12th NCERT/CBSE Matrices Exercise 3.2 Questions 22
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Question (1)

Q1) 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

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

$A + B = \left[ {\begin{array}{*{20}{c}}3&7\\1&7\end{array}} \right]$

(ii)A - B

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

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

$A - B = \left[ {\begin{array}{*{20}{c}}1&1\\5&{ - 3}\end{array}} \right]$
(iii)3A - C

$3A - C = 3\left[ {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{ - 2}&5\\3&4\end{array}} \right]$

$3A - C = \left[ {\begin{array}{*{20}{c}}6&{12}\\9&6\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{ - 2}&5\\3&4\end{array}} \right]$

$3A - C = \left[ {\begin{array}{*{20}{c}}8&7\\6&2\end{array}} \right]$

(iv) AB

$AB = \left[ {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&3\\{ - 2}&5\end{array}} \right]$
Use following formula

$\left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{b_{11}}}&{{b_{12}}}\\{b21}&{{b_{22}}}\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}{{a_{11}} \times {b_{11}} + {a_{12}}{b_{21}}}&{{a_{11}} \times {b_{12}} + {a_{12}}{b_{22}}}\\{{a_{21}} \times {b_{11}} + {a_{22}}{b_{21}}}&{{a_{21}} \times {b_{12}} + {a_{22}}{b_{22}}}\end{array}} \right]$

$AB = \left[ {\begin{array}{*{20}{c}}{2 - 8}&{6 + 20}\\{3 - 4}&{9 + 10}\end{array}} \right]$

$AB = \left[ {\begin{array}{*{20}{c}}{ - 6}&{26}\\{ - 1}&{19}\end{array}} \right]$

(v) BA
$BA = \left[ {\begin{array}{*{20}{c}}1&3\\{ - 2}&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2&4\\3&2\end{array}} \right]$
$BA = \left[ {\begin{array}{*{20}{c}}{2 + 9}&{4 + 6}\\{ - 4 + 15}&{ - 8 + 10}\end{array}} \right]$

$BA = \left[ {\begin{array}{*{20}{c}}{11}&{10}\\{11}&2\end{array}} \right]$

Question (2)

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

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

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

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

Solution

$(i)\left[ {\begin{array}{*{20}{c}}a&b\\{ - b}&a\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}a&b\\b&a\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{2a}&{2b}\\0&{2a}\end{array}} \right]$

$(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}&{ - ab}\end{array}} \right] = $

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


$(iii)\left[ {\begin{array}{*{20}{c}}{ - 1}&4&{ - 6}\\8&5&{16}\\2&8&5\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{12}&7&6\\8&0&5\\3&2&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{11}&{11}&0\\{16}&5&{21}\\5&{10}&9\end{array}} \right]$
$(iv)\left[ {\begin{array}{*{20}{c}}{{{\cos }^2}x}&{{{\sin }^2}x}\\{{{\sin }^2}x}&{{{\cos }^2}x}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{{{\sin }^2}x}&{{{\cos }^2}x}\\{{{\cos }^2}x}&{{{\sin }^2}x}\end{array}} \right]$

$ = \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]$

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

Question (3)

Compute the indicated products.
$(i)\left[ {\begin{array}{*{20}{c}}a&b\\{ - b}&a\end{array}} \right]\left[ {\begin{array}{*{20}{c}}a&{ - b}\\b&a\end{array}} \right]$

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

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

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

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

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

Solution

$(i)\left[ {\begin{array}{*{20}{c}}a&b\\{ - b}&a\end{array}} \right]\left[ {\begin{array}{*{20}{c}}a&{ - b}\\b&a\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{a^2} + {b^2}}&{ - ab + ba}\\{ - ba + ab}&{{b^2} + {a^2}}\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}{{a^2} + {b^2}}&0\\0&{{b^2} + {a^2}}\end{array}} \right]$

$(ii){\left[ {\begin{array}{*{20}{c}}1\\2\\3\end{array}} \right]_{3 \times 1}}{\left[ {\begin{array}{*{20}{c}}2&3&4\end{array}} \right]_{1 \times 3}} = \left[ {\begin{array}{*{20}{c}}2&3&4\\4&6&8\\6&9&{12}\end{array}} \right]$

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

$ = \left[ {\begin{array}{*{20}{c}}{1 - 4}&{2 - 6}&{3 - 2}\\{2 + 6}&{4 + 9}&{6 + 3}\end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}}{ - 3}&{ - 4}&1\\8&{13}&9\end{array}} \right]$

