# NCERT Solutions For Class 12 Maths Chapter 3 Exercise 3.2

## Chapter 3 Ex.3.2 Question 1

Let $$A = \left[ {\begin{array}{*{20}{c}}2&4 \\3&2\end{array}} \right],\;B = \left[ {\begin{array}{*{20}{c}}1&3 \\{ - 2}&5\end{array}} \right],\;C = \left[ {\begin{array}{*{20}{c}}{ - 2}&5 \\3&4\end{array}} \right]$$. Find each of the following:

(i) $$A + B$$

(ii) $$A - B$$

(iii) $$3A - C$$

(iv) $$AB$$

(v) $$BA$$

### Solution

(i) $$A + B$$

\begin{align} &\Rightarrow \left( {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right) + \left( {\begin{array}{*{20}{c}}1&3\\{ - 2}&5\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{2 + 1}&{4 + 3}\\{3 - 2}&{2 + 5}\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}3&7\\1&7\end{array}} \right)\end{align}

(ii) $$A - B$$

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

(iii) $$3A - C$$

\begin{align}&\Rightarrow 3\left( {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{ - 2}&5\\3&4\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{3 \times 2}&{3 \times 4}\\{3 \times 3}&{3 \times 2}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{ - 2}&5\\3&4\end{array}} \right)\\& \Rightarrow \left( {\begin{array}{*{20}{c}}{6 + 2}&{12 - 5}\\{9 - 3}&{6 - 4}\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}8&7\\6&2\end{array}} \right)\end{align}

(iv) $$AB$$

\begin{align} &\Rightarrow \left( {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&3\\{ - 2}&5\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{2\left( 1 \right) + 4\left( { - 2} \right)}&{2\left( 3 \right) + 4\left( 5 \right)}\\{3\left( 1 \right) + 2\left( { - 2} \right)}&{3\left( 3 \right) + 2\left( 5 \right)}\end{array}} \right)\\& \Rightarrow \left( {\begin{array}{*{20}{c}}{2 - 8}&{6 + 20}\\{3 - 4}&{9 + 10}\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{ - 6}&{26}\\{ - 1}&{19}\end{array}} \right)\end{align}

(v) $$BA$$

\begin{align}&\Rightarrow \left( {\begin{array}{*{20}{c}}1&3\\{ - 2}&5\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{1\left( 2 \right) + 3\left( 3 \right)}&{1\left( 4 \right) + 3\left( 2 \right)}\\{ - 2\left( 2 \right) + 5\left( 3 \right)}&{ - 2\left( 4 \right) + 5\left( 2 \right)}\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{2 + 9}&{4 + 6}\\{ - 4 + 15}&{ - 8 + 10}\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{11}&{10}\\{11}&2\end{array}} \right)\end{align}

## Chapter 3 Ex.3.2 Question 2

Compute the following:

(i) $$\left( {\begin{array}{*{20}{c}}a&b\\{ - b}&a\end{array}} \right) + \left( {\begin{array}{*{20}{c}}a&b\\b&a\end{array}} \right)$$

(ii) $$\left( {\begin{array}{*{20}{c}}{{a^2} + {b^2}}&{{b^2} + {c^2}}\\{{a^2} + {c^2}}&{{a^2} + {b^2}}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2ab}&{2bc}\\{ - 2ac}&{ - 2ab}\end{array}} \right)$$

(iii) $$\left( {\begin{array}{*{20}{c}}{ - 1}&4&{ - 6}\\8&5&{16}\\2&8&5\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{12}&7&6\\8&0&5\\3&2&4\end{array}} \right)$$

(iv)$$\left( {\begin{array}{*{20}{c}}{{{\cos }^2}x}&{{{\sin }^2}x}\\{{{\sin }^2}x}&{{{\cos }^2}x}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{{{\sin }^2}x}&{{{\cos }^2}x}\\{{{\cos }^2}x}&{{{\sin }^2}x}\end{array}} \right)$$

### Solution

(i) $$\left( {\begin{array}{*{20}{c}}a&b\\{ - b}&a\end{array}} \right) + \left( {\begin{array}{*{20}{c}}a&b\\b&a\end{array}} \right)$$

\begin{align} &\Rightarrow \left( {\begin{array}{*{20}{c}}{a + a}&{b + b}\\{ - b + b}&{a + a}\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{2a}&{2b}\\0&{2a}\end{array}} \right)\end{align}

(ii) $$\left( {\begin{array}{*{20}{c}}{{a^2} + {b^2}}&{{b^2} + {c^2}}\\{{a^2} + {c^2}}&{{a^2} + {b^2}}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2ab}&{2bc}\\{ - 2ac}&{ - 2ab}\end{array}} \right)$$

\begin{align}& \Rightarrow \left( {\begin{array}{*{20}{c}}{{a^2} + {b^2} + 2ab}&{{b^2} + {c^2} + 2bc}\\{{a^2} + {c^2} - 2ac}&{{a^2} + {b^2} - 2ab}\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{{{\left( {a + b} \right)}^2}}&{{{\left( {b + c} \right)}^2}}\\{{{\left( {a - c} \right)}^2}}&{{{\left( {a - b} \right)}^2}}\end{array}} \right)\end{align}

(iii) $$\left( {\begin{array}{*{20}{c}}{ - 1}&4&{ - 6}\\8&5&{16}\\2&8&5\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{12}&7&6\\8&0&5\\3&2&4\end{array}} \right)$$

\begin{align}& \Rightarrow \left( {\begin{array}{*{20}{c}}{ - 1 + 12}&{4 + 7}&{ - 6 + 6}\\{8 + 8}&{5 + 0}&{16 + 5}\\{2 + 3}&{8 + 2}&{5 + 4}\end{array}} \right)\\& \Rightarrow \left( {\begin{array}{*{20}{c}}{11}&{11}&0\\{16}&5&{21}\\5&{10}&9\end{array}} \right)\end{align}

