NCERT Solutions For Class 12 Maths Chapter 3 Exercise 3.3

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Chapter 3 Ex.3.3 Question 1

Find the transpose of each of the following matrices:

(i) \(\left( {\begin{array}{*{20}{c}}5\\{\frac{1}{2}}\\{ - 1}\end{array}} \right)\)

(ii) \(\left( {\begin{array}{*{20}{c}}1&{ - 1}\\2&3\end{array}} \right)\)

(iii) \(\left( {\begin{array}{*{20}{c}}{ - 1}&5&6\\{\sqrt 3 }&5&6\\2&3&{ - 1}\end{array}} \right)\)

Solution

(i) Let \(A = \left( {\begin{array}{*{20}{c}}5\\{\frac{1}{2}}\\{ - 1}\end{array}} \right)\)

\(\quad Then \;{A^T} = \left( {\begin{array}{*{20}{c}}5&{\frac{1}{2}}&{ - 1}\end{array}} \right)\)

(ii) Let \(A = \left( {\begin{array}{*{20}{c}}1&{ - 1}\\2&3\end{array}} \right)\)

\(\quad Then \;{A^T} = \left( {\begin{array}{*{20}{c}}1&2\\{ - 1}&3\end{array}} \right)\)

(iii) Let \(A = \left( {\begin{array}{*{20}{c}}{ - 1}&5&6\\{\sqrt 3 }&5&6\\2&3&{ - 1}\end{array}} \right)\)

\(\quad Then \;{A^T} = \left( {\begin{array}{*{20}{c}}{ - 1}&{\sqrt 3 }&2\\5&5&3\\6&6&{ - 1}\end{array}} \right)\)

Chapter 3 Ex.3.3 Question 2

If \(A = \left( {\begin{array}{*{20}{c}}{ - 1}&2&3\\5&7&9\\{ - 2}&1&1\end{array}} \right)\) and \(B = \left( {\begin{array}{*{20}{c}}{ - 4}&1&{ - 5}\\1&2&0\\1&3&1\end{array}} \right)\), then verify that

(i) \({\left( {A + B} \right)^\prime } = A' + B'\)

(ii) \({\left( {A - B} \right)^\prime } = A' - B'\)

Solution

It is given that \(A = \left( {\begin{array}{*{20}{c}}{ - 1}&2&3\\5&7&9\\{ - 2}&1&1\end{array}} \right)\) and \(B = \left( {\begin{array}{*{20}{c}}{ - 4}&1&{ - 5}\\1&2&0\\1&3&1\end{array}} \right)\)

Hence, we have \(A' = \left( {\begin{array}{*{20}{c}}{ - 1}&5&{ - 2}\\2&7&1\\3&9&1\end{array}} \right)\) and \(B' = \left( {\begin{array}{*{20}{c}}{ - 4}&1&1\\1&2&3\\{ - 5}&0&1\end{array}} \right)\)

(i) \(\left( {A + B} \right) = \left( {\begin{array}{*{20}{c}}{ - 1}&2&3\\5&7&9\\{ - 2}&1&1\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 4}&1&{ - 5}\\1&2&0\\1&3&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 5}&3&{ - 2}\\6&9&9\\{ - 1}&4&2\end{array}} \right)\)

Hence,

\({\left( {A + B} \right)^\prime } = \left( {\begin{array}{*{20}{c}}{ - 5}&6&{ - 1}\\3&9&4\\{ - 2}&9&2\end{array}} \right)\)

Now,

\[\begin{align}A' + B' = \left( {\begin{array}{*{20}{c}}{ - 1}&5&{ - 2}\\2&7&1\\3&9&1\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 4}&1&1\\1&2&3\\{ - 5}&0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 5}&6&{ - 1}\\3&9&4\\{ - 2}&9&2\end{array}} \right)\end{align}\]

Thus, \({\left( {A + B} \right)^\prime } = A' + B'\).