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

$ = \left[ {\begin{array}{*{20}{c}}{2 + 0 + 12}&{ - 6 + 6 + 0}&{10 + 12 + 20}\\{3 + 0 + 15}&{ - 9 + 8 + 0}&{15 + 16 + 25}\\{4 + 0 + 18}&{ - 12 + 10 + 0}&{20 + 20 + 30}\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}{14}&0&{42}\\{18}&{ - 1}&{56}\\{22}&{ - 2}&{70}\end{array}} \right]$

$\left( v \right){\left[ {\begin{array}{*{20}{c}}2&1\\3&2\\{ - 1}&1\end{array}} \right]_{3 \times 2}}{\left[ {\begin{array}{*{20}{c}}1&0&1\\{ - 1}&2&1\end{array}} \right]_{2 \times 3}}$
$ = \left[ {\begin{array}{*{20}{c}}{2 - 1}&{0 + 2}&{2 + 1}\\{3 - 2}&{0 + 4}&{3 + 2}\\{ - 1 - 1}&{0 + 2}&{ - 1 + 1}\end{array}} \right]$

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

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

$ = \left[ {\begin{array}{*{20}{c}}{6 - 1 + 9}&{ - 9 + 0 + 3}\\{ - 2 + 0 + 6}&{3 + 0 + 2}\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}{14}&{ - 6}\\4&5\end{array}} \right]$

Question (4)

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

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

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

then compute (A+B) and (B-C). Also, verify that A + (B - C) = (A+B) -C

Solution

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

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

LHS = A+(B-C) =

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

RHS = (A+B) -C =

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

∴ A + ( B - C) = (A + B) - C

Question (5)

If $A = \left[ {\begin{array}{*{20}{c}}{\frac{2}{3}}&1&{\frac{5}{3}}\\{\frac{1}{3}}&{\frac{2}{3}}&{\frac{4}{3}}\\{\frac{7}{3}}&2&{\frac{2}{3}}\end{array}} \right]$ and

$B = \left[ {\begin{array}{*{20}{c}}{\frac{2}{5}}&{\frac{3}{5}}&1\\{\frac{1}{5}}&{\frac{2}{5}}&{\frac{4}{5}}\\{\frac{7}{5}}&{\frac{6}{5}}&{\frac{2}{5}}\end{array}} \right]$, then co0mpute 3A - B

Solution

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] = 0$

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] $

$ = \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]$

$ = \left[ {\begin{array}{*{20}{c}}{{{\cos }^2}\theta + {{\sin }^2}\theta }&{\cos \theta \sin \theta - sin\theta \cos \theta }\\{ - \sin \theta \cos \theta + \sin \theta \cos \theta }&{{{\cos }^2}\theta + {{\sin }^2}\theta }\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] = I$

Question (7)

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

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

Solution

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

$X + Y + X - Y = \left[ {\begin{array}{*{20}{c}}7&0\\2&5\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}3&0\\0&3\end{array}} \right]$

$2X = \left[ {\begin{array}{*{20}{c}}{10}&0\\2&8\end{array}} \right]$

$X = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}{10}&0\\2&8\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5&0\\1&4\end{array}} \right]$ $\left[ {\begin{array}{*{20}{c}}5&0\\1&4\end{array}} \right] + Y = \left[ {\begin{array}{*{20}{c}}7&0\\2&5\end{array}} \right]$

$Y = \left[ {\begin{array}{*{20}{c}}7&0\\2&5\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}5&0\\1&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&0\\1&1\end{array}} \right]$


$(ii)2X + 3Y = \left[ {\begin{array}{*{20}{c}}2&3\\4&0\end{array}} \right]$ multiply by 3

$3X + 2Y = \left[ {\begin{array}{*{20}{c}}2&{ - 2}\\{ - 1}&5\end{array}} \right]$ multiply by 2

$6X + 9Y = \left[ {\begin{array}{*{20}{c}}6&9\\{12}&0\end{array}} \right]$

$6X + 4Y = \left[ {\begin{array}{*{20}{c}}4&{ - 4}\\{ - 2}&{10}\end{array}} \right]$

$ \require{cancel} \cancel{6X} + 9Y - \cancel{6X} - 4Y = \left[ {\begin{array}{*{20}{c}}6&9\\{12}&0\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}4&{ - 4}\\{ - 2}&{10}\end{array}} \right]$
$5Y = \left[ {\begin{array}{*{20}{c}}2&{13}\\{14}&{10}\end{array}} \right]$
$Y = \frac{1}{5}\left[ {\begin{array}{*{20}{c}}2&{13}\\{14}&{10}\end{array}} \right]$