(iv)$$\left( {\begin{array}{*{20}{c}}{{{\cos }^2}x}&{{{\sin }^2}x}\\{{{\sin }^2}x}&{{{\cos }^2}x}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{{{\sin }^2}x}&{{{\cos }^2}x}\\{{{\cos }^2}x}&{{{\sin }^2}x}\end{array}} \right)$$

\begin{align}&\Rightarrow \left( {\begin{array}{*{20}{c}}{{{\cos }^2}x + {{\sin }^2}x}&{{{\sin }^2}x + {{\cos }^2}x}\\{{{\sin }^2}x + {{\cos }^2}x}&{{{\cos }^2}x + {{\sin }^2}x}\end{array}} \right)\\&\Rightarrow \left( {\begin{array}{*{20}{c}}1&1\\1&1\end{array}} \right)\end{align}

## Chapter 3 Ex.3.2 Question 3

Compute the indicated products:

(i) $$\left( {\begin{array}{*{20}{c}}a&b\\{ - b}&a\end{array}} \right)\left( {\begin{array}{*{20}{c}}a&{ - b}\\b&a\end{array}} \right)$$

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

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

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

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

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

### Solution

(i) $$\left( {\begin{array}{*{20}{c}}a&b\\{ - b}&a\end{array}} \right)\left( {\begin{array}{*{20}{c}}a&{ - b}\\b&a\end{array}} \right)$$

\begin{align}&\Rightarrow \left( {\begin{array}{*{20}{c}}{a\left( a \right) + b\left( b \right)}&{a\left( { - b} \right) + b\left( a \right)}\\{ - b\left( a \right) + a\left( b \right)}&{ - b\left( { - b} \right) + a\left( a \right)}\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{{a^2} + {b^2}}&{ - ab + ab}\\{ - ab + ab}&{{b^2} + {a^2}}\end{array}} \right)\\&\Rightarrow \left( {\begin{array}{*{20}{c}}{{a^2} + {b^2}}&0\\0&{{a^2} + {b^2}}\end{array}} \right)\end{align}

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

\begin{align} &\Rightarrow \left( {\begin{array}{*{20}{c}}{1\left( 2 \right)}&{1\left( 3 \right)}&{1\left( 4 \right)}\\{2\left( 2 \right)}&{2\left( 3 \right)}&{2\left( 4 \right)}\\{3\left( 2 \right)}&{3\left( 3 \right)}&{3\left( 4 \right)}\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}2&3&4\\4&6&8\\6&9&{12}\end{array}} \right)\end{align}

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

\begin{align}&\Rightarrow \left( {\begin{array}{*{20}{c}}{1\left( 1 \right) - 2\left( 2 \right)}&{1\left( 2 \right) - 2\left( 3 \right)}&{1\left( 3 \right) - 2\left( 1 \right)}\\{2\left( 1 \right) + 3\left( 2 \right)}&{2\left( 2 \right) + 3\left( 3 \right)}&{2\left( 3 \right) + 3\left( 1 \right)}\end{array}} \right)\\& \Rightarrow \left( {\begin{array}{*{20}{c}}{ - 3}&{ - 4}&1\\8&{13}&9\end{array}} \right)\end{align}

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

\begin{align}& \Rightarrow \left( {\begin{array}{*{20}{c}}{2\left( 1 \right) + 3\left( 0 \right) + 4\left( 3 \right)}&{2\left( { - 3} \right) + 3\left( 2 \right) + 4\left( 0 \right)}&{2\left( 5 \right) + 3\left( 4 \right) + 4\left( 5 \right)}\\{3\left( 1 \right) + 4\left( 0 \right) + 5\left( 3 \right)}&{3\left( { - 3} \right) + 4\left( 2 \right) + 5\left( 0 \right)}&{3\left( 5 \right) + 4\left( 4 \right) + 5\left( 5 \right)}\\{4\left( 1 \right) + 5\left( 0 \right) + 6\left( 3 \right)}&{4\left( { - 3} \right) + 5\left( 2 \right) + 6\left( 0 \right)}&{4\left( 5 \right) + 5\left( 4 \right) + 6\left( 5 \right)}\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{14}&0&{42}\\{18}&{ - 1}&{56}\\{22}&{ - 2}&{70}\end{array}} \right)\end{align}

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

\begin{align}&\Rightarrow \left( {\begin{array}{*{20}{c}}{2\left( 1 \right) + 1\left( { - 1} \right)}&{2\left( 0 \right) + 1\left( 2 \right)}&{2\left( 1 \right) + 1\left( 1 \right)}\\{3\left( 1 \right) + 2\left( { - 1} \right)}&{3\left( 0 \right) + 2\left( 2 \right)}&{3\left( 1 \right) + 2\left( 1 \right)}\\{ - 1\left( 1 \right) + 1\left( { - 1} \right)}&{ - 1\left( 0 \right) + 1\left( 2 \right)}&{ - 1\left( 1 \right) + 1\left( 1 \right)}\end{array}} \right)\\&\Rightarrow \left( {\begin{array}{*{20}{c}}1&2&3\\1&4&5\\{ - 2}&2&0\end{array}} \right)\end{align}

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

\begin{align}&\Rightarrow \left( {\begin{array}{*{20}{c}}{3\left( 2 \right) - 1\left( 1 \right) + 3\left( 3 \right)}&{3\left( { - 3} \right) - 1\left( 0 \right) + 3\left( 1 \right)}\\{ - 1\left( 2 \right) + 0\left( 1 \right) + 2\left( 3 \right)}&{ - 1\left( { - 3} \right) + 0\left( 0 \right) + 2\left( 1 \right)}\end{array}} \right)\\&\Rightarrow \left( {\begin{array}{*{20}{c}}{14}&{ - 6}\\4&5\end{array}} \right)\end{align}

## Chapter 3 Ex.3.2 Question 4

If $$A = \left( {\begin{array}{*{20}{c}}1&2&{ - 3}\\5&0&2\\1&{ - 1}&1\end{array}} \right),\;B = \left( {\begin{array}{*{20}{c}}3&{ - 1}&2\\4&2&5\\2&0&3\end{array}} \right),\;C = \left( {\begin{array}{*{20}{c}}4&1&2\\0&3&2\\1&{ - 2}&3\end{array}} \right)$$, then compute $$\left( {A + B} \right)$$ and $$\left( {B - C} \right)$$. Also, verify that $$A + \left( {B - C} \right) = \left( {A + B} \right) - C$$.

### Solution

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

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

Now,

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

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

Hence, $$A + \left( {B - C} \right) = \left( {A + B} \right) - C$$.

