RD Sharma Class 12 Ex 5.4 Solutions Chapter 5 Algebra of Matrices

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter5
Exercise5.4
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 5.4 Solutions Chapter 5 Algebra of Matrices

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 = 2AT

(ii) (A + B)T = AT + BT

(iii) (A − B)T = AT − BT

(iv) (AB)T = BT AT

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 = 2AT

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 = AT + BT

\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 = AT − BT

\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 = BTAT

\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 = BTAT

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 = BTAT

Solution:

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

Assume,

(AB)T = BTAT

\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 AT, BT and verify that

(i) (A + B)T = AT + BT

(ii) (AB)T = BTAT

(iii) (2A)T = 2AT

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 = AT + BT

\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 = BTAT

\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 = 2AT

\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 = BTAT

Solution:

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

Assume,

(AB)T = BTAT

\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 = BTAT

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 = BTAT

⇒ \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 = BTAT

(ii) For the matrices A and B, verify that (AB)T = BTAT, 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 = BTAT

⇒ \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 = BTAT

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}, AT – BT

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 AT – BT.

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 AT – BT

⇒ 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 li, mi, ni ; i = 1, 2, 3 denote the direction cosines of three mutually perpendicular vectors in space, prove that AAT = 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,

li, mi, ni 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,

AAT = I

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