$Y = \left[ {\begin{array}{*{20}{c}}{\frac{2}{5}}&{\frac{{13}}{5}}\\{\frac{{14}}{5}}&2\end{array}} \right]$

$2X + 3\left[ {\begin{array}{*{20}{c}}{\frac{2}{5}}&{\frac{{13}}{5}}\\{\frac{{14}}{5}}&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&3\\4&0\end{array}} \right]$

$2X = \left[ {\begin{array}{*{20}{c}}2&3\\4&0\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{\frac{6}{5}}&{\frac{{39}}{5}}\\{\frac{{42}}{5}}&6\end{array}} \right]$

$2X = \left[ {\begin{array}{*{20}{c}}{\frac{4}{5}}&{ - \frac{{24}}{5}}\\{\frac{{ - 22}}{5}}&{ - 6}\end{array}} \right]$

$X = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}{\frac{4}{5}}&{ - \frac{{24}}{5}}\\{\frac{{ - 22}}{5}}&{ - 6}\end{array}} \right]$

$X = \left[ {\begin{array}{*{20}{c}}{\frac{2}{5}}&{ - \frac{{12}}{5}}\\{\frac{{ - 11}}{5}}&{ - 3}\end{array}} \right]$

Question (8)

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

Solution

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

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

$2X = \left[ {\begin{array}{*{20}{c}}{ - 2}&{ - 2}\\{ - 4}&{ - 2}\end{array}} \right]$

$X = \frac{1}{2}\left[ {\begin{array}{*{20}{c}}{ - 2}&{ - 2}\\{ - 4}&{ - 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 1}\\{ - 2}&{ - 1}\end{array}} \right]$

Question (9)

Find a nd 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

$2\left[ {\begin{array}{*{20}{c}}1&3\\0&x\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}y&0\\1&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5&6\\1&8\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}2&6\\0&{2x}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}y&0\\1&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5&6\\1&8\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}{2 + y}&6\\1&{2x + 2}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}5&6\\1&8\end{array}} \right]$

⇒ 2 +y = 5
∴ y = 3
2x +2 = 8
2x = 6
x = 3
∴ x = 3, y = 3

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

$2\left[ {\begin{array}{*{20}{c}}x&z\\y&t\end{array}} \right] + 3\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\0&2\end{array}} \right] = 3\left[ {\begin{array}{*{20}{c}}3&5\\4&6\end{array}} \right]$

$ \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]$

$\left[ {\begin{array}{*{20}{c}}{2x + 3}&{2z - 3}\\{2y}&{2t + 6}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}9&{15}\\{12}&{18}\end{array}} \right]$

⇒ 2x + 3 = 9
x = 3
2z - 3 = 15
z = 9
2y = 12
y = 6
2t + 6 = 18
t = 6

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 the values of x and y

Solution

$x\left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right] + y\left[ {\begin{array}{*{20}{c}}{ - 1}\\1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{10}\\5\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}{2x}\\{3y}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - y}\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{10}\\5\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}{2x - y}\\{3y + 3}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{10}\\5\end{array}} \right]$

⇒ 2x - y = 10 ---(1)
3x + y = 5 ---(2)
add (1) and (2)
2x - y +3x +y = 10 +5
5x = 15
x = 3
Replace x = 3 in (1)
6 - y = 10
y = -4
x = 3 and y = -4

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 the values of x, y, z and w

Solution

$3\left[ {\begin{array}{*{20}{c}}x&y\\z&w\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}x&6\\{ - 1}&{2w}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}4&{x + y}\\{z + w}&3\end{array}} \right]$

$\left[ {\begin{array}{*{20}{c}}{3x}&{3y}\\{3z}&{3w}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{x + 4}&{6 + x + y}\\{ - 1 + z + w}&{2w + 3}\end{array}} \right]$


⇒ 3x = x + 4
2x = 4
x = 2
3y = 6 + x + y
3y = 6 + 2 + y
y = 4
3w = 2w +3
w = 3
3z = -1 +z + w
2z - w = -1
Replacing w = 3 2z - 3 = -1
z = 1
x = 2, y = 4, z = 1, w = 3

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(x)(Fy) = F(x+y)

Solution

LHS = F(x) F(y) $ = \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]$
Use following formula

$\left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{b_{11}}}&{{b_{12}}}\\{b21}&{{b_{22}}}\end{array}} \right]$


$ = \left[ {\begin{array}{*{20}{c}}{\cos x\cos y - \sin x\sin y}&{ - \sin x\sin y - \sin x\cos y}&{0 + 0 + 0}\\{\sin x\cos y + \cos x\sin y}&{ - \sin x\sin y + \cos x\cos y}&{0 + 0 + 0}\\{0 + 0 + 0}&{0 + 0 + 0}&{0 + 0 + 1}\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(x+y)
=R.H.S.