## Chapter 3 Ex.3.2 Question 5

If $$A = \left( {\begin{array}{*{20}{c}}{\frac{2}{3}}&1&{\fracabc{3}}\\{\frac{1}{3}}&{\frac{2}{3}}&{\frac{4}{3}}\\{\frac{7}{3}}&2&{\frac{2}{3}}\end{array}} \right)$$ and B = \left( {\begin{array}{*{20}{c}}{\frac{2}{5}}&{\frac{3}{5}}&1\\{\frac{1}{5}}&{\frac{2}{5}}&{\frac{4}{5}}\\{\frac{7}{5}}&{\frac \begin{align}3A - 5B &= 3\left( {\begin{array}{*{20}{c}}{\frac{2}{3}}&1&{\frac{5}{3}}\\{\frac{1}{3}}&{\frac{2}{3}}&{\frac{4}{3}}\\{\frac{7}{3}}&2&{\frac{2}{3}}\end{array}} \right) - 5\left( {\begin{array}{*{20}{c}}{\frac{2}{5}}&{\frac{3}{5}}&1\\{\frac{1}{5}}&{\frac{2}{5}}&{\frac{4}{5}}\\{\frac{7}{5}}&{\frac{6}{5}}&{\frac{2}{5}}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}2&3&5\\1&2&4\\7&6&2\end{array}} \right) - \left( {\begin{array}{*{20}{c}}2&3&5\\1&2&4\\7&6&2\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right)\end{align} {5}}&{\frac{2}{5}}\end{array}} \right), then compute $$3A - 5B$$.

{6}

## Chapter 3 Ex.3.2 Question 6

Simplify $$\cos \theta \left( {\begin{array}{*{20}{c}}{\cos \theta }&{\sin \theta }\\{ - \sin \theta }&{\cos \theta }\end{array}} \right) + \sin \theta \left( {\begin{array}{*{20}{c}}{\sin \theta }&{ - \cos \theta }\\{\cos \theta }&{\sin \theta }\end{array}} \right)$$.

### Solution

$$\cos \theta \left( {\begin{array}{*{20}{c}}{\cos \theta }&{\sin \theta }\\{ - \sin \theta }&{\cos \theta }\end{array}} \right) + \sin \theta \left( {\begin{array}{*{20}{c}}{\sin \theta }&{ - \cos \theta }\\{\cos \theta }&{\sin \theta }\end{array}} \right)$$

\begin{align}& \Rightarrow \left( {\begin{array}{*{20}{c}}{{{\cos }^2}\theta }&{\cos \theta \sin \theta }\\{ - \sin \theta \cos \theta }&{{{\cos }^2}\theta }\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{{{\sin }^2}\theta }&{ - \sin \theta \cos \theta }\\{\sin \theta \cos \theta }&{{{\sin }^2}\theta }\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{{{\cos }^2}\theta + {{\sin }^2}\theta }&{\sin \theta \cos \theta - \sin \theta \cos \theta }\\{ - \sin \theta \cos \theta + \sin \theta \cos \theta }&{{{\cos }^2}\theta + {{\sin }^2}\theta }\end{array}} \right)\\& \Rightarrow \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\end{align}

## Chapter 3 Ex.3.2 Question 7

Find $$X$$ and $$Y$$, if

(i) $$X + Y = \left( {\begin{array}{*{20}{c}}7&0\\2&5\end{array}} \right)\,\,{\rm{and }}X - Y = \left( {\begin{array}{*{20}{c}}3&0\\0&3\end{array}} \right)$$

(ii) $$2X + 3Y = \left( {\begin{array}{*{20}{c}}2&3\\4&0\end{array}} \right)\,\,{\rm{and }}3X + 2Y = \left( {\begin{array}{*{20}{c}}2&{ - 2}\\{ - 1}&5\end{array}} \right)$$

### Solution

$$X + Y = \left( {\begin{array}{*{20}{c}}7&0\\2&5\end{array}} \right)\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)$$

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

Adding equations $$\left( 1 \right)$$ and $$\left( 2 \right)$$,

\begin{align}2X &= \left( {\begin{array}{*{20}{c}}7&0\\2&5\end{array}} \right) + \left( {\begin{array}{*{20}{c}}3&0\\0&3\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{10}&0\\2&8\end{array}} \right)\\X &= \frac{1}{2}\left( {\begin{array}{*{20}{c}}{10}&0\\2&8\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}5&0\\1&4\end{array}} \right)\end{align}

Now,

\begin{align}&\Rightarrow X + Y = \left( {\begin{array}{*{20}{c}}7&0\\2&5\end{array}} \right)\\ &\Rightarrow Y = \left( {\begin{array}{*{20}{c}}7&0\\2&5\end{array}} \right) - \left( {\begin{array}{*{20}{c}}5&0\\1&4\end{array}} \right)\\ &\Rightarrow Y = \left( {\begin{array}{*{20}{c}}2&0\\1&1\end{array}} \right)\end{align}

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

$$3X + 2Y = \left( {\begin{array}{*{20}{c}}2&{ - 2}\\{ - 1}&5\end{array}} \right)\;\;\;\;\;\;\; \ldots \left( 2 \right)$$

Multiplying equation $$\left( 1 \right)$$by 2,

\begin{align}&2\left( {2X + 3Y} \right) = 2\left( {\begin{array}{*{20}{c}}2&3\\4&0\end{array}} \right)\\&4X + 6Y = \left( {\begin{array}{*{20}{c}}4&6\\8&0\end{array}} \right)\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}

Multiplying equation $$\left( 2 \right)$$ by 3,

\begin{align}&3\left( {3X + 2Y} \right) = 3\left( {\begin{array}{*{20}{c}}2&{ - 2}\\{ - 1}&5\end{array}} \right)\\&9X + 6Y = \left( {\begin{array}{*{20}{c}}6&{ - 6}\\{ - 3}&{15}\end{array}} \right)\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}

