrix} 1 & 0 \\ 2 & -4 \end{bmatrix} \right)^T=\begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix}^T-\begin{bmatrix} 1 & 0 \\ 2 & -4 \end{bmatrix}^T\\ \begin{bmatrix} 2-1 & -3-0 \\ -7-2 & 5+4 \end{bmatrix}^T=\begin{bmatrix} 2 & -7 \\ -3 & 5 \end{bmatrix}-\begin{bmatrix} 1 & 2 \\ 0 & -4 \end{bmatrix}\\ \begin{bmatrix} 1 & -3 \\ -9 & 9 \end{bmatrix}^T=\begin{bmatrix} 1 & -9 \\ -3 & 9 \end{bmatrix}\\ \begin{bmatrix} 1 & -9 \\ -3 & 9 \end{bmatrix}=\begin{bmatrix} 1 & -9 \\ -3 & 9 \end{bmatrix}

L.H.S = R.H.S

Hence, proved

(iv) Given: A = \begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix}    and B = \begin{bmatrix} 1 & 0 \\ 2 & -4 \end{bmatrix}

Assume,

(AB)^T = B^TA^T

\left(\begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 2 & -4 \end{bmatrix}\right)^T=\begin{bmatrix} 1 & 0 \\ 2 & -4 \end{bmatrix}^T\begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix}^T\\ \begin{bmatrix} 2-6 & 0+12 \\ -7+10 & 0-20 \end{bmatrix}^T=\begin{bmatrix} 1 & 2 \\ 0 & -4 \end{bmatrix}\begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix}\\ \begin{bmatrix} -4 & 3 \\ 12 & -20 \end{bmatrix}=\begin{bmatrix} -4 & 3 \\ 12 & -20 \end{bmatrix}  

Therefore, (AB)^T = B^TA^T

Hence, proved.

Question 2: A = \begin{bmatrix}3\\5\\2\end{bmatrix}  and B = \begin{bmatrix}1&0&4\end{bmatrix} Verify that (AB)^T = B^TA^T

Solution:

Given: A = \begin{bmatrix}3\\5\\2\end{bmatrix}  and B = \begin{bmatrix}1&0&4\end{bmatrix}

Assume,

(AB)^T = B^TA^T

\left(\begin{bmatrix}3\\5\\2\end{bmatrix}\begin{bmatrix}1&0&4\end{bmatrix}\right)^T=\begin{bmatrix}1&0&4\end{bmatrix}^T\begin{bmatrix}3\\5\\2\end{bmatrix}^T\\ \begin{bmatrix}3&0&12\\5&0&20\\2&0&8\end{bmatrix}^T=\begin{bmatrix}1\\0\\4\end{bmatrix}\begin{bmatrix}3&5&2\end{bmatrix}\\ \begin{bmatrix}3&5&2\\0&0&0\\12&20&8\end{bmatrix}=\begin{bmatrix}3&5&2\\0&0&0\\12&20&8\end{bmatrix}

L.H.S = R.H.S

Hence proved

Question 3: Let A = \begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}  and B = \begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}

Find A^T, B^T and verify that

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

(ii) (AB)^T = B^TA^T

(iii) (2A)^T = 2A^T

Solution:

(i) Given: A = \begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}

and B = \begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}

Assume

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

\left(\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}+\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}\right)^T=\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}+\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}^T\\ \begin{bmatrix}1+1&-1+2&0+3\\2+2&1+1&3+3\\1+0&2+1&1+1\end{bmatrix}^T=\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}+\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}\\ \begin{bmatrix}2&1&3\\4&2&6\\1&3&2\end{bmatrix}^T=\begin{bmatrix}1+1&2+2&1+0\\-1+2&1+1&2+1\\0+3&3+3&1+1\end{bmatrix}\\ \begin{bmatrix}2&4&1\\1&2&3\\3&6&2\end{bmatrix}=\begin{bmatrix}2&4&1\\1&2&3\\3&6&2\end{bmatrix}

L.H.S = R.H.S

Hence proved

(ii) Given: A = \begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}  and B = \begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}

