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Maths 5 Marks Questions

Determinants · Uttarakhand Board (UBSE) STD 12 Science (Uttarakhand - English Medium). 17 questions.

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Question 15 Marks
Solve the system of equations $ \frac{2}{x} + \frac{3}{y} + \frac{{10}}{z} = 4  , \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{{ 20}}{z} = 2$
Answer
Let $ \frac{1}{x} = u,\frac{1}{y} = v$ and $ \frac{1}{z} = w$
$2u + 3v + 10w = 4$
$4u – 6v+ 5w = 1$
$6u + 9v – 20w = 2$
$ A = \left[ {\begin{array}{*{20}{c}} 2&3&{10} \\ 4&{ - 6}&5 \\ 6&9&{ - 20} \end{array}} \right],$
$X= \left[ {\begin{array}{*{20}{c}} u \\ v \\ w \end{array}} \right]$
$ B = \left[ {\begin{array}{*{20}{c}} 4 \\ 1 \\ 2 \end{array}} \right]$
Now, $ \begin{array}{l}\vert A\vert=\begin{vmatrix}2&3&10\\4&-6&5\\6&9&-20\end{vmatrix}\\\end{array}$
$= 2[120 - 45] -3[-80 - 30] +10[36 + 36]$
$= 150 + 330 + 720$
$= 1200 \neq0$
$\Rightarrow A$ is non $-$ singular and hence $A^{-1}$ exists.
Now,$ A_{11} = 75, A_{12} = 110, A_{13} = 72$
$A_{21} = 150, A_{22} = -100, A_{23} = 0$
$A_{31} = 75, A_{32}$
$= 30, A_{33}$
$= -2$
$\therefore \text{ adj} A = \left[ {\begin{array}{*{20}{c}} {75}&{150}&{75} \\ {110}&{ - 100}&{30} \\ {72}&0&{ - 24} \end{array}} \right]$
$ {A^{ - 1}} = \frac{1}{{\left| A \right|}}(\text{adj} A)$
$= \frac{1}{{1200}}\left[ {\begin{array}{*{20}{c}} {75}&{150}&{75} \\ {110}&{ - 100}&{30} \\ {72}&0&{ - 24} \end{array}} \right]$
$ X= {A^{ - 1}}B$
$= \frac{1}{{1200}}\left[ {\begin{array}{*{20}{c}} {600} \\ {400} \\ {240} \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {\frac{1}{2}} \\ {\frac{1}{3}} \\ {\frac{1}{5}} \end{array}} \right]$
$ \left[ {\begin{array}{*{20}{c}} y \\ v \\ w \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{1}{2}} \\ {\frac{1}{3}} \\ {\frac{1}{5}} \end{array}} \right]$
$ u = \frac{1}{2},v = \frac{1}{3},w = \frac{1}{5}$
$ \frac{1}{x} = \frac{1}{2},\frac{1}{y} = \frac{1}{3},\frac{1}{z} = \frac{1}{5}$
$x = 2, y = 3, z = 5$
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Question 25 Marks
Examine the consistency of the system of equation $3x - y - 2z = 2;\,\,2y - z = - 1;3x - 5y = 3$
Answer
Matrix form of given equations is AX = B

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

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

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

$\Rightarrow \left| A \right| = 3\left( {0 - 5} \right) - \left( { - 1} \right)\left( {0 + 3} \right) + \left( { - 2} \right)\left( {0 - 6} \right)$$= 3\left( { - 5} \right) + 3 + 12 = - 15 + 15 = 0$

Now $\left( {adj.A} \right) = \left[ {\begin{array}{*{20}{c}} { - 5}&{10}&5 \\ { - 3}&6&3 \\ { - 6}&{12}&6 \end{array}} \right]$

And $\left( {adj.A} \right)B = \left[ {\begin{array}{*{20}{c}} { - 5}&{10}&5 \\ { - 3}&6&3 \\ { - 6}&{12}&6 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 3 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 10 - 10 + 15} \\ { - 6 - 6 + 9} \\ { - 12 - 12 + 18} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 5} \\ { - 3} \\ { - 6} \end{array}} \right] \ne 0$

Therefore, given equations are inconsistent.