(ii) \(\left( {A - B} \right) = \left( {\begin{array}{*{20}{c}}{ - 1}&2&3\\5&7&9\\{ - 2}&1&1\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{ - 4}&1&{ - 5}\\1&2&0\\1&3&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&1&8\\4&5&9\\{ - 3}&{ - 2}&0\end{array}} \right)\)

Hence,

\({\left( {A - B} \right)^\prime } = \left( {\begin{array}{*{20}{c}}3&4&{ - 3}\\1&5&{ - 2}\\8&9&0\end{array}} \right)\)

Now,

\[\begin{align}A' - B' = \left( {\begin{array}{*{20}{c}}{ - 1}&5&{ - 2}\\2&7&1\\3&9&1\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{ - 4}&1&1\\1&2&3\\{ - 5}&0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}3&4&{ - 3}\\1&5&{ - 2}\\8&9&0\end{array}} \right)\end{align}\]

Thus, \({\left( {A - B} \right)^\prime } = A' - B'\).

Chapter 3 Ex.3.3 Question 3

If \(A' = \left( {\begin{array}{*{20}{c}}3&4\\{ - 1}&2\\0&1\end{array}} \right)\) and \(B = \left( {\begin{array}{*{20}{c}}{ - 1}&2&1\\1&2&3\end{array}} \right)\), then verify that

(i) \({\left( {A + B} \right)^\prime } = A' + B'\)

(ii) \({\left( {A - B} \right)^\prime } = A' - B'\)

Solution

It is known that \(A = {\left( {A'} \right)^\prime }\)

Hence,

\(A = \left( {\begin{array}{*{20}{c}}3&{ - 1}&0\\4&2&1\end{array}} \right)\) and \(B' = \left( {\begin{array}{*{20}{c}}{ - 1}&1\\2&2\\1&3\end{array}} \right)\)

(i) \(A + B = \left( {\begin{array}{*{20}{c}}3&{ - 1}&0\\4&2&1\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 1}&2&1\\1&2&3\end{array}} \right) = \left( {\begin{array}{*{20}{c}}2&1&1\\5&4&4\end{array}} \right)\)

Therefore,

\({\left( {A + B} \right)^\prime } = \left( {\begin{array}{*{20}{c}}2&5\\1&4\\1&4\end{array}} \right)\)

Now,

\(A' + B' = \left( {\begin{array}{*{20}{c}}3&4\\{ - 1}&2\\0&1\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 1}&1\\2&2\\1&3\end{array}} \right) = \left( {\begin{array}{*{20}{c}}2&5\\1&4\\1&4\end{array}} \right)\)

Hence, \({\left( {A + B} \right)^\prime } = A' + B'\).

(ii) \(A - B = \left( {\begin{array}{*{20}{c}}3&{ - 1}&0\\4&2&1\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{ - 1}&2&1\\1&2&3\end{array}} \right) = \left( {\begin{array}{*{20}{c}}4&{ - 3}&{ - 1}\\3&0&{ - 2}\end{array}} \right)\)

Therefore,

\({\left( {A - B} \right)^\prime } = \left( {\begin{array}{*{20}{c}}4&3\\{ - 3}&0\\{ - 1}&{ - 2}\end{array}} \right)\)

Now,

\(A' - B' = \left( {\begin{array}{*{20}{c}}3&4\\{ - 1}&2\\0&1\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{ - 1}&1\\2&2\\1&3\end{array}} \right) = \left( {\begin{array}{*{20}{c}}4&3\\{ - 3}&0\\{ - 1}&{ - 2}\end{array}} \right)\)

Hence, \({\left( {A - B} \right)^\prime } = A' - B'\).

Chapter 3 Ex.3.3 Question 4

If \(A' = \left( {\begin{array}{*{20}{c}}{ - 2}&3\\1&2\end{array}} \right)\) and \(B = \left( {\begin{array}{*{20}{c}}{ - 1}&0\\1&2\end{array}} \right)\), then find \({\left( {A + 2B} \right)^\prime }\).

Solution

It is known that \(A = \left( {A'} \right)'\).