Assume,

\left(\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}\right)^T=\begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}^T\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}^T\\ \begin{bmatrix}1-2+0&2-1+0&3-3+0\\2+2+0&4+1+3&6+3+3\\1+4+0&2+2+1&3+6+1\end{bmatrix}^T=\begin{bmatrix}1&2&0\\2&1&1\\3&3&1\end{bmatrix}\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}\\ \begin{bmatrix}-1&1&0\\4&8&12\\5&5&10\end{bmatrix}^T=\begin{bmatrix}1-2+0&2+2+0&1+4+0\\2-1+0&4+1+3&2+2+1\\3-3+0&6+3+3&3+6+1\end{bmatrix}\\ \begin{bmatrix}-1&4&5\\1&8&5\\0&12&10\end{bmatrix}=\begin{bmatrix}-1&4&5\\1&8&5\\0&12&10\end{bmatrix}

(AB)^T = B^TA^T

\begin{bmatrix}-1&4&5\\1&8&5\\0&12&10\end{bmatrix}=\begin{bmatrix}-1&4&5\\1&8&5\\0&12&10\end{bmatrix}

L.H.S =R.H.S

Hence proved

(iii) Given: A = \begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}  and B = \begin{bmatrix}1&2&3\\2&1&3\\0&1&1\end{bmatrix}

Assume,

(2A)^T = 2A^T

\left(2\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}\right)^T=2\begin{bmatrix}1&-1&0\\2&1&3\\1&2&1\end{bmatrix}^T\\ \begin{bmatrix}2&-2&0\\4&2&6\\2&4&2\end{bmatrix}^T=2\begin{bmatrix}1&2&1\\-1&1&2\\0&3&1\end{bmatrix}\\ \begin{bmatrix}2&4&2\\-2&2&4\\0&6&2\end{bmatrix}=\begin{bmatrix}2&4&2\\-2&2&4\\0&6&2\end{bmatrix}

L.H.S = R.H.S

Hence proved

Question 4: if A = \begin{bmatrix}-2\\4\\5\end{bmatrix}, B = \begin{bmatrix}1&3&-6\end{bmatrix}, verify that (AB)^T = B^TA^T

Solution:

Given: A = \begin{bmatrix}-2\\4\\5\end{bmatrix}  and B = \begin{bmatrix}1&3&-6\end{bmatrix}

Assume,

(AB)^T = B^TA^T

\left(\begin{bmatrix}-2\\4\\5\end{bmatrix}\begin{bmatrix}1&3&-6\end{bmatrix}\right)^T=\begin{bmatrix}1&3&-6\end{bmatrix}^T\begin{bmatrix}-2\\4\\5\end{bmatrix}^T\\ \begin{bmatrix}-2&-6&-12\\4&12&-24\\-5&15&-30\end{bmatrix}^T=\begin{bmatrix}1\\3\\-6\end{bmatrix}\begin{bmatrix}-2&4&5\end{bmatrix}\\ \begin{bmatrix}-2&4&5\\-6&12&15\\-12&-24&-30\end{bmatrix}=\begin{bmatrix}-2&4&5\\-6&12&15\\-12&-24&-30\end{bmatrix}

L.H.S = R.H.S

Hence proved

Question 5: If A = \begin{bmatrix}2&4&-1\\-1&0&2\end{bmatrix}and B = \begin{bmatrix}3&4\\-1&2\\2&1\end{bmatrix}, find (AB)T

Solution:

Given: A = \begin{bmatrix}2&4&-1\\-1&0&2\end{bmatrix}  and B = \begin{bmatrix}3&4\\-1&2\\2&1\end{bmatrix}

Here we have to find (AB)^T

\left(\begin{bmatrix}2&4&-1\\-1&0&2\end{bmatrix}\begin{bmatrix}3&4\\-1&2\\2&1\end{bmatrix}\right)^T\\ \begin{bmatrix}6-4-2&8+8-1\\-3-0+4&-4+0+2\end{bmatrix}^T\\ \begin{bmatrix}0&15\\1&-2\end{bmatrix}^T\\ \begin{bmatrix}0&1\\15&-2\end{bmatrix}

Hence,

(AB)^T = \begin{bmatrix}0&1\\15&-2\end{bmatrix}

Question 6: 

(i) For two matrices A and B, A=\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix},\ B=\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix},  verify that (AB)^T = B^TA^T

Solution:

Given,

A=\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix},\ B=\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}

(AB)^T = B^TA^T

⇒ \left(\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix}\begin{bmatrix}1&-1\\0&2\\5&0\end{bmatrix}\right)^T=A=\begin{bmatrix}1&-1&\\0&2\\5&0\end{bmatrix}^T\begin{bmatrix}2&1&3\\4&1&0\end{bmatrix}^T

