12th NCERT/CBSE Matrices Exercise 3.3 Questions 12
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Question (1)

Find the transpose of each of the following matrices:
$\left( i \right)\left[ {\begin{array}{*{20}{c}}5\\{\frac{1}{2}}\\{ - 1}\end{array}} \right]$

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

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

Solution

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

\[\text{Transpose} \quad A' = \left[ {\begin{array}{*{20}{c}}5&{\frac{1}{2}}&{ - 1}\end{array}} \right]\]

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

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

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

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

Question (2)

\[ \text{If} \quad A = \left[ {\begin{array}{*{20}{c}}{ - 1}&2&3\\5&7&9\\{ - 2}&1&1\end{array}} \right]\]

\[ \text{and} B = \left[ {\begin{array}{*{20}{c}}{ - 4}&1&{ - 5}\\1&2&0\\1&3&1\end{array}} \right]\], then verify that

(i) (A + B)' = A' + B'     (ii) (A-B)' = A' - B'

Solution

\[A = \left[ {\begin{array}{*{20}{c}}{ - 1}&2&3\\5&7&9\\{ - 2}&1&1\end{array}} \right] \Rightarrow A' = \left[ {\begin{array}{*{20}{c}}{ - 1}&5&{ - 2}\\2&7&1\\3&9&1\end{array}} \right]\]

\[B = \left[ {\begin{array}{*{20}{c}}{ - 4}&1&{ - 5}\\1&2&0\\1&3&1\end{array}} \right] \Rightarrow B' = \left[ {\begin{array}{*{20}{c}}{ - 4}&1&1\\1&2&3\\{ - 5}&0&1\end{array}} \right]\]

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

\[ = {\left[ {\begin{array}{*{20}{c}}{ - 5}&3&{ - 2}\\6&9&9\\1&4&2\end{array}} \right]^1} = \left[ {\begin{array}{*{20}{c}}{ - 5}&6&1\\3&9&4\\{ - 2}&9&2\end{array}} \right]\]

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

\[ = \left[ {\begin{array}{*{20}{c}}{ - 5}&6&{ - 1}\\3&9&4\\{ - 2}&9&2\end{array}} \right]\] = LHS

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

\[ = {\left[ {\begin{array}{*{20}{c}}3&1&8\\4&{ - 5}&9\\{ - 3}&{ - 2}&0\end{array}} \right]^1}\]

\[ = \left[ {\begin{array}{*{20}{c}}3&4&{ - 3}\\1&5&{ - 2}\\8&9&0\end{array}} \right]\]

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

\[ = \left[ {\begin{array}{*{20}{c}}3&4&{ - 3}\\1&5&{ - 2}\\8&9&0\end{array}} \right]\]

= LHS
∴ (A-B)' = A' - B'

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
(a) (A + B)' = A' + B' (ii) (A-B)' = A' - B'

Solution

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

\[ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}3&{ - 1}&0\\4&2&1\end{array}} \right]\]

\[B = \left[ {\begin{array}{*{20}{c}}{ - 1}&2&1\\1&2&3\end{array}} \right]\]

\[ \Rightarrow B' = \left[ {\begin{array}{*{20}{c}}{ - 1}&1\\2&2\\1&3\end{array}} \right]\]

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

\[ = {\left[ {\begin{array}{*{20}{c}}2&1&1\\5&4&4\end{array}} \right]^|}\]

\[ = \left[ {\begin{array}{*{20}{c}}2&5\\1&4\\1&4\end{array}} \right]\]

RHS = A' + B'
\[ = \left[ {\begin{array}{*{20}{c}}3&4\\{ - 1}&2\\0&1\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 1}&1\\2&2\\1&3\end{array}} \right]\]

\[ = \left[ {\begin{array}{*{20}{c}}2&5\\1&4\\1&4\end{array}} \right] = LHS\]