Therefore,

\(A = \left( {\begin{array}{*{20}{c}}{ - 2}&1\\3&2\end{array}} \right)\)

Now,

\[\begin{align}A + 2B &= \left( {\begin{array}{*{20}{c}}{ - 2}&1\\3&2\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}{ - 1}&0\\1&2\end{array}} \right)\\ \;\; &= \left( {\begin{array}{*{20}{c}}{ - 2}&1\\3&2\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 2}&0\\2&4\end{array}} \right) \\ &= \left( {\begin{array}{*{20}{c}}{ - 4}&1\\5&6 \end{array}} \right)\end{align}\]

Chapter 3 Ex.3.3 Question 5

For the matrices \(A\) and \(B\), verify that \({\left( {AB} \right)^\prime } = B'A'\) where

(i) \(A = \left[ {\begin{array}{*{20}{c}}1\\{ - 4}\\3\end{array}} \right],\;B = \left[ {\begin{array}{*{20}{c}}{ - 1}&2&1\end{array}} \right]\)

(ii) \(A = \left[ {\begin{array}{*{20}{c}}0\\1\\2\end{array}} \right],\;B = \left[ {\begin{array}{*{20}{c}}1&5&7\end{array}} \right]\)

Solution

(i) It is given that \(A = \left[ {\begin{array}{*{20}{c}}1\\{ - 4}\\3\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{ - 1}&2&1\end{array}} \right]\)

Hence,

\[\begin{align}AB &= \left[ {\begin{array}{*{20}{c}}1\\{ - 4}\\3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 1}&2&1\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1}&2&1\\4&{ - 8}&{ - 4}\\{ - 3}&6&3\end{array}} \right]\end{align}\]

Therefore,

\({\left( {AB} \right)^\prime } = \left[ {\begin{array}{*{20}{c}}{ - 1}&4&{ - 3}\\2&{ - 8}&6\\1&{ - 4}&3\end{array}} \right]\)

Now,

\(A' = \left[ {\begin{array}{*{20}{c}}1&{ - 4}&3\end{array}} \right]\) and \(B' = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\\1\end{array}} \right]\)

Hence,

\[\begin{align}B'A' &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\\1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ - 4}&3\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1}&4&{ - 3}\\2&{ - 8}&6\\1&{ - 4}&3\end{array}} \right]\end{align}\]

Thus, \({\left( {AB} \right)^\prime } = B'A'\)

(ii) It is given that \(A = \left[ {\begin{array}{*{20}{c}}0\\1\\2\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}1&5&7\end{array}} \right]\)

Hence,

\[\begin{align}AB &= \left[ {\begin{array}{*{20}{c}}0\\1\\2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&5&7\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}0&0&0\\1&5&7\\2&{10}&{14}\end{array}} \right]\end{align}\]

Therefore,

\({\left( {AB} \right)^\prime } = \left[ {\begin{array}{*{20}{c}}0&1&2\\0&5&{10}\\0&7&{14}\end{array}} \right]\)

Now,

\(A' = \left[ {\begin{array}{*{20}{c}}0&1&2\end{array}} \right]\) and \(B' = \left[ {\begin{array}{*{20}{c}}1\\5\\7\end{array}} \right]\)

Therefore,

\[\begin{align}B'A' &= \left[ {\begin{array}{*{20}{c}}1\\5\\7\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0&1&2\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}0&1&2\\0&5&{10}\\0&7&{14}\end{array}} \right]\end{align}\]

Thus, \({\left( {AB} \right)^\prime } = B'A'\).

Chapter 3 Ex.3.3 Question 6

If

(i) \(A = \left( {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right)\), then verify \(A'A = I\)

(ii) \(A = \left( {\begin{array}{*{20}{c}}{\sin \alpha }&{\cos \alpha }\\{ - \cos \alpha }&{\sin \alpha }\end{array}} \right)\), then verify \(A'A = I\)

Solution

(i) It is given that \(A = \left( {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right)\)

Therefore,

\(A' = \left( {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right)\)

Now,

\[ \begin{align}A'A &= \left( {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }&\\{\sin \alpha }&{\cos \alpha }\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{\cos \alpha \cos \alpha + \left( { - \sin \alpha } \right)\left( { - \sin \alpha } \right)}&{\sin \alpha \cos \alpha + \left( { - \sin \alpha } \right)\cos \alpha }\\{\sin \alpha \cos \alpha + \cos \alpha \left( { - \sin \alpha } \right)}&{\sin \alpha \sin \alpha + \cos \alpha \cos \alpha }\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{{{\cos }^2}\alpha + {{\sin }^2}\alpha }&{\sin \alpha \cos \alpha - \sin \alpha \cos \alpha }\\{\sin \alpha \cos \alpha - \sin \alpha \cos \alpha }&{{{\sin }^2}\alpha + {{\cos }^2}\alpha }\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\&= I\end{align}\]