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Question 35 Marks
Examine the consistency of the system of equation x + y + z = 1; 2x + 3y + 2z = 2; ax + ay + 2az = 4
Answer
Matrix form of given equations is AX = B
$\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 2&3&2 \\ a&a&{2a} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ 4 \end{array}} \right]$
Here, $A = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 2&3&2 \\ a&a&{2a} \end{array}} \right]$
$\therefore \left| A \right| = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 2&3&2 \\ a&a&{2a} \end{array}} \right]$
$\Rightarrow$ |A| = 1(6a - 2a) - 1(4a - 2a) + 1(2a - 3a) $ = 4a - 2a - a \ne 0$
Therefore, Unique solution and hence equations are consistent.
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Question 45 Marks
The cost of 4kg onion, 3kg wheat and 2kg rice is Rs. 60. The cost of 2kg onion, 4kg wheat and 6kg rice is Rs. 90. The cost of 6kg onion 2kg wheat and 3kg rice is Rs. 70. Find the cost of each item per kg by matrix method.
Answer
Let cost of 1kg onion = x

cost of 1kg wheat = y

cost of 1kg rise = z

By the question ,we have,

4x + 3y + 2z = 60

2x + 4y + 6z = 90

6x + 2y + 3z = 70

$A = \left[ {\begin{array}{*{20}{c}} 4&3&2 \\ 2&4&6 \\ 6&2&3 \end{array}} \right]B = \left[ {\begin{array}{*{20}{c}} {60} \\ {90} \\ {70} \end{array}} \right]X = \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right]$

$\left| A \right| = \left|{\begin{array}{*{20}{c}} 4&3&2 \\ 2&4&6 \\ 6&2&3 \end{array}} \right|= 50 \ne 0$

$Now,A_{11}=0,A_{12}=30,A_{13}=-20$

$A_{21}=-5,A_{22}=0,A_{23}=10$

$A_{31}=10,A_{32}=-20,A_{33}=10$

$\therefore adjA = \left[ {\begin{array}{*{20}{c}} 0&{ - 5}&{10} \\ {30}&0&{ - 20} \\ { - 20}&{10}&{10} \end{array}} \right]$

${A^{ - 1}} = \frac{1}{{\left| A \right|}}(adjA) = \frac{1}{{50}}\left[ {\begin{array}{*{20}{c}} 0&{ - 5}&{10} \\ {30}&0&{ - 20} \\ { - 20}&{10}&{10} \end{array}} \right]$

$X = {A^{ - 1}}B$

$\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5 \\ 8 \\ 8 \end{array}} \right]$