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

\[ = {\left[ {\begin{array}{*{20}{c}}4&{ - 3}&{ - 1}\\3&0&{ - 2}\end{array}} \right]^|}\]

\[ = \left[ {\begin{array}{*{20}{c}}4&3\\{ - 3}&0\\{ - 1}&{ - 2}\end{array}} \right]\]

RHS = A' - B'

\[ = \left[ {\begin{array}{*{20}{c}}3&4\\{ - 1}&2\\0&1\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{ - 1}&1\\2&2\\1&3\end{array}} \right]\]

\[ = \left[ {\begin{array}{*{20}{c}}4&3\\{ - 3}&2\\0&1\end{array}} \right] = LHS\] \[ = \left[ {\begin{array}{*{20}{c}}4&3\\{ - 3}&2\\0&1\end{array}} \right] = LHS\]

Question (4)

If  $ = \left[ {\begin{array}{*{20}{c}}4&3\\{ - 3}&2\\0&1\end{array}} \right] = LHS$   and   $B = \left[ {\begin{array}{*{20}{c}}{ - 1}&0\\1&2\end{array}} \right]$, then find (A + 2B)'

Solution

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

Question (5)

For the matrices A and B, verify that (AB)' = 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

\[(i)A = \left[ {\begin{array}{*{20}{c}}1\\{ - 4}\\3\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}{ - 1}&2&1\end{array}} \right]\] \[\left( {AB} \right) = \left[ {\begin{array}{*{20}{c}}1\\{ - 4}\\3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 1}&2&1\end{array}} \right]\] \[\left( {AB} \right) = \left[ {\begin{array}{*{20}{c}}{ - 1}&2&1\\4&{ - 8}&{ - 4}\\{ - 3}&6&3\end{array}} \right]\] \[LHS = \left( {AB} \right)' = \left[ {\begin{array}{*{20}{c}}{ - 1}&4&{ - 3}\\2&{ - 8}&6\\1&{ - 4}&3\end{array}} \right]\] \[B'A' = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\\1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ - 4}&3\end{array}} \right]\] \[B'A' = \left[ {\begin{array}{*{20}{c}}{ - 1}&4&{ - 3}\\2&{ - 8}&6\\1&{ - 4}&3\end{array}} \right] = LHS\] \[(ii)A = \left[ {\begin{array}{*{20}{c}}0\\1\\2\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}1&5&7\end{array}} \right]\] \[AB = \left[ {\begin{array}{*{20}{c}}0\\1\\2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&5&7\end{array}} \right]\] \[ = \left[ {\begin{array}{*{20}{c}}0&0&0\\1&5&7\\2&{10}&{14}\end{array}} \right]\] \[LHS = \left( {AB} \right)' = \left[ {\begin{array}{*{20}{c}}0&1&2\\0&5&{10}\\0&7&{14}\end{array}} \right]\] \[RHS = B'A' = \left[ {\begin{array}{*{20}{c}}1\\5\\7\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0&1&2\end{array}} \right]\] \[ = \left[ {\begin{array}{*{20}{c}}0&1&2\\0&5&{10}\\0&7&{14}\end{array}} \right] = LHS\]

Question (6)

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

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

Solution

\[\left( i \right)A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]\] \[ \Rightarrow A' = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]\] \[LHS = A'A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]\] \[ = \left[ {\begin{array}{*{20}{c}}{{{\cos }^2}\alpha + {{\sin }^2}\alpha }&{\cos \alpha \sin \alpha - \sin \alpha \cos \alpha }\\{\sin \alpha \cos \alpha - \sin \alpha \cos \alpha }&{{{\sin }^2}\alpha + {{\cos }^2}\alpha }\end{array}} \right]\] \[ = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = I = RHS\]
\[(ii)A = \left[ {\begin{array}{*{20}{c}}{\sin \alpha }&{\cos \alpha }\\{ - \cos \alpha }&{sin\alpha }\end{array}} \right]\] \[ \Rightarrow A' = \left[ {\begin{array}{*{20}{c}}{\sin \alpha }&{ - \cos \alpha }\\{\cos \alpha }&{sin\alpha }\end{array}} \right]\] \[A'A = \left[ {\begin{array}{*{20}{c}}{\sin \alpha }&{ - \cos \alpha }\\{\cos \alpha }&{sin\alpha }\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\sin \alpha }&{\cos \alpha }\\{ - \cos \alpha }&{sin\alpha }\end{array}} \right]\] \[A'A = \left[ {\begin{array}{*{20}{c}}{{{\sin }^2}\alpha + {{\cos }^2}\alpha }&{sin\alpha \cos \alpha - \cos \alpha }\\{\cos \alpha \cos \alpha - \cos \alpha \cos \alpha }&{{{\cos }^2}\alpha + si{n^2}\alpha }\end{array}} \right]\] \[ = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = I\]