Thus, \(A'A = I\)

(ii) It is given that \(A = \left( {\begin{array}{*{20}{c}} {\sin \alpha }&{\cos \alpha }\\ { - \cos \alpha }&{\sin \alpha } \end{array}} \right)\)

Therefore,

\(A' = \left( {\begin{array}{*{20}{c}}{\sin \alpha }&{ - \cos \alpha }\\{\cos \alpha }&{\sin \alpha }\end{array}} \right)\)

Now,

\[\begin{align}A'A &= \left( {\begin{array}{*{20}{c}}{\sin \alpha }&{ - \cos \alpha }\\{\cos \alpha }&{\sin \alpha }\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\sin \alpha }&{\cos \alpha }\\{ - \cos \alpha }&{\sin \alpha }\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{\sin \alpha \sin \alpha + \left( { - \cos \alpha } \right)\left( { - \cos \alpha } \right)}&{\sin \alpha \cos \alpha + \left( { - \cos \alpha } \right)\sin \alpha }\\{\sin \alpha \cos \alpha + \sin \alpha \left( { - \cos \alpha } \right)}&{\sin \alpha \sin \alpha + \cos \alpha \cos \alpha }\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{{{\cos }^2}\alpha + {{\sin }^2}\alpha }&{\sin \alpha \cos \alpha - \sin \alpha \cos \alpha }\\{\sin \alpha \cos \alpha - \sin \alpha \cos \alpha }&{{{\sin }^2}\alpha + {{\cos }^2}\alpha }\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\ &= I\end{align}\]

Thus, \(A'A = I\)

Chapter 3 Ex.3.3 Question 7

Show that the matrix \(A = \left( {\begin{array}{*{20}{c}}1&{ - 1}&5\\{ - 1}&2&1\\5&1&3\end{array}} \right)\) is a symmetric matrix.

Show that the matrix \(A = \left( {\begin{array}{*{20}{c}}0&1&{ - 1}\\{ - 1}&0&1\\1&{ - 1}&0\end{array}} \right)\) is a skew symmetric matrix.

Solution

\(A = \left( {\begin{array}{*{20}{c}}1&{ - 1}&5\\{ - 1}&2&1\\5&1&3\end{array}} \right)\)

Now,

\[\begin{align}A' &= \left( {\begin{array}{*{20}{c}}1&{ - 1}&5\\{ - 1}&2&1\\5&1&3\end{array}} \right)\\ &= A\end{align}\]

Hence, A is a symmetric matrix.

\(A = \left( {\begin{array}{*{20}{c}}0&1&{ - 1}\\{ - 1}&0&1\\1&{ - 1}&0\end{array}} \right)\)

\[\begin{align}A' &= \left( {\begin{array}{*{20}{c}}0&{ - 1}&1\\1&0&{ - 1}\\{ - 1}&1&0\end{array}} \right)\\ &= - \left( {\begin{array}{*{20}{c}}0&1&{ - 1}\\{ - 1}&0&1\\1&{ - 1}&0\end{array}} \right)\\& = - A\end{align}\]

Hence, \(A\) is a skew symmetric matrix.

Chapter 3 Ex.3.3 Question 8

For the matrix , verify that

\(\left( {A + A'} \right)\)is a symmetric matrix.

\(\left( {A - A'} \right)\)is a skew symmetric matrix.

\(A = \left( {\begin{array}{*{20}{c}}1&5\\6&7\end{array}} \right)\)

Solution

It is given that \(A = \left( {\begin{array}{*{20}{c}}1&5\\6&7\end{array}} \right)\)

Hence, \(A' = \left( {\begin{array}{*{20}{c}}1&6\\5&7\end{array}} \right)\)\(\left( {A + A'} \right) = \left( {\begin{array}{*{20}{c}}1&5\\6&7\end{array}} \right) + \left( {\begin{array}{*{20}{c}}1&6\\5&7\end{array}} \right) = \left( {\begin{array}{*{20}{c}}2&{11}\\{11}&{14}\end{array}} \right)\)

Therefore,

\[\begin{align}{\left( {A + A'} \right)^\prime } = \left( {\begin{array}{*{20}{c}}2&{11}\\{11}&{14}\end{array}} \right)\\ = \left( {A + A'} \right)\end{align}\]

Thus, \(\left( {A + A'} \right)\) is a symmetric matrix.