x = 5, y = 8, z = 8

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Question 55 Marks
If $A = \left[ {\begin{array}{*{20}{c}} 2&{ - 3}&5 \\ 3&2&{ - 4} \\ 1&1&{ - 2} \end{array}} \right]$ find $A^{-1},$ using $A^{-1}$ solve the system of equations
$2x - 3y + 5z = 11$
$3x + 2y - 4z = -5$
$x + y - 2z = -3$
Answer
A = $\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&5 \\ 3&2&{ - 4} \\ 1&1&{ - 2} \end{array}} \right]$$\therefore$ $|A| = 2(-4 + 4) + 3(-6 +4)+ 5(3 - 2) = 0 -6 +5 = -1$
Now, $A_{11} = 0, A_{12}= 2, A_{13}= 1$
$A_{21}= -1, A_{22} = -9, A_{23}= -5$
$A_{31}= 2, A_{32}= 23,A_{33} = 13$
$\therefore {A^{ - 1}} = \frac{1}{{|A|}}(adjA) = - \left[ {\begin{array}{*{20}{c}} 0&{ - 1}&2 \\ 2&{ - 9}&{23} \\ 1&{ - 5}&{13} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&1&{ - 2} \\ { - 2}&9&{ - 23} \\ { - 1}&5&{ - 13} \end{array}} \right]....(1)$
Now, the given system of equations can be written in the form of $AX = B$, where
$A = \left[ {\begin{array}{*{20}{c}} 2&{ - 3}&5 \\ 3&2&{ - 4} \\ 1&1&{ - 2} \end{array}} \right]\;,X = \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right]\;and\;B = \left[ {\begin{array}{*{20}{c}} {11} \\ { - 5} \\ { - 3} \end{array}} \right]$
The solution of the system of equations is given by $X = A^{-1}B$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&1&{ - 2} \\ { - 2}&9&{ - 23} \\ { - 1}&5&{ - 13} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {11} \\ { - 5} \\ { - 3} \end{array}} \right]$ [Using (1) ]
$ = \left[ {\begin{array}{*{20}{c}} {0 - 5 + 6} \\ { - 22 - 45 + 69} \\ { - 11 - 25 + 39} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ 3 \end{array}} \right]$
Hence,$ x = 1, y = 2$ and $z = 3.​​​​​​$
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Question 65 Marks
Solve the system of linear equation, using matrix method $x - y + 2z = 7; 3x + 4y - 5z = -5; 2x - y + 3z = 12$
Answer
Matrix form of given equations is $AX = B \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&4&{ - 5} \\ 2&{ - 1}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 7 \\ { - 5} \\ {12} \end{array}} \right]$
Here $A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&4&{ - 5} \\ 2&{ - 1}&3 \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}} 7 \\ { - 5} \\ {12} \end{array}} \right]$
$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&4&{ - 5} \\ 2&{ - 1}&3 \end{array}} \right|$
$ =1(12-5)-(-1)(9+10)+2(-3-8) $
$ =7+19-22=4 \neq 0 $
$ A_{11}=7, A_{12}=-19, A_{13}=-11 $
$ A_{21}=1, A_{22}=-1, A_{23}=-1 $
$ A_{31}=-3, A_{32}=11, A_{33}=7$
adj A=$\left[ {\begin{array}{*{20}{c}} 7&1&{ - 3} \\ { - 19}&{ - 1}&{11} \\ { - 11}&{ - 1}&7 \end{array}} \right]$
Therefore, solution is unique and $X = {A^{ - 1}}B = \frac{1}{{\left| A \right|}}\left( {adj.A} \right)B$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{c}} 7&1&{ - 3} \\ { - 19}&{ - 1}&{11} \\ { - 11}&{ - 1}&7 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 7 \\ { - 5} \\ {12} \end{array}} \right]$
$= \frac{1}{4}\left[ {\begin{array}{*{20}{c}} {49 - 5 - 36} \\ { - 133 + 5 + 132} \\ { - 77 + 5 + 84} \end{array}} \right]$
$= \frac{1}{4}\left[ {\begin{array}{*{20}{c}} 8 \\ 4 \\ {12} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \\ 1 \\ 3 \end{array}} \right]$
Therefore, $x = 2, y = 1$ and $z = 3$
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Question 75 Marks
Solve the system of linear equation, using matrix method 2x + 3y + 3z = 5; x - 2y + z = - 4; 3x - y - 2z = 3
Answer
Matrix form of given equations is AX = B

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

Here $\left| A \right| = \left[ {\begin{array}{*{20}{c}} 2&3&3 \\ 1&{ - 2}&1 \\ 3&{ - 1}&{ - 2} \end{array}} \right],X\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}} 5 \\ { - 4} \\ 3 \end{array}} \right]$

$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 2&3&3 \\ 1&{ - 2}&1 \\ 3&{ - 1}&{ - 2} \end{array}} \right|$