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

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

$ = \left[ {\begin{array}{*{20}{c}}{10 - 3}&{5 - 4}\\{12 - 21}&{6 - 28}\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}7&1\\{ - 9}&{ - 22}\end{array}} \right]$

$RHS = \left[ {\begin{array}{*{20}{c}}2&1\\3&4\end{array}} \right]\left[ {\begin{array}{*{20}{c}}5&{ - 1}\\6&7\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}{10 + 6}&{ - 2 + 7}\\{15 + 12}&{ - 3 + 28}\end{array}} \right]$

≠ LHS

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

$ = \left[ {\begin{array}{*{20}{c}}{ - 1 + 0 + 6}&{1 - 2 + 9}&{0 + 2 + 12}\\{0 + 0 + 0}&{0 - 1 + 0}&{0 + 1 + 0}\\{ - 1 + 0 + 0}&{1 - 1 + 0}&{0 + 1 + 0}\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}5&8&{14}\\0&{ - 1}&1\\{ - 1}&0&1\end{array}} \right]$

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

=$\left[ {\begin{array}{*{20}{c}}{ - 1 + 0 + 0}&{ - 2 + 1 + 0}&{ - 3 + 0 + 0}\\{0 + 0 + 1}&{0 - 1 + 1}&{0 + 0 + 0}\\{2 + 0 + 4}&{4 + 3 + 4}&{6 + 0 + 0}\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 1}&{ - 3}\\1&0&0\\6&{11}&6\end{array}} \right]$

≠ LHS

Question (15)

Find A2 - 5A + 6I
If $A = \left[ {\begin{array}{*{20}{c}}2&0&1\\2&1&3\\1&{ - 1}&0\end{array}} \right]$

Solution

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

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

A2 - 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] + \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&{ - 3}\\{ - 1}&{ - 1}&{ - 10}\\{ - 5}&4&4\end{array}} \right]$

Question (16)

If $A = \left[ {\begin{array}{*{20}{c}}1&0&2\\0&2&1\\2&0&3\end{array}} \right]$,
prove that A3 -6A2 + 7A +2I = 0

Solution

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

${A^2} = \left[ {\begin{array}{*{20}{c}}5&0&8\\2&4&5\\8&0&{13}\end{array}} \right]$

A3 = A2 A

${A^3} = \left[ {\begin{array}{*{20}{c}}5&0&8\\2&4&5\\8&0&{13}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&2\\0&2&1\\2&0&3\end{array}} \right]$

${A^3} = \left[ {\begin{array}{*{20}{c}}{21}&0&{34}\\{12}&8&{23}\\{34}&0&{55}\end{array}} \right]$

LHS = A3 - 6A2 + 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}}0&0&0\\0&0&0\\0&0&0\end{array}} \right] = 0$

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 A2 = kA -2I

Solution

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

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

A2 = kA - 4I
$\left[ {\begin{array}{*{20}{c}}1&{ - 2}\\4&{ - 4}\end{array}} \right] = k\left[ {\begin{array}{*{20}{c}}3&{ - 2}\\4&{ - 2}\end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$

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

⇒ 3k - 2 = 1
k = 1
-2 = 2k
k = 1

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

LHS = I + A
$I + A = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0&{ - \tan \frac{\alpha }{2}}\\{\tan \frac{\alpha }{2}}&0\end{array}} \right]$

$I + A = \left[ {\begin{array}{*{20}{c}}1&{ - \tan \frac{\alpha }{2}}\\{\tan \frac{\alpha }{2}}&1\end{array}} \right]$

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

$ = \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]$

$\cos \alpha + \sin \alpha \tan \frac{\alpha }{2} = \cos \alpha + 2\sin \frac{\alpha }{2} \cancel{\cos \frac{\alpha }{2}} \cdot \frac{{\sin \frac{\alpha }{2}}}{\cancel{{\cos \frac{\alpha }{2}}}}$

$ = \cos \alpha + 2{\sin ^2}\frac{\alpha }{2}$

$ = 1 - \cancel{2{\sin ^2}\frac{\alpha }{2}} + \cancel{ 2{\sin ^2}\frac{\alpha }{2}} = 1$ $ - \sin \alpha + \cos \alpha \tan \frac{\alpha }{2} = $
$ = - 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2} + \cos \alpha \frac{{\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}$