From $$\left( 3 \right)$$ and $$\left( 4 \right)$$,

$$\left( {4X + 6Y} \right) - \left( {9X + 6Y} \right) = \left( {\begin{array}{*{20}{c}}4&6\\8&0\end{array}} \right) - \left( {\begin{array}{*{20}{c}}6&{ - 6}\\{ - 3}&{15}\end{array}} \right)$$

\begin{align} - 5X &= \left( {\begin{array}{*{20}{c}}{4 - 6}&{6 + 6}\\{8 + 3}&{0 - 15}\end{array}} \right)\\ - 5X &= \left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{11}&{ - 15}\end{array}} \right)\\X &= \frac{{ - 1}}{5}\left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{11}&{ - 15}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{\frac{2}{5}}&{\frac{{ - 12}}{5}}\\{\frac{{ - 11}}{5}}&3\end{array}} \right)\end{align}

Now,

\begin{align}&\Rightarrow 2X + 3Y = \left( {\begin{array}{*{20}{c}}2&3\\4&0\end{array}} \right)\\ &\Rightarrow 2\left( {\begin{array}{*{20}{c}}{\frac{2}{5}}&{\frac{{ - 12}}{5}}\\{\frac{{ - 11}}{5}}&3\end{array}} \right) + 3Y = \left( {\begin{array}{*{20}{c}}2&3\\4&0\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{\frac{4}{5}}&{\frac{{ - 24}}{5}}\\{\frac{{ - 22}}{5}}&6\end{array}} \right) + 3Y = \left( {\begin{array}{*{20}{c}}2&3\\4&0\end{array}} \right)\\ &\Rightarrow 3Y = \left( {\begin{array}{*{20}{c}}2&3\\4&0\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{\frac{4}{5}}&{\frac{{ - 24}}{5}}\\{\frac{{ - 22}}{5}}&6\end{array}} \right)\\ &\Rightarrow 3Y = \left( {\begin{array}{*{20}{c}}{\frac{6}{5}}&{\frac{{39}}{5}}\\{\frac{{42}}{5}}&{ - 6}\end{array}} \right)\\ &\Rightarrow Y = \frac{1}{3}\left( {\begin{array}{*{20}{c}}{\frac{6}{5}}&{\frac{{39}}{5}}\\{\frac{{42}}{5}}&{ - 6}\end{array}} \right)\\ &\Rightarrow Y = \left( {\begin{array}{*{20}{c}}{\frac{2}{5}}&{\frac{{13}}{5}}\\{\frac{{14}}{5}}&{ - 2}\end{array}} \right)\end{align}

## Chapter 3 Ex.3.2 Question 8

Find , if $$Y = \left( {\begin{array}{*{20}{c}}3&2 \\1&4\end{array}} \right)$$ and $$2X + Y = \left( {\begin{array}{*{20}{c}}1&0 \\{ - 3}&2\end{array}} \right)$$.

### Solution

\begin{align}&2X + Y = \left( {\begin{array}{*{20}{c}}1&0 \\{ - 3}&2\end{array}} \right) \hfill \\&\Rightarrow 2X + \left( {\begin{array}{*{20}{c}}3&2 \\1&4\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0 \\{ - 3}&2\end{array}} \right) \hfill \\&\Rightarrow 2X = \left( {\begin{array}{*{20}{c}}1&0 \\{ - 3}&2\end{array}} \right) - \left( {\begin{array}{*{20}{c}}3&2 \\1&4\end{array}} \right) \hfill \\&\Rightarrow 2X = \left( {\begin{array}{*{20}{c}}{ - 2}&{ - 2} \\{ - 4}&{ - 2}\end{array}} \right) \hfill \\&\Rightarrow X = \frac{1}{2}\left( {\begin{array}{*{20}{c}}{ - 2}&{ - 2} \\{ - 4}&{ - 2}\end{array}} \right) \hfill \\&\Rightarrow X = \left( {\begin{array}{*{20}{c}}{ - 1}&{ - 1} \\{ - 2}&{ - 1}\end{array}} \right) \hfill \\\end{align}

## Chapter 3 Ex.3.2 Question 9

Find and $$y$$, if $$2\left( {\begin{array}{*{20}{c}}1&3 \\0&x\end{array}} \right) + \left( {\begin{array}{*{20}{c}}y&0 \\1&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}5&6 \\1&8\end{array}} \right)$$.

### Solution

\begin{align} &\Rightarrow 2\left( {\begin{array}{*{20}{c}}1&3\\0&x\end{array}} \right) + \left( {\begin{array}{*{20}{c}}y&0\\1&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}5&6\\1&8\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}2&6\\0&{2x}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}y&0\\1&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}5&6\\1&8\end{array}} \right)\\& \Rightarrow \left( {\begin{array}{*{20}{c}}{2 + y}&6\\1&{2x + 2}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}5&6\\1&8\end{array}} \right)\end{align}

Comparing the corresponding elements of these two matrices,

\begin{align}2 + y &= 5\\ \Rightarrow y &= 3\\\\&2x + 2 = 8\\& \Rightarrow x = 3\end{align}

Therefore, $$x = 3$$ and $$y = 3$$.

## Chapter 3 Ex.3.2 Question 10

Solve the equation for $$x,y,z$$ and $$t$$ if $$2\left( {\begin{array}{*{20}{c}}x&z\\y&t\end{array}} \right) + 3\left( {\begin{array}{*{20}{c}}1&{ - 1}\\0&2\end{array}} \right) = 3\left( {\begin{array}{*{20}{c}}3&5\\4&6\end{array}} \right)$$.

### Solution

\begin{align} &\Rightarrow 2\left( {\begin{array}{*{20}{c}}x&z\\y&t\end{array}} \right) + 3\left( {\begin{array}{*{20}{c}}1&{ - 1}\\0&2\end{array}} \right) = 3\left( {\begin{array}{*{20}{c}}3&5\\4&6\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{2x}&{2z}\\{2y}&{2t}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}3&{ - 3}\\0&6\end{array}} \right) = \left( {\begin{array}{*{20}{c}}9&{15}\\{12}&{18}\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{2x + 3}&{2z - 3}\\{2y}&{2t + 6}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}9&{15}\\{12}&{18}\end{array}} \right)\end{align}

Comparing the corresponding elements of these two matrices,

\begin{align}&2x + 3 = 9\\&\Rightarrow 2x = 6\\ &\Rightarrow x = 3\end{align}

\begin{align}&2y = 12\\ &\Rightarrow y = 6\end{align}

\begin{align}&2z - 3 = 15\\ &\Rightarrow 2z = 18\\ &\Rightarrow z = 9\end{align}

\begin{align}&2t + 6 = 18\\ &\Rightarrow 2t = 12\\ &\Rightarrow t = 6\end{align}

Therefore, $$x = 3,y = 6,z = 9$$and $$t = 6$$.