= 2(4 + 1) - 3( - 2 - 3) + 3( - 1 + 6)

$ = 10 + 15 + 15 = 40 \ne 0$

Therefore, solution is unique and $X = {A^{ - 1}}B = \frac{1}{{\left| A \right|}}\left( {adj.A} \right)B$

$\Rightarrow \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \frac{1}{{40}}\left[ {\begin{array}{*{20}{c}} 5&3&9 \\ 5&{ - 13}&1 \\ 5&{11}&{ - 7} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5 \\ { - 4} \\ 3 \end{array}} \right]$

$= \frac{1}{{40}}\left[ {\begin{array}{*{20}{c}} {25 - 12 + 27} \\ {25 + 52 + 3} \\ {25 - 44 - 21} \end{array}} \right]$

$= \frac{1}{{40}}\left[ {\begin{array}{*{20}{c}} {40} \\ {80} \\ { - 40} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right]$

Therefore, x = 1, y = 2 and z = - 1 .

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Question 85 Marks
Solve the system of linear equation, using matrix method x - y + z = 4; 2x + y - 3z = 0; x + y + z = 2
Answer
Matrix form of given equations is AX = B

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

Here $A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&1 \\ 2&1&{ - 3} \\ 1&1&1 \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}} 4 \\ 0 \\ 2 \end{array}} \right]$

$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 1&{ - 1}&1 \\ 2&1&{ - 3} \\ 1&1&1 \end{array}} \right|$

= 1(1 + 3) - ( - 1)(2 + 3) + 1(2 - 1)

$ = 4 + 5 + 1 = 10 \ne 0$

Therefore, solution is unique and $X = {A^{ - 1}}B = \frac{1}{{\left| A \right|}}\left( {adj.A} \right)B$

$\Rightarrow \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} 4&2&2 \\ { - 5}&0&5 \\ 1&2&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 4 \\ 0 \\ 2 \end{array}} \right]$

$= \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} {16 + 0 + 4} \\ { - 20 + 0 + 10} \\ {4 +0 + 6} \end{array}} \right]$

$= \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} {20} \\ { - 10} \\ {10} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \\ { - 1} \\ 1 \end{array}} \right]$

Therefore, x = 2, y = - 1 and z = 1

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Question 95 Marks
Solve the system of linear equation, using matrix method 2x + y + z = 1; $x - 2y - z = \frac{3}{2};\,\,3y - 5z = 9$
Answer
Matrix form of given equations is AX = B

$\Rightarrow \left[ {\begin{array}{*{20}{c}} 2&1&1 \\ 1&{ - 2}&{ - 1} \\ 0&3&{ - 5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{3}{2}} \\ 9 \end{array}} \right]$

Here $A = \left[ {\begin{array}{*{20}{c}} 2&1&1 \\ 1&{ - 2}&{ - 1} \\ 0&3&{ - 5} \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right]$and $B = \left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{3}{2}} \\ 9 \end{array}} \right]$

Here,$A_{11}=13,A_{12}=5,A_{13}=3\\ A_{21}=8,A_{22}=-10,A_{23}=-6\\ A_{31}=1,A_{32}=3,A_{33}=-5$

adj A=$\left[ {\begin{array}{*{20}{c}} 13&8&1 \\ 5&{ - 10}&{ 3} \\ 3&-6&{ - 5} \end{array}} \right]$

$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 2&1&1 \\ 1&{ - 2}&{ - 1} \\ 0&3&{ - 5} \end{array}} \right| $ = 2(10 + 3) - 1(3 - 0) = 26 + 5 + 3 $ = 34 \ne 0$

Therefore, solution is unique and $X = {A^{ - 1}}B = \frac{1}{{\left| A \right|}}\left( {adj.A} \right)B$

$\Rightarrow \left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}} {13}&8&1 \\ 5&{ - 10}&3 \\ 3&{ - 6}&{ - 5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{3}{2}} \\ 9 \end{array}} \right]$