\(\left( {A - A'} \right) = \left( {\begin{array}{*{20}{c}}1&5\\6&7\end{array}} \right) - \left( {\begin{array}{*{20}{c}}1&6\\5&7\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0&{ - 1}\\{ - 1}&0\end{array}} \right)\)

Therefore,

\[\begin{align}{\left( {A - A'} \right)^\prime } = \left( {\begin{array}{*{20}{c}}0&1\\{ - 1}&0\end{array}} \right)\\ = - \left( {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right)\\ = - \left( {A - A'} \right)\end{align}\]

Thus, \(\left( {A - A'} \right)\) is a skew symmetric matrix.

Chapter 3 Ex.3.3 Question 9

Find \(\frac{1}{2}\left( {A + A'} \right)\) and \(\frac{1}{2}\left( {A - A'} \right)\), when \(A = \left( {\begin{array}{*{20}{c}}0&a&b\\{ - a}&0&c\\{ - b}&{ - c}&0\end{array}} \right)\).

Solution

It is given that \(A = \left( {\begin{array}{*{20}{c}}0&a&b\\{ - a}&0&c\\{ - b}&{ - c}&0\end{array}} \right)\)

Hence,

\(A' = \left( {\begin{array}{*{20}{c}}0&{ - a}&{ - b}\\a&0&{ - c}\\b&c&0\end{array}} \right)\)

Now,

\[\begin{align}\left( {A + A'} \right) = \left( {\begin{array}{*{20}{c}}0&a&b\\{ - a}&0&c\\{ - b}&{ - c}&0\end{array}} \right) + \left( {\begin{array}{*{20}{c}}0&{ - a}&{ - b}\\a&0&{ - c}\\b&c&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right)\end{align}\]

Therefore,

\(\frac{1}{2}\left( {A + A'} \right) = \left( {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right)\)

Now,

\[\begin{align}\left( {A - A'} \right) = \left( {\begin{array}{*{20}{c}}0&a&b\\{ - a}&0&c\\{ - b}&{ - c}&0\end{array}} \right) - \left( {\begin{array}{*{20}{c}}0&{ - a}&{ - b}\\a&0&{ - c}\\b&c&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}0&{2a}&{2b}\\{ - 2a}&0&{2c}\\{ - 2b}&{ - 2c}&0\end{array}} \right)\end{align}\]

Thus,

\[\begin{align}\frac{1}{2}\left( {A - A'} \right) = \left( {\begin{array}{*{20}{c}}0&{2a}&{2b}\\{ - 2a}&0&{2c}\\{ - 2b}&{ - 2c}&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}0&a&b\\{ - a}&0&c\\{ - b}&{ - c}&0\end{array}} \right)\end{align}\]

Chapter 3 Ex.3.3 Question 10

Express the following as the sum of a symmetric and skew symmetric matrix:

(i) \(\left( {\begin{array}{*{20}{c}}3&5\\1&{ - 1}\end{array}} \right)\)

(ii) \(\left( {\begin{array}{*{20}{c}}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right)\)

(iii)\(\left( {\begin{array}{*{20}{c}}3&3&{ - 1}\\{ - 2}&{ - 2}&1\\{ - 4}&{ - 5}&2\end{array}} \right)\)

(iv) \(\left( {\begin{array}{*{20}{c}}1&5\\{ - 1}&2\end{array}} \right)\)

Solution

(i) Let \(A = \left( {\begin{array}{*{20}{c}}3&5\\1&{ - 1}\end{array}} \right)\)

Hence,

\(A' = \left( {\begin{array}{*{20}{c}}3&1\\5&{ - 1}\end{array}} \right)\)

Now,

\[\begin{align}\left( {A + A'} \right) &= \left( {\begin{array}{*{20}{c}}3&5\\1&{ - 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}3&1\\5&{ - 1}\end{array}} \right)\\&= \left( {\begin{array}{*{20}{c}}6&6\\6&{ - 2}\end{array}} \right)\end{align}\]

