Matrices and Determinants Exercise: 5.1 Class 11 Basic Mathematics Solution [NEB UPDATED]

XI Basic Mathematics Solution Exercise 5.1

Exercise 5.1

1) If A = $\left( {\begin{array}{*{20}{c}}4&{ - 5}\\3&6\end{array}} \right)$ and B = $\left( {\begin{array}{*{20}{c}}2&2\\{ - 1}&{ - 2}\end{array}} \right)$, Find A^T,,B^T(AB)^T  show that (AB)^T=B^TA^T.

Solution:

Given, A = $\left( {\begin{array}{*{20}{c}}4&{ - 5}\\3&6\end{array}} \right)$ and B = $\left( {\begin{array}{*{20}{c}}2&2\\{ - 1}&{ - 2}\end{array}} \right)$

A^T = $\left( {\begin{array}{*{20}{c}}4&3\\{ - 5}&6\end{array}} \right)$ and 

B^T = $\left( {\begin{array}{*{20}{c}}2&{ - 1}\\2&{ - 2}\end{array}} \right)$

AB = $\left( {\begin{array}{*{20}{c}}4&{ - 5}\\3&6\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&3\\{ - 1}&{ - 2}\end{array}} \right)$ =$\left( {\begin{array}{*{20}{c}}{8 + 5}&{12 + 10}\\{6 - 6}&{9 - 12}\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}{13}&{22}\\0&{ - 3}\end{array}} \right)$

(AB)^T = $\left( {\begin{array}{*{20}{c}}{13}&0\\{22}&{ - 3}\end{array}} \right)$

B^T.A^T = $\left( {\begin{array}{*{20}{c}}2&{ - 1}\\3&{ - 2}\end{array}} \right)\left( {\begin{array}{*{20}{c}}4&3\\{ - 5}&6\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{8 + 5}&{6 - 6}\\{12 + 10}&{9 - 12}\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}{13}&0\\{22}&{ - 3}\end{array}} \right)$.

 

b) If  A = $\left( {\begin{array}{*{20}{c}}1&2&3\\0&1&4\\2&3&0\end{array}} \right)$ and B = $\left( {\begin{array}{*{20}{c}}{ - 1}&2&1\\3&0&{ - 1}\\2&1&0\end{array}} \right)$ , Find A',B', (AB)'and B'A'.

Solution:

Given, A = $\left( {\begin{array}{*{20}{c}}1&2&3\\0&1&4\\2&3&0\end{array}} \right)$ and A’ 

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

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

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

AB = $\left( {\begin{array}{*{20}{c}}1&2&3\\0&1&4\\2&3&0\end{array}} \right){\rm{\: }}\left( {\begin{array}{*{20}{c}}{ - 1}&2&1\\3&0&{ - 1}\\2&1&0\end{array}} \right)$ 

= $\left( {\begin{array}{*{20}{c}}{ - 1 + 6 + 6}&{2 + 0 + 3}&{1 - 2 + 0}\\{0 + 3 + 8}&{0 + 0 + 4}&{0 - 1 + 0}\\{ - 2 + 9 + 0}&{4 + 0 + 0}&{2 - 3 + 0}\end{array}} \right)$ 

=  $\left( {\begin{array}{*{20}{c}}{11}&5&{ - 1}\\{11}&4&{ - 1}\\7&4&{ - 1}\end{array}} \right)$

(AB)’ =  $\left( {\begin{array}{*{20}{c}}{11}&{11}&7\\5&4&4\\{ - 1}&{ - 1}&{ - 1}\end{array}} \right)$

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

=$ \left( {\begin{array}{*{20}{c}}{ - 1 + 6 + 6}&{0 + 3 + 8}&{ - 2 + 9 + 0}\\{2 + 0 + 3}&{0 + 0 + 4}&{4 + 0 + 0}\\{1 - 2 + 0}&{0 - 1 + 0}&{2 - 3 + 0}\end{array}} \right)$ 

= $\left( {\begin{array}{*{20}{c}}{11}&{11}&7\\5&4&4\\{ - 1}&{ - 1}&{ - 1}\end{array}} \right)$.