$= \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}} {13 + 12 + 9} \\ {5 - 15 + 27} \\ {3 - 9 - 45} \end{array}} \right]$

$ = \frac{1}{{34}}\left[ {\begin{array}{*{20}{c}} {34} \\ {17} \\ { - 51} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\ {\frac{1}{2}} \\ {\frac{{ - 3}}{2}} \end{array}} \right]$

Therefore, $x = 1,y = \frac{1}{2}$ and $z = \frac{3}{2}$

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Question 105 Marks
Find the inverse of the matrix (if it exists) given $\left[ {\begin{array}{*{20}{c}} 2&1&3 \\ 4&{ - 1}&0 \\ { - 7}&2&1 \end{array}} \right]$
Answer
Let $A = \left[ {\begin{array}{*{20}{c}} 2&1&3 \\ 4&{ - 1}&0 \\ { - 7}&2&1 \end{array}} \right]$

$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 2&1&3 \\ 4&{ - 1}&0 \\ { - 7}&2&1 \end{array}} \right| = 2\left\{ {\left( { - 1} \right) - \left( 4 \right)} \right\} + 3\left( {8 - 7} \right) = - 3 \ne 0$

$\therefore {A^{ - 1}}$ exists.

${A_{11}} = + \left| {\begin{array}{*{20}{c}} { - 1}&0 \\ 2&1 \end{array}} \right| = + \left( { - 1 - 0} \right) = - 1,{A_{12}} = - \left| {\begin{array}{*{20}{c}} 4&0 \\ { - 7}&1 \end{array}} \right| = - \left( {4 - 0} \right) = - 4$

${A_{13}} = + \left| {\begin{array}{*{20}{c}} 4&{ - 1} \\ { - 7}&2 \end{array}} \right| = + \left( {8 - 7} \right) = 1,{A_{21}} = - \left| {\begin{array}{*{20}{c}} 1&3 \\ 2&1 \end{array}} \right| = - \left( {1 - 6} \right) = 5$

${A_{22}} = + \left| {\begin{array}{*{20}{c}} 2&3 \\ { - 7}&1 \end{array}} \right| = + \left( {2 + 21} \right) = 23,{A_{23}} = - \left| {\begin{array}{*{20}{c}} 2&1 \\ { - 7}&2 \end{array}} \right| = - \left( {4 + 7} \right) = 11$

${A_{31}} = + \left| {\begin{array}{*{20}{c}} 1&3 \\ { - 1}&0 \end{array}} \right| = + \left( {0 + 3} \right) = 3,{A_{32}} = - \left| {\begin{array}{*{20}{c}} 2&3 \\ 4&0 \end{array}} \right| = - \left( {0 - 12} \right) = 12$

${A_{33}} = + \left| {\begin{array}{*{20}{c}} 2&1 \\ 4&{ - 1} \end{array}} \right| = + \left( { - 2 - 4} \right) = - 6$

$\therefore adj.A = \left[ {\begin{array}{*{20}{c}} { - 1}&4&1 \\ 5&{23}&{ - 11} \\ 3&{12}&{ - 6} \end{array}} \right]' = \left[ {\begin{array}{*{20}{c}} { - 1}&4&1 \\ 5&{23}&{ - 11} \\ 3&{12}&{ - 6} \end{array}} \right]$

$\therefore {A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A = \frac{{ - 1}}{3} \left[ {\begin{array}{*{20}{c}} { - 1}&5&2 \\ { - 4}&{23}&{12} \\ 1&{ - 11}&{ - 6} \end{array}} \right]$

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Question 115 Marks
Find the inverse of the matrix (if it exists) given $\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 3&3&0 \\ 5&2&{ - 1} \end{array}} \right]$
Answer
Let $A = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 3&3&0 \\ 5&2&{ - 1} \end{array}} \right]$

$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 1&0&0 \\ 3&3&0 \\ 5&2&{ - 1} \end{array}} \right|$ = 1(-3 - 0) - 0 + 0 $ = - 3 \ne 0$

$\therefore {A^{ - 1}}$ exists.