Let

 \[\begin{align}P &= \frac{1}{2}\left( {A + A'} \right)\\ &= \frac{1}{2}\left( {\begin{array}{*{20}{c}}6&6\\6&{ - 2}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}3&3\\3&{ - 1}\end{array}} \right)\end{align}\]

Now,

\[\begin{align}P' = \left( {\begin{array}{*{20}{c}}3&3\\3&{ - 1}\end{array}} \right)\\ = P\end{align}\]

Thus, \(P = \frac{1}{2}\left( {A + A'} \right)\)is a symmetric matrix.

Now,

\[\begin{align}\left( {A - A'} \right) &= \left( {\begin{array}{*{20}{c}}3&5\\1&{ - 1}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}3&1\\5&{ - 1}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}0&4\\{ - 4}&0\end{array}} \right)\end{align}\]

Let

\[\begin{align}Q &= \frac{1}{2}\left( {A - A'} \right)\\ &= \frac{1}{2}\left( {\begin{array}{*{20}{c}}0&4\\{ - 4}&0\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}0&2\\{ - 2}&0\end{array}} \right)\end{align}\]

Now,

\[\begin{align}Q' &= \left( {\begin{array}{*{20}{c}}0&{ - 2}\\2&0\end{array}} \right)\\ &= - Q\end{align}\]

Thus, \(Q = \frac{1}{2}\left( {A - A'} \right)\)is a skew symmetric matrix.

Representing \(A\) as the sum of \(P\) and \(Q\):

\[\begin{align}P + Q &= \left( {\begin{array}{*{20}{c}}3&3\\3&{ - 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}0&2\\{ - 2}&0\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}3&5\\1&{ - 1}\end{array}} \right)\\ &= A\end{align}\]

(ii) Let \(A = \left( {\begin{array}{*{20}{c}}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right)\)

Hence,

\(A' = \left( {\begin{array}{*{20}{c}}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right)\)

Now,

\(\left( {A + A'} \right) = \left( {\begin{array}{*{20}{c}}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right) + \left( {\begin{array}{*{20}{c}}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{12}&{ - 4}&4\\{ - 4}&6&{ - 2}\\4&{ - 2}&6\end{array}} \right)\)

Let

\[\begin{align}P& = \frac{1}{2}\left( {A + A'} \right)\\ &= \frac{1}{2}\left( {\begin{array}{*{20}{c}}{12}&{ - 4}&4\\{ - 4}&6&{ - 2}\\4&{ - 2}&6\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right)\end{align}\]

Now,

\[\begin{align}P' &= \left( {\begin{array}{*{20}{c}}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right)\\ &= P\end{align}\]

Thus, \(P = \frac{1}{2}\left( {A + A'} \right)\)is a symmetric matrix.

Now,

\[\begin{align}\left( {A - A'} \right) &= \left( {\begin{array}{*{20}{c}}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right) - \left( {\begin{array}{*{20}{c}}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right)\end{align}\]

Let

\[\begin{align}Q &= \frac{1}{2}\left( {A - A'} \right)\\ &= \frac{1}{2}\left( {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right)\end{align}\]

Now,

\[\begin{align}Q' &= \left( {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right)\\ &= - Q\end{align}\]

Thus, \(Q = \frac{1}{2}\left( {A - A'} \right)\)is a skew symmetric matrix.

Representing \(A\) as the sum of \(P\) and \(Q\):

\[\begin{align}P + Q &= \left( {\begin{array}{*{20}{c}}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right) + \left( {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right)\\ &= A\end{align}\]

(iii) Let \(A = \left( {\begin{array}{*{20}{c}}3&3&{ - 1}\\{ - 2}&{ - 2}&1\\{ - 4}&{ - 5}&2\end{array}} \right)\)

Hence,

\(A' = \left( {\begin{array}{*{20}{c}}3&{ - 2}&{ - 4}\\3&{ - 2}&{ - 5}\\{ - 1}&1&2\end{array}} \right)\)