Question (7)

(i) 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.

(ii) 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

\[(i)A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&5\\{ - 1}&2&1\\5&1&3\end{array}} \right]\] \[A' = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&5\\{ - 1}&2&1\\5&1&3\end{array}} \right] = A\] ⇒ A is a symmetric matrix

\[A' = \left[ {\begin{array}{*{20}{c}}0&{ - 1}&0\\1&0&{ - 1}\\{ - 1}&1&0\end{array}} \right] = - \left[ {\begin{array}{*{20}{c}}0&1&{ - 1}\\{ - 1}&0&1\\1&{ - 1}&0\end{array}} \right] = - A\] ⇒ A is skew symmetric matrix

Question (8)

For the matrix $A = \left[ {\begin{array}{*{20}{c}}1&5\\6&7\end{array}} \right]$, verify that
(i) (A + A') is a symmetric matrix
(ii) (A - A') is a skew symmetric matrix

Solution

\[A = \left[ {\begin{array}{*{20}{c}}1&5\\6&7\end{array}} \right] \Rightarrow A' = \left[ {\begin{array}{*{20}{c}}1&6\\5&7\end{array}} \right]\] \[(i)A - A' = \left[ {\begin{array}{*{20}{c}}1&5\\6&7\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}1&6\\5&7\end{array}} \right]\] \[(i)A - A' = \left[ {\begin{array}{*{20}{c}}2&{11}\\{11}&{14}\end{array}} \right]\] \[\left( {A + A'} \right)' = \left[ {\begin{array}{*{20}{c}}2&{11}\\{11}&{14}\end{array}} \right] = A + A'\] ⇒ (A + A') is symmetric matrix

\[\left( {ii} \right)A - A' = \left[ {\begin{array}{*{20}{c}}1&5\\6&7\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}1&6\\5&7\end{array}} \right]\] \[A - A' = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]\] \[\left( {A - A'} \right)' = \left[ {\begin{array}{*{20}{c}}0&1\\{ - 1}&0\end{array}} \right] = - \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right] = - \left( {A - A'} \right)\] ⇒ ( A - A') is skew symmetric matrix

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

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

Question (10)

Express the following matrics as the sum of a symmetric and a skew symmetric matrix:
$\left( i \right)\left[ {\begin{array}{*{20}{c}}3&5\\1&{ - 1}\end{array}} \right]$     $\left( {ii} \right)\left[ {\begin{array}{*{20}{c}}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right]$