${A_{11}} = + \left| {\begin{array}{*{20}{c}} 3&0 \\ 2&{ - 1} \end{array}} \right| = + \left( { - 3 - 0} \right) = - 3,$ ${A_{12}} = - \left| {\begin{array}{*{20}{c}} 3&0 \\ 5&{ - 1} \end{array}} \right| = - \left( { - 3 - 0} \right) = 3$

${A_{13}} = + \left| {\begin{array}{*{20}{c}} 3&3 \\ 5&2 \end{array}} \right| = + \left( {6 - 15} \right) = - 9,$ ${A_{21}} = - \left| {\begin{array}{*{20}{c}} 0&0 \\ 2&{ - 1} \end{array}} \right| = - \left( {0 - 0} \right) = 0$

${A_{22}} = + \left| {\begin{array}{*{20}{c}} 1&0 \\ 5&{ - 1} \end{array}} \right| = + \left( { - 1 - 0} \right) = - 1,$ ${A_{23}} = - \left| {\begin{array}{*{20}{c}} 1&0 \\ 5&2 \end{array}} \right| = - \left( {2 - 0} \right) = - 2$

${A_{31}} = + \left| {\begin{array}{*{20}{c}} 0&0 \\ 3&0 \end{array}} \right| = + \left( {0 - 0} \right) = 0,$ ${A_{32}} = - \left| {\begin{array}{*{20}{c}} 1&0 \\ 3&0 \end{array}} \right| = - \left( {0 - 0} \right) = 0$

${A_{33}} = + \left| {\begin{array}{*{20}{c}} 1&0 \\ 3&3 \end{array}} \right| = + \left( {3 - 0} \right) = 3$

$\therefore adj.A = \left[ {\begin{array}{*{20}{c}} { - 3}&3&{ - 9} \\ 0&{ - 1}&{ - 2} \\ 0&0&3 \end{array}} \right]' $ $= \left[ {\begin{array}{*{20}{c}} { - 3}&0&0 \\ 3&{ - 1}&0 \\ { - 9}&{ - 2}&3 \end{array}} \right]$

$\therefore {A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A = \frac{{ - 1}}{3} \left[ {\begin{array}{*{20}{c}} { - 3}&0&0 \\ 3&{ - 1}&0 \\ { - 9}&{ - 2}&3 \end{array}} \right]$

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Question 125 Marks
Find the inverse of the matrix (if it exists) given $\left[ {\begin{array}{*{20}{c}} 1&2&3 \\ 0&2&4 \\ 0&0&5 \end{array}} \right]$
Answer
Let $A = \left[ {\begin{array}{*{20}{c}} 1&2&3 \\ 0&2&4 \\ 0&0&5 \end{array}} \right]$

$\therefore |A| = \left| {\begin{array}{*{20}{c}} 1&2&3 \\ 0&2&4 \\ 0&0&5 \end{array}} \right| $ = 1(10 - 0) - 2(0 - 0) + 3(0 - 0) $ = 10 \ne 0$

$\therefore {A^{ - 1}}$ exists.

${A_{11}} = + \left[ {\begin{array}{*{20}{c}} 2&4 \\ 0&5 \end{array}} \right] = + \left( {10 - 0} \right) = 10,$ ${A_{12}} = - \left| {\begin{array}{*{20}{c}} 0&4 \\ 0&5 \end{array}} \right| = - \left( {0 - 0} \right) = 0$

${A_{13}} = + \left[ {\begin{array}{*{20}{c}} 0&2 \\ 0&0 \end{array}} \right] = + \left( {0 - 0} \right) = 0,$ ${A_{21}} = - \left| {\begin{array}{*{20}{c}} 2&3 \\ 0&5 \end{array}} \right| = - \left( {10 - 0} \right) = - 10$