Now,

\[\begin{align}\left( {A + A'} \right) &= \left( {\begin{array}{*{20}{c}}3&3&{ - 1}\\{ - 2}&{ - 2}&1\\{ - 4}&{ - 5}&2\end{array}} \right) + \left( {\begin{array}{*{20}{c}}3&{ - 2}&{ - 4}\\3&{ - 2}&{ - 5}\\{ - 1}&1&2\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}6&1&{ - 5}\\1&{ - 4}&{ - 4}\\{ - 5}&{ - 4}&4\end{array}} \right)\end{align}\]

Let

\[\begin{align}P&= \frac{1}{2}\left( {A + A'} \right)\\ &= \frac{1}{2}\left( {\begin{array}{*{20}{c}}6&1&{ - 5}\\1&{ - 4}&{ - 4}\\{ - 5}&{ - 4}&4\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}3&{\frac{1}{2}}&{\frac{{ - 5}}{2}}\\{\frac{1}{2}}&{ - 2}&{ - 2}\\{\frac{{ - 5}}{2}}&{ - 2}&2\end{array}} \right)\end{align}\]

Now,

\[\begin{align}P' = \left( {\begin{array}{*{20}{c}}3&{\frac{1}{2}}&{\frac{{ - 5}}{2}}\\{\frac{1}{2}}&{ - 2}&{ - 2}\\{\frac{{ - 5}}{2}}&{ - 2}&2\end{array}} \right)\\ = P\end{align}\]

Thus, \(P = \frac{1}{2}\left( {A + A'} \right)\) is a symmetric matrix.

Now,

\[\begin{align}\left( {A - A'} \right) &= \left( {\begin{array}{*{20}{c}}3&3&{ - 1}\\{ - 2}&{ - 2}&1\\{ - 4}&{ - 5}&2\end{array}} \right) - \left( {\begin{array}{*{20}{c}}3&{ - 2}&{ - 4}\\3&{ - 2}&{ - 5}\\{ - 1}&1&2\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}0&5&3\\{ - 5}&0&6\\{ - 3}&{ - 6}&0\end{array}} \right)\end{align}\]

Let

\[\begin{align}Q &= \frac{1}{2}\left( {A - A'} \right)\\ &= \frac{1}{2}\left( {\begin{array}{*{20}{c}}0&5&3\\{ - 5}&0&6\\{ - 3}&{ - 6}&0\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}0&{\frac{5}{2}}&{\frac{3}{2}}\\{\frac{{ - 5}}{2}}&0&3\\{\frac{{ - 3}}{2}}&{ - 3}&0\end{array}} \right)\end{align}\]

Now,

\[\begin{align}Q' &= \left( {\begin{array}{*{20}{c}}0&{\frac{{ - 5}}{2}}&{\frac{{ - 3}}{2}}\\{\frac{5}{2}}&0&{ - 3}\\{\frac{3}{2}}&3&0\end{array}} \right)\\ &= - Q\end{align}\]

Thus, \(Q = \frac{1}{2}\left( {A - A'} \right)\) is a skew symmetric matrix.

Representing \(A\) as the sum of \(P\) and \(Q\):

\[\begin{align}P + Q &= \left( {\begin{array}{*{20}{c}}3&{\frac{1}{2}}&{\frac{{ - 5}}{2}}\\{\frac{1}{2}}&{ - 2}&{ - 2}\\{\frac{{ - 5}}{2}}&{ - 2}&2\end{array}} \right) + \left( {\begin{array}{*{20}{c}}0&{\frac{5}{2}}&{\frac{3}{2}}\\{\frac{{ - 5}}{2}}&0&3\\{\frac{{ - 3}}{2}}&{ - 3}&0\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}3&3&{ - 1}\\{ - 2}&{ - 2}&1\\{ - 4}&{ - 5}&2\end{array}} \right)\\&= A\end{align}\]

(iv) Let \(A = \left( {\begin{array}{*{20}{c}}1&5\\{ - 1}&2\end{array}} \right)\)

Hence,

\(A' = \left( {\begin{array}{*{20}{c}}1&{ - 1}\\5&2\end{array}} \right)\)