 

2.a) If A = $\left( {\begin{array}{*{20}{c}}1&2\\{ - 3}&6\\0&1\end{array}} \right)$, B = $\left( {\begin{array}{*{20}{c}}0&3\\5&7\\1&{ - 4}\end{array}} \right)$and k=3; compute A', B'and veryfy that:

(i)(A’)’= A

(ii)(A + B)’= A’ + B’

(ii)(A + B)’= A’ + B’

Solution:

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

So, A’ = $\left( {\begin{array}{*{20}{c}}1&{ - 3}&0\\2&6&1\end{array}} \right)$

B = $\left( {\begin{array}{*{20}{c}}0&3\\5&7\\1&{ - 4}\end{array}} \right)$

So, B’ = $\left( {\begin{array}{*{20}{c}}0&5&1\\3&7&{ - 4}\end{array}} \right)$ 

(i)(A’)’= A

A’ = $\left( {\begin{array}{*{20}{c}}1&{ - 3}&0\\2&6&1\end{array}} \right)$

So, (A’)’ = $\left( {\begin{array}{*{20}{c}}1&2\\{ - 3}&6\\0&1\end{array}} \right)$ = A. 

(ii)(A + B)’= A’ + B’

A + B = $\left( {\begin{array}{*{20}{c}}1&2\\{ - 3}&6\\0&1\end{array}} \right) + \left( {\begin{array}{*{20}{c}}0&3\\5&7\\1&{ - 4}\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}{1 + 0}&{2 + 3}\\{ - 3 + 5}&{6 + 7}\\{0 + 1}&{1 - 4}\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}1&5\\2&{13}\\1&{ - 3}\end{array}} \right)$

Or, (A + B)’ = $\left( {\begin{array}{*{20}{c}}1&2&1\\5&{13}&{ - 3}\end{array}} \right)$

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

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

= $\left( {\begin{array}{*{20}{c}}1&2&1\\5&{13}&{ - 3}\end{array}} \right)$So, (A + B)’ = A’ + B’. 

(iii)(kA)’ = kA’

kA = 3$\left( {\begin{array}{*{20}{c}}1&2\\{ - 3}&6\\0&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&6\\{ - 9}&{18}\\0&3\end{array}} \right)$

So, (kA)’ = $\left( {\begin{array}{*{20}{c}}3&{ - 9}&0\\6&{18}&3\end{array}} \right)$

Or, kA’ = 3$\left( {\begin{array}{*{20}{c}}{ - 1}&3&0\\2&6&1\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}3&{ - 9}&0\\6&{18}&3\end{array}} \right)$

So, (kA)’ = kA’

 

c)If A = $\frac{1}{3}\left( {\begin{array}{*{20}{c}}1&2&2\\2&1&{ - 2}\\{ - 2}&2&{ - 1}\end{array}} \right)$ , verify that A.A^T = A^T.A = I where I is a unit matrix of order 3.

Soln:

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

A.A^T = $\frac{1}{9}\left( {\begin{array}{*{20}{c}}1&2&2\\2&1&{ - 2}\\{ - 2}&2&{ - 1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&2&{ - 2}\\2&1&2\\2&{ - 2}&{ - 1}\end{array}} \right)$

= $\frac{1}{9}\left( {\begin{array}{*{20}{c}}{1 + 4 + 4}&{2 + 2 - 4}&{ - 2 + 4 - 2}\\{2 + 2 - 4}&{4 + 1 + 4}&{ - 4 + 2 + 2}\\{ - 2 + 4 - 2}&{ - 4 + 2 + 2}&{4 + 4 + 1}\end{array}} \right)$

= $\frac{1}{9}\left( {\begin{array}{*{20}{c}}9&0&0\\0&9&0\\0&0&9\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right) = $ I