⇒ \begin{bmatrix}2+0+15&-2+20\\4+0+0&-4+2+0\end{bmatrix}^T=\begin{bmatrix}1&0&5\\-1&2&0\end{bmatrix}\begin{bmatrix}2&4\\1&1\\3&0\end{bmatrix}

⇒ \begin{bmatrix}17&0\\4&-2\end{bmatrix}^T=\begin{bmatrix}2+0+15&4+0+0\\-2+2+0&-4+2+0\end{bmatrix}

⇒ \begin{bmatrix}17&4\\0&-2\end{bmatrix}=\begin{bmatrix}17&4\\0&-2\end{bmatrix}

⇒ L.H.S = R.H.S

Hence,

(AB)^T = B^TA^T

(ii) For the matrices A and B, verify that (AB)^T = B^TA^T, where

A=\begin{bmatrix}1&3\\2&4\end{bmatrix},\ B=\begin{bmatrix}1&4\\2&5\end{bmatrix}

Solution:

Given,

A=\begin{bmatrix}1&3\\2&4\end{bmatrix},\ B=\begin{bmatrix}1&4\\2&5\end{bmatrix}

(AB)^T = B^TA^T

⇒ \left(\begin{bmatrix}1&3\\2&4\end{bmatrix}\begin{bmatrix}1&4\\2&5\end{bmatrix}\right)^T=\begin{bmatrix}1&4&\\2&5\end{bmatrix}^T\begin{bmatrix}1&3\\2&4\end{bmatrix}^T

⇒ \begin{bmatrix}1+6&4+15\\2+8&8+20\end{bmatrix}^T=\begin{bmatrix}1&2\\4&5\end{bmatrix}\begin{bmatrix}1&2\\3&4\end{bmatrix}

⇒ \begin{bmatrix}7&19\\10&28\end{bmatrix}^T=\begin{bmatrix}1+6&2+8\\4+15&8+20\end{bmatrix}

⇒ \begin{bmatrix}7&10\\19&28\end{bmatrix}=\begin{bmatrix}7&10\\19&28\end{bmatrix}

⇒ L.H.S = R.H.s

So,

(AB)^T = B^TA^T

Question 7: Find A^T=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}\ and\ B=\begin{bmatrix}-1&2&1\\1&2&3\end{bmatrix}, A^T - B^T

Solution:

Given that A^T=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}\ and\ B=\begin{bmatrix}-1&2&1\\1&2&3\end{bmatrix}

We need to find A^T - B^T.

Given that, B=\begin{bmatrix}-1&2&1\\1&2&3\end{bmatrix}

 B^T=\begin{bmatrix}-1&2&1\\1&2&3\end{bmatrix}^T=\begin{bmatrix}-1&1\\2&2\\1&3\end{bmatrix}

Let us find A^T - B^T

⇒ A^T-B^T=\begin{bmatrix}3&4\\-1&2\\0&1\end{bmatrix}-\begin{bmatrix}-1&1\\2&2\\1&3\end{bmatrix}

⇒ A^T-B^T=\begin{bmatrix}3+1&4-1\\-1-2&2-2\\0-1&1-3\end{bmatrix}

⇒ A^T-B^T=\begin{bmatrix}4&3\\-3&0\\-1&-2\end{bmatrix}

Question 8: If A=\begin{bmatrix}cos\alpha&sin\alpha\\-sin\alpha&cos\alpha\end{bmatrix}, then verify that A'A = 1

Solution:

A=\begin{bmatrix}cos\alpha&sin\alpha\\-sin\alpha&cos\alpha\end{bmatrix}

\therefore A'=\begin{bmatrix}cos\alpha&-sin\alpha\\sin\alpha&cos\alpha\end{bmatrix}

A'A=\begin{bmatrix}cos\alpha&-sin\alpha\\sin\alpha&cos\alpha\end{bmatrix}\begin{bmatrix}cos\alpha&sin\alpha\\-sin\alpha&cos\alpha\end{bmatrix}

⇒ \begin{bmatrix}(cos\alpha) (cos\alpha)+(-sin\alpha)(-sin\alpha)&(cos\alpha)(sin\alpha)+(-sin\alpha)(cos\alpha)\\(sin\alpha)(cos\alpha)+(cos\alpha)(-sin\alpha)&(sin\alpha)(sin\alpha)+(cos\alpha)(cos\alpha)\end{bmatrix}