## Chapter 3 Ex.3.2 Question 11

If $$x\left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right) + y\left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{10}\\5\end{array}} \right)$$, find values of $$x$$ and $$y$$.

### Solution

\begin{align} &\Rightarrow x\left( {\begin{array}{*{20}{c}}2\\3\end{array}} \right) + y\left( {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{10}\\5\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{2x}\\{3x}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - y}\\y\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{10}\\5\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{2x - y}\\{3x + y}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{10}\\5\end{array}} \right)\end{align}

Comparing the corresponding elements of these two matrices,

\begin{align}2x - y = 10\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\3x + y = 5\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

By adding these two equations, we get

\begin{align}&5x = 15\\ &\Rightarrow x = 3\end{align}

Now, putting this value in $$\left( 2 \right)$$

\begin{align}&\Rightarrow 3x + y = 5\\&\Rightarrow y = 5 - 3x\\&\Rightarrow y = 5 - 3\left( 3 \right)\\ &\Rightarrow y = 5 - 9\\&\Rightarrow y = - 4 \end{align}

Therefore, $$x = 3$$ and $$y = 4$$.

## Chapter 3 Ex.3.2 Question 12

Given $$3\left( {\begin{array}{*{20}{c}}x&y\\z&w\end{array}} \right) = \left( {\begin{array}{*{20}{c}}x&6\\{ - 1}&{2w}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}4&{x + y}\\{z + w}&3\end{array}} \right)$$ , find values of $$w,x,y$$ and $$z$$.

### Solution

\begin{align}&\Rightarrow 3\left( {\begin{array}{*{20}{c}}x&y\\z&w\end{array}} \right) = \left( {\begin{array}{*{20}{c}}x&6\\{ - 1}&{2w}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}4&{x + y}\\{z + w}&3\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{3x}&{3y}\\{3z}&{3w}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{x + 4}&{6 + x + y}\\{ - 1 + z + w}&{2w + 3}\end{array}} \right)\end{align}

Comparing the corresponding elements of these two matrices,

\begin{align} &\Rightarrow 3x = x + 4\\ &\Rightarrow 2x = 4\\ &\Rightarrow x = 2\end{align}

\begin{align} &\Rightarrow 3y = 6 + x + y\\ &\Rightarrow 2y = 6 + x\\ &\Rightarrow 2y = 6 + 2\\ &\Rightarrow 2y = 8\\ &\Rightarrow y = 4\end{align}

\begin{align} &\Rightarrow 3w = 2w + 3\\ &\Rightarrow w = 3\end{align}

\begin{align} &\Rightarrow 3z = - 1 + z + w\\ &\Rightarrow 2z = w - 1\\ &\Rightarrow 2z = 3 - 1\\ &\Rightarrow 2z = = 2\\ &\Rightarrow z = 1\end{align}

Therefore,$$x = 2,y = 4,z = 1$$ and $$w = 3$$

## Chapter 3 Ex.3.2 Question 13

If $$F\left( x \right) = \left( {\begin{array}{*{20}{c}}{\cos x}&{ - \sin x}&0\\{\sin x}&{\cos x}&0\\0&0&1\end{array}} \right)$$, show that $$F\left( x \right)F\left( y \right) = F\left( {x + y} \right)$$.

### Solution

It is given that $$F\left( x \right) = \left( {\begin{array}{*{20}{c}}{\cos x}&{ - \sin x}&0\\{\sin x}&{\cos x}&0\\0&0&1\end{array}} \right)$$

Then, $$F\left( y \right) = \left( {\begin{array}{*{20}{c}}{\cos y}&{ - \sin y}&0\\{\sin y}&{\cos y}&0\\0&0&1\end{array}} \right)$$

Now,

$$F\left( {x + y} \right) = \left( {\begin{array}{*{20}{c}}{\cos \left( {x + y} \right)}&{ - \sin \left( {x + y} \right)}&0\\{\sin \left( {x + y} \right)}&{\cos \left( {x + y} \right)}&0\\0&0&1\end{array}} \right)$$

\begin{align}F\left( x \right)F\left( y \right) &= \left( {\begin{array}{*{20}{c}}{\cos x}&{ - \sin x}&0\\{\sin x}&{\cos x}&0\\0&0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\cos y}&{ - \sin y}&0\\{\sin y}&{\cos y}&0\\0&0&1\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{\cos x\cos y - \sin x\sin y + 0}&{ - \cos x\sin y - \sin x\cos y + 0}&0\\{\sin x\cos y + \cos x\sin y + 0}&{ - \sin x\sin y + \cos x\cos y + 0}&0\\0&0&0\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{\cos \left( {x + y} \right)}&{ - \sin \left( {x + y} \right)}&0\\{\sin \left( {x + y} \right)}&{\cos \left( {x + y} \right)}&0\\0&0&1\end{array}} \right)\\ &= F\left( {x + y} \right)\end{align}

Therefore, $$F\left( x \right)F\left( y \right) = F\left( {x + y} \right)$$

## Chapter 3 Ex.3.2 Question 14

Show that

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

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

### Solution

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

\begin{align}\left( {\begin{array}{*{20}{c}}5&{ - 1}\\6&7\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&1\\3&4\end{array}} \right) &= \left( {\begin{array}{*{20}{c}}{5\left( 2 \right) - 1\left( 3 \right)}&{5\left( 1 \right) - 1\left( 4 \right)}\\{6\left( 2 \right) + 7\left( 3 \right)}&{6\left( 1 \right) + 7\left( 4 \right)}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{10 - 3}&{5 - 4}\\{12 + 21}&{6 + 28}\end{array}} \right)\\& = \left( {\begin{array}{*{20}{c}}7&1\\{33}&{34}\end{array}} \right)\end{align}

\begin{align}\left( {\begin{array}{*{20}{c}}2&1\\3&4\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&{ - 1}\\6&7\end{array}} \right) &= \left( {\begin{array}{*{20}{c}}{2\left( 5 \right) + 1\left( 6 \right)}&{2\left( { - 1} \right) + 1\left( 7 \right)}\\{3\left( 5 \right) + 4\left( 6 \right)}&{3\left( { - 1} \right) + 4\left( 7 \right)}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{10 + 6}&{ - 2 + 7}\\{15 + 24}&{ - 3 + 28}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{16}&5\\{39}&{25}\end{array}} \right)\end{align}