Now,

\[\begin{align}\left( {A + A'} \right) &= \left( {\begin{array}{*{20}{c}}1&5\\{ - 1}&2\end{array}} \right) + \left( {\begin{array}{*{20}{c}}1&{ - 1}\\5&2\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}2&4\\4&4\end{array}} \right)\end{align}\]

Let

\[\begin{align}P& = \frac{1}{2}\left( {A + A'} \right)\\ &= \frac{1}{2}\left( {\begin{array}{*{20}{c}}2&4\\4&4\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}1&2\\2&2\end{array}} \right)\end{align}\]

Now,

\[\begin{align}P' &= \left( {\begin{array}{*{20}{c}}1&2\\2&2\end{array}} \right)\\ &= P\end{align}\]

Thus, \(P = \frac{1}{2}\left( {A + A'} \right)\) is a symmetric matrix.

Now,

\[\begin{align}\left( {A - A'} \right) &= \left( {\begin{array}{*{20}{c}}1&5\\{ - 1}&2\end{array}} \right) - \left( {\begin{array}{*{20}{c}}1&{ - 1}\\5&2\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}0&6\\{ - 6}&0\end{array}} \right)\end{align}\]

Let

\[\begin{align}Q &= \frac{1}{2}\left( {A - A'} \right)\\ &= \frac{1}{2}\left( {\begin{array}{*{20}{c}}0&6\\{ - 6}&0\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}0&3\\{ - 3}&0\end{array}} \right)\end{align}\]

Now,

\[\begin{align}Q' &= \left( {\begin{array}{*{20}{c}}0&{ - 3}\\3&0\end{array}} \right)\\ = - Q\end{align}\]

Thus, \(Q = \frac{1}{2}\left( {A - A'} \right)\) is a skew symmetric matrix.

Representing \(A\) as the sum of \(P\) and \(Q\):

\[\begin{align}P + Q &= \left( {\begin{array}{*{20}{c}}1&2\\2&2\end{array}} \right) + \left( {\begin{array}{*{20}{c}}0&3\\{ - 3}&0\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}1&5\\{ - 1}&2\end{array}} \right)\\&= A\end{align}\]

Chapter 3 Ex.3.3 Question 11

If \(A,B\) are symmetric matrices of the same order, then \(AB - BA\) is a

(A) Skew symmetric matrix (B) Symmetric matrix

(C) Zero matrix (D) Identity matrix

Solution

If \(A\)and \(B\)are symmetric matrices of the same order, then

\(A' = A\) and \(B' = B\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\)

Now consider,

\[\begin{align} {\left( {AB - BA} \right)^\prime } &= {\left( {AB} \right)^\prime } - {\left( {BA} \right)^\prime }\;\;\;\;\;\;\;\;\;\;\;\left[ {\because {{\left( {A - B} \right)}^\prime } = A' - B'} \right] \\ &= B'A' - A'B\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {\because {{\left( {AB} \right)}^\prime } = B'A'} \right] \\ &= BA - AB\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {from\;\left( 1 \right)} \right] \\ &= - \left( {AB - BA} \right) \\ \end{align} \]

Therefore,

\({\left( {AB - BA} \right)^\prime } = - \left( {AB - BA} \right)\)

Thus, \(AB - BA\) is a skew symmetric matrix.

The Correct option is A.

\(A + A' = I \frac{\pi }{6}\)

Chapter 3 Ex.3.3 Question 12

If \(A = \left( {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right)\), then , if the value of \(\alpha \) is:

(A) \(\frac{\pi }{6}\)

(B) \(\frac{\pi }{3}\)

(C) \(\pi \)

(D) \(\frac{{3\pi }}{2}\)

Solution

It is given that \(A = \left( {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right)\)

Hence,

\(A' = \left( {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right)\)

Now,

\(A + A' = I\)

Therefore,

\[\begin{align}\left( {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{2\cos \alpha }&0\\0&{2\cos \alpha }\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\end{align}\]

Comparing the corresponding elements of the two matrices, we have:

\[\begin{align}&\Rightarrow 2\cos \alpha = 1\\&\Rightarrow \cos \alpha = \frac{1}{2}\\&\Rightarrow \alpha = {\cos ^{ - 1}}\frac{1}{2}\\&\Rightarrow \alpha = \frac{\pi }{3}\end{align}\]

Thus, the correct option is \(B\).