$ = \sin \frac{\alpha }{2}\left[ { - 2\cos \frac{\alpha }{2} + \frac{{\cos \alpha }}{{\cos \frac{\alpha }{2}}}} \right]$

$ = \sin \frac{\alpha }{2}\left[ {\frac{{ - 2\cos^2 \frac{\alpha }{2} + \cos \alpha }}{{\cos \frac{\alpha }{2}}}} \right]$

$ = \frac{{\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}\left[ { - 2{{\cos }^2}\frac{\alpha }{2} - 1 + 2{{\cos }^2}\frac{\alpha }{2}} \right]$

$ = - \tan \frac{\alpha }{2}$

\[RHS = \left[ {\begin{array}{*{20}{c}}1&{ - \tan \frac{\alpha }{2}}\\{\tan \frac{\alpha }{2}}&1\end{array}} \right]\] =LHS

Question (19)

A trust fund has Rs30,000 that must be invested in two different types of bonds. th first bond pays 50% interest per year, and the cond bond pays 7% intrerest per year. Using matrix multiplication, determine how to divide Rs. 30,000 among the two types pf bonds. if the trust fund must obtain an annual total interest of
a) Rs 1800    (b) Rs2000

Solution

Let Rs. x be invested in 5% and Rs(30,000 - x) is invested in 7%
(i) I = 1800
$\left[ {\begin{array}{*{20}{c}}x&{30,000 - x}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\frac{5}{{100}}}\\{\frac{7}{{100}}}\end{array}} \right] = \left[ {1800} \right]$

$\left[ {\frac{{5x}}{{100}} + \frac{{7\left( {30,000 - x} \right)}}{{100}}} \right] = \left[ {1800} \right]$

$\left[ {\frac{{5x}}{{100}} + \frac{{2,10,000}}{{100}} - \frac{{7x}}{{100}}} \right] = \left[ {1800} \right]$

$\left[ {2100 - \frac{{2x}}{{100}}} \right] = \left[ {1800} \right]$

$ \Rightarrow 2100 - \frac{{2x}}{{100}} = 1800$

$\frac{{2x}}{{100}} = 300$

x = 15,000
so, 15,000 is invested in each type of bond

(ii) I =2000
$\left[ {\begin{array}{*{20}{c}}x&{30,000 - x}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\frac{5}{{100}}}\\{\frac{7}{{100}}}\end{array}} \right] = \left[ {2000} \right]$

$\left[ {\begin{array}{*{20}{c}}{2100}&{\frac{{ - 2x}}{{100}}}\end{array}} \right] = \left[ {2000} \right]$

$ \Rightarrow 2100 - \frac{{2x}}{{100}} = 2000$

$100 = \frac{{2x}}{{100}}$

x = 5,000
so Rs 5000 is invested in 5% and 25000 is invested in 7%

Question (20)

The bookshop of a particular has 10 dozen chemistry book, 8 dozen physics books, 10 dozen economics books. their selling prices are Rs 80, Rs 60 and Rs40 each respectively. Find the total amount the bookshop will receive from selling all the books using algebra.

Solution

1 dozen = 12 books
${\left[ {\begin{array}{*{20}{c}}{120}&{96}&{120}\end{array}} \right]_{1 \times 1}}{\left[ {\begin{array}{*{20}{c}}{80}\\{60}\\{40}\end{array}} \right]_{3 \times 1}}$

$ = \left[ {9600 + 5760 + 4800} \right]$

$ = \left[ {20160} \right]$

He will get Rs20160 by selling nooks

Assume X, Y, Z and P are matrices of order 2×n, 3×k, 2×p, n ×3 and p ×k respectively. Choose the correct answer in Exercise 21 and 22

Question (21)

The restriction on n, k 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

py + wy = [p]p×k [y]3×k + [w]n×3 [y]3×k
⇒ k = 3
⇒ [py]p×k + [wy[n×k
p = n
so k = 3 nad p = n
so A is correct answer

Question (22)

In n = p, then the order of the matrix 7X-5Z is :
(A) p ×2     (B) 2 × n
(C) n × 3     (D) p   n

Solution

n = p
7× (-5z) = 7[x]2×n - 5[z]2×p = 2× n
order of 7x - 5z = 2×n
B is correct answer
Exercise 3.1 ⇐
⇒ Exercise 3.3