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

Solution

\[\left( i \right)A = \left[ {\begin{array}{*{20}{c}}3&5\\1&{ - 1}\end{array}} \right],A' = \left[ {\begin{array}{*{20}{c}}3&1\\5&{ - 1}\end{array}} \right]\] \[\frac{1}{2}\left( {A + A'} \right) = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}6&6\\6&{ - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&3\\3&{ - 1}\end{array}} \right] \Rightarrow \text{symmetric} \] \[\frac{1}{2}\left( {A - A'} \right) = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}0&4\\{ - 4}&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0&2\\{ - 2}&{ 0}\end{array}} \right]\] \[\therefore A = \left[ {\begin{array}{*{20}{c}}3&3\\3&{ - 1}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}0&2\\{ - 2}&0\end{array}} \right]\] \[\left( {ii} \right)A = \left[ {\begin{array}{*{20}{c}}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right],A' = \left[ {\begin{array}{*{20}{c}}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right]\] \[\frac{1}{2}\left( {A + A'} \right) = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}{12}&{ - 4}&4\\{ - 4}&6&{ - 2}\\4&{ - 2}&6\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right]\] \[\frac{1}{2}\left( {A - A'} \right) = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] = 0\] \[A = \left[ {\begin{array}{*{20}{c}}6&{ - 2}&2\\{ - 2}&3&{ - 1}\\2&{ - 1}&3\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right]\]
\[\left( {iii} \right)A = \left[ {\begin{array}{*{20}{c}}3&3&{ - 1}\\{ - 2}&{ - 2}&1\\{ - 4}&{ - 5}&2\end{array}} \right],A' = \left[ {\begin{array}{*{20}{c}}3&{ - 2}&{ - 4}\\3&{ - 2}&{ - 5}\\{ - 1}&1&2\end{array}} \right]\] \[\frac{1}{2}\left( {A + A'} \right) = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}6&1&{ - 5}\\1&{ - 4}&{ - 4}\\{ - 5}&{ - 4}&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3&{\frac{1}{2}}&{\frac{{ - 5}}{2}}\\{\frac{1}{2}}&{ - 2}&{ - 2}\\{\frac{{ - 3}}{2}}&{ - 2}&2\end{array}} \right]\] \[\frac{1}{2}\left( {A - A'} \right) = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}0&5&3\\5&0&6\\{ - 3}&{ - 6}&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0&{\frac{5}{2}}&{\frac{3}{2}}\\{\frac{{ - 5}}{2}}&0&3\\{\frac{{ - 3}}{2}}&{ - 3}&0\end{array}} \right]\] \[A = \left[ {\begin{array}{*{20}{c}}3&{\frac{1}{2}}&{\frac{{ - 5}}{2}}\\{\frac{1}{2}}&{ - 2}&{ - 2}\\{\frac{{ - 5}}{2}}&{ - 2}&2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}0&{\frac{5}{2}}&{\frac{3}{2}}\\{\frac{{ - 5}}{2}}&0&3\\{\frac{{ - 3}}{2}}&{ - 3}&0\end{array}} \right]\] \[\left( {iv} \right)A = \left[ {\begin{array}{*{20}{c}}1&5\\{ - 1}&2\end{array}} \right],A' = \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\5&2\end{array}} \right]\] \[\frac{1}{2}\left( {A + A'} \right) = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}2&4\\4&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&2\\2&2\end{array}} \right]\] \[\frac{1}{2}\left( {A - A'} \right) = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}0&6\\{ - 6}&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0&3\\{ - 3}&0\end{array}} \right]\] \[A = \left[ {\begin{array}{*{20}{c}}1&2\\2&2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}0&3\\{ - 3}&0\end{array}} \right]\]

Choose the correct answer in the Exercises 11 and 12

Question (11)

If A, B are symmetric matrices of same order, then AB - BA is a
(A) Skew symmetric matrix     (B) Symmetric matrix
(C) Zero matrix   nbsp; (D) Identity matrix

Solution

A, B are symmetric matrix of same order then AB-BA is skew fymmetric
So A is correct answer

Question (12)

$ If A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$, the A + A' = 1, if the value of α is
$\left( A \right)\frac{\pi }{6}$     $\left( B \right)\frac{\pi }{3}$
$\left( C \right)\pi $     $\left( D \right)\frac{{3\pi }}{2}$

Solution

\[A = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]\] \[A + A' = I\] \[\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]\left[ { - \begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[ \Rightarrow \left[ {\begin{array}{*{20}{c}}{2\cos \alpha }&0\\0&{2\cos \alpha }\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] \[ \Rightarrow 2\cos \alpha = 1\] \[\cos \alpha = \frac{1}{2}\] \[\alpha = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right)\] \[\alpha = \frac{\pi }{3}\] So B is correct answer
Exercise 3.2⇐
⇒Exercise 3.4