Thus, $$\left( {\begin{array}{*{20}{c}}5&{ - 1}\\6&7\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&1\\3&4\end{array}} \right) \ne \left( {\begin{array}{*{20}{c}}2&1\\3&4\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&{ - 1}\\6&7\end{array}} \right)$$

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

\begin{align}\left( {\begin{array}{*{20}{c}}1&2&3\\0&1&0\\1&1&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&1&0\\0&{ - 1}&1\\2&3&4\end{array}} \right) &= \left( {\begin{array}{*{20}{c}}{1\left( { - 1} \right) + 2\left( 0 \right) + 3\left( 2 \right)}&{1\left( 1 \right) + 2\left( { - 1} \right) + 3\left( 3 \right)}&{1\left( 0 \right) + 2\left( 1 \right) + 3\left( 4 \right)}\\{0\left( { - 1} \right) + 1\left( 0 \right) + 0\left( 2 \right)}&{0\left( 1 \right) + 1\left( { - 1} \right) + 0\left( 3 \right)}&{0\left( 0 \right) + 1\left( 1 \right) + 0\left( 4 \right)}\\{1\left( { - 1} \right) + 1\left( 0 \right) + 0\left( 2 \right)}&{1\left( 1 \right) + 1\left( { - 1} \right) + 0\left( 3 \right)}&{1\left( 0 \right) + 1\left( 1 \right) + 0\left( 4 \right)}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}5&8&{14}\\0&{ - 1}&1\\{ - 1}&0&1\end{array}} \right)\end{align}

\begin{align}\left( {\begin{array}{*{20}{c}}{ - 1}&1&0\\0&{ - 1}&1\\2&3&4\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&2&3\\0&1&0\\1&1&0\end{array}} \right) &= \left( {\begin{array}{*{20}{c}}{ - 1\left( 1 \right) + 1\left( 0 \right) + 0\left( 1 \right)}&{ - 1\left( 2 \right) + 1\left( 1 \right) + 0\left( 1 \right)}&{ - 1\left( 3 \right) + 1\left( 0 \right) + 0\left( 0 \right)}\\{0\left( 1 \right) + \left( { - 1} \right)\left( 0 \right) + 1\left( 1 \right)}&{0\left( 2 \right) + \left( { - 1} \right)\left( 1 \right) + 1\left( 1 \right)}&{0\left( 3 \right) + - 1\left( 0 \right) + 1\left( 0 \right)}\\{2\left( 1 \right) + 3\left( 0 \right) + 4\left( 1 \right)}&{2\left( 2 \right) + 3\left( 1 \right) + 4\left( 1 \right)}&{2\left( 3 \right) + 3\left( 0 \right) + 4\left( 0 \right)}\end{array}} \right)\\&= \left( {\begin{array}{*{20}{c}}{ - 1}&{ - 1}&{ - 3}\\1&0&0\\6&{11}&6\end{array}} \right)\end{align}

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

## Chapter 3 Ex.3.2 Question 15

Find $${A^2} - 5A + 6I$$, if $$A = \left( {\begin{array}{*{20}{c}}2&0&1\\2&1&3\\1&{ - 1}&0\end{array}} \right)$$

### Solution

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

Therefore,

\begin{align}{A^2} - 5A + 6I &= \left( {\begin{array}{*{20}{c}}5&{ - 1}&2\\9&{ - 2}&5\\0&{ - 1}&{ - 2}\end{array}} \right) - 5\left( {\begin{array}{*{20}{c}}2&0&1\\2&1&3\\1&{ - 1}&0\end{array}} \right) + 6\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}5&{ - 1}&2\\9&{ - 2}&5\\0&{ - 1}&{ - 2}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{10}&0&5\\{10}&5&{15}\\5&{ - 5}&0\end{array}} \right) + \left( {\begin{array}{*{20}{c}}6&0&0\\0&6&0\\0&0&6\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{5 - 10}&{ - 1 - 0}&{2 - 5}\\{9 - 10}&{ - 2 - 5}&{5 - 15}\\{0 - 5}&{ - 1 + 5}&{ - 2 - 0}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}6&0&0\\0&6&0\\0&0&6\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{ - 5}&{ - 1}&{ - 3}\\{ - 1}&{ - 7}&{ - 10}\\{ - 5}&4&{ - 2}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}6&0&0\\0&6&0\\0&0&6\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}1&{ - 1}&{ - 3}\\{ - 1}&{ - 1}&{ - 10}\\{ - 5}&4&4\end{array}} \right)\end{align}

## Chapter 3 Ex.3.2 Question 16

If $$A = \left( {\begin{array}{*{20}{c}}1&0&2\\0&2&1\\2&0&3\end{array}} \right)$$, prove that $${A^3} - 6{A^2} + 7A + 2I = 0$$.

### Solution

\begin{align}{A^2} &= A.A\\ &= \left( {\begin{array}{*{20}{c}}1&0&2\\0&2&1\\2&0&3\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0&2\\0&2&1\\2&0&3\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{1 + 0 + 4}&{0 + 0 + 0}&{2 + 0 + 6}\\{0 + 0 + 2}&{0 + 4 + 0}&{0 + 2 + 3}\\{2 + 0 + 6}&{0 + 0 + 0}&{4 + 0 + 9}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}5&0&8\\2&4&5\\8&0&{13}\end{array}} \right)\end{align}

Now,

\begin{align}{A^3} &= {A^2}.A\\ &= \left( {\begin{array}{*{20}{c}}5&0&8\\2&4&5\\8&0&{13}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0&2\\0&2&1\\2&0&3\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{5 + 0 + 16}&{0 + 0 + 0}&{10 + 0 + 24}\\{2 + 0 + 10}&{0 + 8 + 0}&{4 + 4 + 15}\\{8 + 0 + 26}&{0 + 0 + 0}&{16 + 0 + 39}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{21}&0&{34}\\{12}&8&{23}\\{34}&0&{55}\end{array}} \right)\end{align}