⇒ \begin{bmatrix}cos^2\alpha+sin^2\alpha&sin\alpha cos\alpha-sin\alpha cos\alpha\\sin\alpha cos\alpha-sin\alpha cos\alpha&sin^2\alpha+cos^2\alpha\end{bmatrix}

⇒ \begin{bmatrix}1&0\\0&1\end{bmatrix}=I

Hence,we have verified that A'A = I

Question 9: A=\begin{bmatrix}sin\alpha&cos\alpha\\-cos\alpha&sin\alpha\end{bmatrix}, then verify that A'A = I

Solution:

A=\begin{bmatrix}sin\alpha&cos\alpha\\-cos\alpha&sin\alpha\end{bmatrix}\\ \therefore\ A'=\begin{bmatrix}sin\alpha&-cos\alpha\\cos\alpha&sin\alpha\end{bmatrix}\\ A'A=\begin{bmatrix}sin\alpha&-cos\alpha\\cos\alpha&sin\alpha\end{bmatrix}\begin{bmatrix}sin\alpha&cos\alpha\\-cos\alpha&sin\alpha\end{bmatrix}

=\begin{bmatrix}(sin\alpha)(sin\alpha)+(-cos\alpha)(-cos\alpha)&(sin\alpha)(cos\alpha)+(-cos\alpha)(sin\alpha)\\(cos\alpha)(sin\alpha)+(sin\alpha)(-cos\alpha)&(cos\alpha)(cos\alpha)+(sin\alpha)(sin\alpha)\end{bmatrix}

=\begin{bmatrix}sin^2\alpha+cos^2\alpha&sin\alpha cos\alpha-sin\alpha cos\alpha\\sin\alpha cos\alpha-sin\alpha cos\alpha&cos^2\alpha+sin^2\alpha\end{bmatrix}

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

Hence, we have verified that A'A = I

Question 10: If l_i, m_i, n_i ; i = 1, 2, 3 denote the direction cosines of three mutually perpendicular vectors in space, prove that AA^T = I,

Where A=\begin{bmatrix}l_1&m_1&n_1\\l_2&m_2&n_2\\l_3&m_3&n_3\end{bmatrix}

Solution:

Given,

l_i, m_i, n_i are direction cosines of three mutually perpendicular vectors

⇒  \left.\begin{aligned}  l_1l_2+m_1m_2+n_1n_2=0\\  l_2l_3+m_2m_3+n_2n_3=0\\  l_1l_3+m_1m_3+n_1n_3=0 \end{aligned}\right\}\ \ \ \ \ \ --- (A)

And,

 \left.\begin{aligned}  l_1^2+m_1^2+n_1^2=1\\  l_2^2+m_2^2+n_2^2=1\\  l_3^2+m_3^2+n_3^2=1 \end{aligned}\right\}\ \ \ \ \ ---(B)

Given,

A=\begin{bmatrix}l_1&m_1&n_1\\ l_2&m_2&n_2\\ l_3&m_3&n_3\end{bmatrix}

AA^T=\begin{bmatrix}l_1&m_1&n_1\\ l_2&m_2&n_2\\ l_3&m_3&n_3\end{bmatrix}\begin{bmatrix}l_1&m_1&n_1\\ l_2&m_2&n_2\\ l_3&m_3&n_3\end{bmatrix}^T\\ =\begin{bmatrix}l_1&m_1&n_1\\ l_2&m_2&n_2\\ l_3&m_3&n_3\end{bmatrix}\begin{bmatrix}l_1&l_2&l_3\\ m_1&m_2&m_3\\ n_1&n_2&n_3\end{bmatrix}

\begin{bmatrix}l_1^2+m_1^2+n_1^2&l_1l_2+m_1m_2+n_1n_2&l_1l_3+m_1m_3+n_1n_3\\ l_1l_2+m_1m_2+n_1n_2&l_2^2+m_2^2+n_2^2&l_2l_3+m_2m_3+n_2n_3\\ l_1l_3+m_1m_3+n_1n_3&l_3l_2+m_3m_2+n_3n_2&l_3^2+m_3^2+n_3^2\end{bmatrix}

\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}\ \ \ \ \ \ \ [using\ (A)\ and\ (B)]

= I

Hence,

AA^T = I

Summary

Exercise 5.4 covers the following key topics:

Elementary row operations

Elementary column operations

Properties of elementary operations

Equivalence of matrices

Row-reduced echelon form

Applications of elementary operations in solving systems of linear equations

These concepts are essential for more advanced topics in linear algebra and have practical applications in fields such as computer graphics, economics, and engineering.