A^T.A = $\frac{1}{9}\left( {\begin{array}{*{20}{c}}1&2&{ - 2}\\2&1&2\\2&{ - 2}&{ - 1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&2&2\\2&1&{ - 2}\\{ - 2}&2&{ - 1}\end{array}} \right)$

= $\frac{1}{9}\left( {\begin{array}{*{20}{c}}{1 + 4 + 4}&{2 + 2 - 4}&{ - 2 - 4 + 2}\\{2 + 2 - 4}&{4 + 1 + 4}&{4 - 2 - 2}\\{2 - 4 + 2}&{4 - 2 - 2}&{4 + 4 + 1}\end{array}} \right)$

= $\frac{1}{9}\left( {\begin{array}{*{20}{c}}9&0&0\\0&9&0\\0&0&9\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right) = $ I

So, A.A^T = A^T.A = I

 

3) If A = $\left( {\begin{array}{*{20}{c}}2&4&3\\2&3&4\\5&2&6\end{array}} \right)$, Find A^T

a)Show that the sum of the given matrix and its transpose is a symmetric matrix.

Solution:

Given:

A = $\left( {\begin{array}{*{20}{c}}2&4&3\\2&3&4\\5&2&6\end{array}} \right)$

Then A^T = $\left( {\begin{array}{*{20}{c}}2&2&5\\4&3&2\\3&4&6\end{array}} \right)$

Now, A + A^T = $\left( {\begin{array}{*{20}{c}}2&4&3\\2&3&4\\5&2&6\end{array}} \right) + \left( {\begin{array}{*{20}{c}}2&2&5\\4&3&2\\3&4&6\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}4&6&8\\6&6&6\\8&6&{12}\end{array}} \right)$

Which is a symmetric matrix. 

b) Show that the difference of the given matrix and its transpose is a sjew-symmetric matrix.

Now, A – A^T = $\left( {\begin{array}{*{20}{c}}2&4&3\\2&3&4\\5&2&6\end{array}} \right) - \left( {\begin{array}{*{20}{c}}2&2&5\\4&3&2\\3&4&6\end{array}} \right)$ = $\left( {\begin{array}{*{20}{c}}0&2&{ - 2}\\{ - 2}&0&2\\2&{ - 2}&0\end{array}} \right)$

4. a) If A= $\left( {\begin{array}{*{20}{c}}4&{x+2}\\2x-1&0\end{array}} \right)$ and A=A^T , Find the value of x.

Solution: 

A= $\left( {\begin{array}{*{20}{c}}4&{x+2}\\2x-1&0\end{array}} \right)$ and 

A^T = $\left( {\begin{array}{*{20}{c}}4&{2x-1}\\x+2&0\end{array}} \right)$ 

And, A=A^T

^i.e. $\left( {\begin{array}{*{20}{c}}4&{x+2}\\2x-1&0\end{array}} \right)$=$\left( {\begin{array}{*{20}{c}}4&{2x-1}\\x+2&0\end{array}} \right)$ 

or, x+2=2x-1

or,-x=-3

or,x=3

Thus, x=3.

b) If A=$\left( {\begin{array}{*{20}{c}}0&{2y-3}\\1-y&0\end{array}} \right)$ and A=-A^T, Find the value of y.

Solution:

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

A^T=$\left( {\begin{array}{*{20}{c}}0&{1-y}\\2y-3&0\end{array}} \right)$ and 

-A^T=$\left( {\begin{array}{*{20}{c}}0&{y-1}\\3-2y&0\end{array}} \right)$ 

And, A=-A^T

$\left( {\begin{array}{*{20}{c}}0&{2y-3}\\1-y&0\end{array}} \right)$ =$\left( {\begin{array}{*{20}{c}}0&{y-1}\\3-2y&0\end{array}} \right)$ 

i.e. 2y-3=y-1

or, y=2

Thus, y=2