Therefore,

\begin{align}{A^3} - 6{A^2} + 7A + 2I &= \left( {\begin{array}{*{20}{c}}{21}&0&{34}\\{12}&8&{23}\\{34}&0&{55}\end{array}} \right) - 6\left( {\begin{array}{*{20}{c}}5&0&8\\2&4&5\\8&0&{13}\end{array}} \right) + 7\left( {\begin{array}{*{20}{c}}1&0&2\\0&2&1\\2&0&3\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{21}&0&{34}\\{12}&8&{23}\\{34}&0&{55}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{30}&0&{48}\\{12}&{24}&{30}\\{48}&0&{78}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}7&0&{14}\\0&{14}&7\\{14}&0&{21}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}2&0&0\\0&2&0\\0&0&2\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{21 + 7 + 2}&{0 + 0 + 0}&{34 + 14 + 0}\\{12 + 0 + 0}&{8 + 14 + 2}&{23 + 7 + 0}\\{34 + 14 + 0}&{0 + 0 + 0}&{55 + 21 + 2}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{30}&0&{48}\\{12}&{24}&{30}\\{48}&0&{78}\end{array}} \right)\end{align}

\begin{align} &= \left( {\begin{array}{*{20}{c}}{30}&0&{48}\\{12}&{24}&{30}\\{48}&0&{78}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{30}&0&{48}\\{12}&{24}&{30}\\{48}&0&{78}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}0&0&0\\0&0&0\\0&0&0\end{array}} \right) = 0\end{align}

Hence, $${A^3} - 6{A^2} + 7A + 2I = 0$$.

## Chapter 3 Ex.3.2 Question 17

If $$A = \left( {\begin{array}{*{20}{c}}3&{ - 2}\\4&{ - 2}\end{array}} \right)$$ and $$I = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)$$, find $$k$$ so that $${A^2} = kA - 2I$$.

### Solution

\begin{align}{A^2}&= A.A\\& = \left( {\begin{array}{*{20}{c}}3&{ - 2}\\4&{ - 2}\end{array}} \right)\left( {\begin{array}{*{20}{c}}3&{ - 2}\\4&{ - 2}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{3\left( 3 \right) + \left( { - 2} \right)\left( 4 \right)}&{3\left( { - 2} \right) + \left( { - 2} \right)\left( { - 2} \right)}\\{4\left( 3 \right) + \left( { - 2} \right)\left( 4 \right)}&{4\left( { - 2} \right) + \left( { - 2} \right)\left( { - 2} \right)}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}1&{ - 2}\\4&{ - 4}\end{array}} \right)\end{align}

Now,

\begin{align} &\Rightarrow {A^2} = kA - 2I\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}1&{ - 2}\\4&{ - 4}\end{array}} \right) = k\left( {\begin{array}{*{20}{c}}3&{ - 2}\\4&{ - 2}\end{array}} \right) - 2\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\& \Rightarrow \left( {\begin{array}{*{20}{c}}1&{ - 2}\\4&{ - 4}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{3k}&{ - 2k}\\{4k}&{ - 2k}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}2&0\\0&2\end{array}} \right)\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}1&{ - 2}\\4&{ - 4}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{3k - 2}&{ - 2k}\\{4k}&{ - 2k - 2}\end{array}} \right)\end{align}

Comparing the corresponding elements, we have:

\begin{align}&3k - 2 = 1\\ &\Rightarrow 3k = 3\\ &\Rightarrow k = 1\end{align}

Therefore, the value of $$k = 1$$.

## Chapter 3 Ex.3.2 Question 18

If $$A = \left( {\begin{array}{*{20}{c}}0&{ - \tan \frac{\alpha }{2}}\\{\tan \frac{\alpha }{2}}&0\end{array}} \right)$$ and I is the identity matrix of order $$2$$, show that$$I + A = \left( {I - A} \right)\left( {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right)$$

### Solution

\begin{align}LHS &= I + A\\ &= \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right) + \left( {\begin{array}{*{20}{c}}0&{ - \tan \frac{\alpha }{2}}\\{\tan \frac{\alpha }{2}}&0\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}1&{ - \tan \frac{\alpha }{2}}\\{\tan \frac{\alpha }{2}}&1\end{array}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

\begin{align}RHS &= \left( {I - A} \right)\left( {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right)\\ &= \left( {\left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right) - \left( {\begin{array}{*{20}{c}}0&{ - \tan \frac{\alpha }{2}}\\{\tan \frac{\alpha }{2}}&0\end{array}} \right)} \right)\left( {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}1&{\tan \frac{\alpha }{2}}\\{ - \tan \frac{\alpha }{2}}&1\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 + \sin \alpha \tan \frac{\alpha }{2}}&{ - \sin \alpha + \cos \alpha \tan \frac{\alpha }{2}}\\{ - \cos \alpha \tan \frac{\alpha }{2} + \sin \alpha }&{\sin \alpha \tan \frac{\alpha }{2} + \cos \alpha }\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{1 - 2{{\sin }^2}\frac{\alpha }{2} + 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\tan \frac{\alpha }{2}}&{ - 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2} + \left( {2{{\cos }^2}\frac{\alpha }{2} - 1} \right)\tan \frac{\alpha }{2}}\\{ - \left( {2{{\cos }^2}\frac{\alpha }{2} - 1} \right)\tan \frac{\alpha }{2} + 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}&{2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\tan \frac{\alpha }{2} + 1 - 2{{\sin }^2}\frac{\alpha }{2}}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{1 - 2{{\sin }^2}\frac{\alpha }{2} + 2{{\sin }^2}\frac{\alpha }{2}}&{ - 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2} + 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2} - \tan \frac{\alpha }{2}}\\{ - 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2} + \tan \frac{\alpha }{2} + 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}&{2{{\sin }^2}\frac{\alpha }{2} + 1 - 2{{\sin }^2}\frac{\alpha }{2}}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}1&{ - \tan \frac{\alpha }{2}}\\{\tan \frac{\alpha }{2}}&1\end{array}} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

Thus, from $$\left( 1 \right)$$ and $$\left( 2 \right)$$, we get

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

## Chapter 3 Ex.3.2 Question 19

A trust fund has $$₹30000$$ that must be invested in two different types of bonds. The first bond pays $$5\%$$ interest per year, and the second bond pays $$7\%$$ interest per year. Using matrix multiplication, determine how to divide $$₹30000$$ among the two types of bonds. If the trust fund must obtain an annual total interest of:

$$₹ 1800$$

$$₹ 2000$$

### Solution

Let ₹$$x$$ be invested in the first bond. Then, the sum of money invested in the second bond will be $$₹$$$$\left( {30000 - x} \right)$$.