${A_{22}} = + \left| {\begin{array}{*{20}{c}} 1&3 \\ 0&5 \end{array}} \right| = + \left( {5 - 0} \right) = 5,$ ${A_{23}} = - \left| {\begin{array}{*{20}{c}} 1&2 \\ 0&0 \end{array}} \right| = - \left( {0 - 0} \right) = 0$

${A_{31}} = + \left| {\begin{array}{*{20}{c}} 2&3 \\ 2&4 \end{array}} \right| = + \left( {8 - 6} \right) = 2,$ ${A_{32}} = - \left| {\begin{array}{*{20}{c}} 1&3 \\ 0&4 \end{array}} \right| = - \left( {4 - 0} \right) = - 4$

${A_{33}} = + \left| {\begin{array}{*{20}{c}} 1&2 \\ 0&2 \end{array}} \right| = + \left( {2 - 0} \right) = 2$

$\therefore adj.A = \left| {\begin{array}{*{20}{c}} {10}&0&0 \\ { - 10}&5&0 \\ 2&{ - 4}&2 \end{array}} \right|'$

$= \left| {\begin{array}{*{20}{c}} {10}&{ - 10}&2 \\ 0&5&{ - 4} \\ 0&0&2 \end{array}} \right|$

$\therefore {A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A=\frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} {10}&{ - 10}&2 \\ 0&5&{ - 4} \\ 0&0&2 \end{array}} \right]$