It is given that the first bond pays $$5\%$$ interest per year and the second bond pays $$7\%$$ interest per year.

Therefore, in order to obtain an annual total interest of ₹1800, we have:

\begin{align}&\left[ {\begin{array}{*{20}{c}}x&{\left( {30000 - x} \right)}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}\begin{array}{l}\frac{5}{{100}}\\\end{array}\\{\frac{7}{{100}}}\end{array}} \right] = 1800{\rm{ }}\left[ {{\rm{S}}{\rm{.I for 1 year}} = \frac{{{\rm{Principal \times Rate}}}}{{{\rm{100}}}}} \right]\\& \Rightarrow \frac{{5x}}{{100}} + \frac{{7\left( {30000 - x} \right)}}{{100}} = 1800\\ &\Rightarrow 5x + 210000 - 7x = 180000\\ &\Rightarrow 210000 - 2x = 180000\\& \Rightarrow 2x = 210000 - 180000\\ &\Rightarrow 2x = 30000\\&\Rightarrow x = 15000\end{align}

Thus, in order to obtain an annual total interest of ₹1800, the trust fund should invest ₹15000 in the first bond and the remaining ₹15000 in the second bond.

Let ₹$$x$$ be invested in the first bond. Then, the sum of money invested in the second bond will be ₹$$\left( {30000 - x} \right)$$.

Therefore, in order to obtain an annual total interest of ₹2000, we have:

\begin{align}&\left[ {\begin{array}{*{20}{c}}x&{\left( {30000 - x} \right)}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}\begin{array}{l}\frac{5}{{100}}\\\end{array}\\{\frac{7}{{100}}}\end{array}} \right] = 2000{\rm{ }}\\ &\Rightarrow \frac{{5x}}{{100}} + \frac{{7\left( {30000 - x} \right)}}{{100}} = 2000\\ &\Rightarrow 5x + 210000 - 7x = 200000\\ &\Rightarrow 210000 - 2x = 200000\\ &\Rightarrow 2x = 210000 - 200000\\ &\Rightarrow 2x = 10000\\ &\Rightarrow x = 5000\end{align}

Thus, in order to obtain an annual total interest of ₹1800, the trust fund should invest ₹5000 in the first bond and the remaining ₹25000 in the second bond.

## Chapter 3 Ex.3.2 Question 20

The bookshop of a particular school has $$10$$ dozen chemistry books, $$8$$ dozen physics books, $$10$$ dozen economics books. Their selling prices are ₹, ₹$$60$$ and ₹$$40$$ each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

### Solution

The total amount of money that will be received from the sale of all these books can be represented in the form of a matrix as:$$80$$

\begin{align}12\left[ {\begin{array}{*{20}{c}}{10}&8&{10}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{80}\\{60}\\{40}\end{array}} \right] &= 12\left[ {10\left( {80} \right) + 8\left( {60} \right) + 10\left( {40} \right)} \right]\\ &= 12\left( {800 + 480 + 400} \right)\\ &= 12\left( {1680} \right)\\ &= 20160\end{align}

Thus, the bookshop will receive ₹$$20160$$ from the sale of all these books.

## Chapter 3 Ex.3.2 Question 21

Assume $$X,Y,Z,W$$ and $$P$$ are matrices of order $$2 \times n,\;3 \times k,\;2 \times p,\;n \times 3$$ and $$p \times k$$ respectively. The restriction on and $$p$$ so that $$PY + WY$$ will be defined are:

(A) $$k = 3,\;p = n$$ (B) $$k$$ is arbitrary,$$p = 2$$

(C) $$p$$ is arbitrary, $$k = 3$$ (D) $$k = 2,\;p = 3$$

### Solution

Matrices $$P$$ and are of the orders $$p \times k$$ and $$3 \times k$$ respectively.

Therefore, matrix $$PY$$will be defined if $$k~=\text{3}$$.

Consequently, $$PY$$ will be of the order $$p \times k$$.

Matrices $$W$$ and $$Y$$ are of the orders $$n \times 3$$ and $$3 \times k$$ respectively.

Since the number of columns in $$W$$ is equal to the number of rows in $$Y$$, matrix $$WY$$ is well-defined and is of the order $$n \times k$$.

Matrices $$PY$$ and $$WY$$ can be added only when their orders are the same.

However, $$PY$$ is of the order $$p \times k$$ and $$WY$$ is of the order $$n \times k$$.

Therefore, we must have $$p = n$$.

Thus, $$k = 3$$ and $$p = n$$ are the restrictions on $$n,k$$ and $$p$$ so that $$PY + WY$$will be defined.

The correct option is A.

## Chapter 3 Ex.3.2 Question 22

Assume $$X,Y,Z,W$$ and $$P$$ are matrices of order $$2 \times n,\;3 \times k,\;2 \times p,\;n \times 3$$ and $$p \times k$$ respectively. If $$n = p$$, then the order of the matrix $$7X - 5Z$$ is:

(A) $$p \times 2$$ (B) $$2 \times n$$

(C) $$n \times 3$$ (D) $$p \times n$$

### Solution

Matrix $$X$$ is of the order $$2 \times n$$.

Therefore, matrix $$7X$$ is also of the same order.

Matrix $$Z$$ is of the order $$2 \times p$$, i.e., $$2 \times n$$ [Since $$n = p$$]

Therefore, matrix $$5Z$$ is also of the same order.

Now, both the matrices $$7X$$ and 5Z are of the order $$2 \times n$$.

Thus, matrix $$7X - 5Z$$ is well-defined and is of the order $$2 \times n$$.

The correct option is B.

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