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Question 135 Marks
If $A = \left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&{ - 1}&2 \end{array}} \right]$, verify that $A^3 - 6A^2 + 9A - 4I = 0$ and hence find $A^{-1}$
Answer
Given: $A = \left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&-1&2 \end{array}} \right]$$\therefore {A^2} = \left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&-1&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&{ - 1}&2 \end{array}} \right]$
$\Rightarrow {A^2}=\left[ {\begin{array}{*{20}{c}} {4 + 1 + 1}&{ - 2 - 2 - 1}&{2 + 1 + 2} \\ { - 2 - 2 - 1}&{1 + 4 + 1}&{ - 1 - 2 - 2} \\ {2 + 1 + 2}&{ - 1 - 2 - 2}&{1 + 1 + 4} \end{array}} \right]$ $=\left[ {\begin{array}{*{20}{c}} 6&{ - 5}&5 \\ { - 5}&6&{ - 5} \\ 5&{ - 5}&6 \end{array}} \right]$
Now $A^3 = A^2.A = \left[ {\begin{array}{*{20}{c}} 6&{ - 5}&5 \\ { - 5}&6&{ - 5} \\ 5&{ - 5}&6 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&{ - 1}&2 \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {12 + 5 + 5}&{ - 6 - 10 - 5}&{6 + 5 + 10} \\ { - 10 - 6 - 5}&{5 + 12 + 5}&{ - 5 - 6 - 10} \\ {10 + 5 + 6}&{ - 5 - 10 - 6}&{5 + 5 + 12} \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {22}&{ - 21}&{21} \\ { - 21}&{22}&{ - 21} \\ {21}&{ - 21}&{22} \end{array}} \right]$
L.H.S. $ = {A^3} - 6{A^2} + 9A - 4I$
$= \left[ {\begin{array}{*{20}{c}} {22}&{ - 21}&{21} \\ { - 21}&{22}&{ - 21} \\ {21}&{ - 21}&{22} \end{array}} \right] - 6\left[ {\begin{array}{*{20}{c}} 6&{ - 5}&5 \\ { - 5}&6&{ - 5} \\ 5&{ - 5}&6 \end{array}} \right] $ $+ 9\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&{ - 1}&2 \end{array}} \right] - 4\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {22}&{ - 21}&{21} \\ { - 21}&{22}&{ - 21} \\ {21}&{ - 21}&{22} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} {36}&{ - 30}&{30} \\ { - 30}&{36}&{ - 30} \\ {30}&{ - 30}&{36} \end{array}} \right] $ $+ \left[ {\begin{array}{*{20}{c}} {18}&{ - 9}&9 \\ { - 9}&{18}&{ - 9} \\ 9&{ - 9}&{18} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 4&0&0 \\ 0&4&0 \\ 0&0&4 \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {22 - 36}&{ - 21 + 30}&{21 - 30} \\ { - 21 + 30}&{22 - 36}&{ - 21 + 30} \\ {21 - 30}&{ - 21 + 30}&{22 - 36} \end{array}} \right] $ $+ \left[ {\begin{array}{*{20}{c}} {18 - 4}&{ - 9 - 0}&{9 - 0} \\ { - 9 - 0}&{18 - 4}&{ - 9 - 0} \\ {9 - 0}&{ - 9 - 0}&{18 - 4} \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} { - 14}&9&{ - 9} \\ 9&{ - 14}&9 \\ { - 9}&9&{14} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} { - 14}&9&{ - 9} \\ 9&{ - 14}&9 \\ { - 9}&9&{14} \end{array}} \right] $ $= \left[ {\begin{array}{*{20}{c}} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{array}} \right] = 0$= R.H.S.
Now, to find ${A^{ - 1}}$, multiplying ${A^3} - 6{A^2} + 9A - 4{I^{ - 1}} = 0.{A^{ - 1}}$by ${A^{ - 1}}$
$\Rightarrow {A^3}{A^{ - 1}} - 6{A^2}{A^{ - 1}} + 9A{A^{ - 1}} - 4I.{A^{ - 1}} = 0{A^{ - 1}}$
$ \Rightarrow {A^2} - 6A + 9I - 4{A^{ - 1}} = 0$
$\Rightarrow 4{A}^{-1} = {A^2} - 6A + 9I$
$\Rightarrow 4{A}^{-1} = \left[ {\begin{array}{*{20}{c}} 6&{ - 5}&5 \\ { - 5}&6&{ - 5} \\ 5&{ - 5}&6 \end{array}} \right]$ $ - 6\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ { - 1}&2&{ - 1} \\ 1&{ - 1}&2 \end{array}} \right] + 9\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right]$
$\Rightarrow 4{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 6&{ - 5}&5 \\ { - 5}&6&{ - 5} \\ 5&{ - 5}&6 \end{array}} \right] $ $- \left[ {\begin{array}{*{20}{c}} {12}&{ - 6}&6 \\ { - 6}&{12}&{ - 6} \\ 6&{ - 6}&{12} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 9&0&0 \\ 0&9&0 \\ 0&0&9 \end{array}} \right]$
$\Rightarrow 4{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {6 - 12 + 9}&{ - 5 + 6 + 0}&{5 - 6 + 0} \\ { - 5 + 6 + 0}&{6 - 12 + 9}&{ - 5 + 6 + 0} \\ {5 - 6 + 0}&{ - 5 + 6 + 0}&{6 - 12 + 9} \end{array}} \right] $ $= \left[ {\begin{array}{*{20}{c}} 3&1&{ - 1} \\ 1&3&1 \\ { - 1}&1&3 \end{array}} \right]$
$ \Rightarrow {A^{ - 1}} = \frac{1}{4}\left[ {\begin{array}{*{20}{c}} 3&1&{ - 1} \\ 1&3&1 \\ { - 1}&1&3 \end{array}} \right]$
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Question 145 Marks
For the matrix $A = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right]$, show that $A^3 - 6A^2 + 5A + 11I = 0.$ Hence find $A^{-1}.$
Answer
Given: $A = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right]$$\therefore {A^2} = \left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right]$
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}} {1 + 1 + 2}&{1 + 2 - 1}&{1 - 3 + 3} \\ {1 + 2 - 6}&{1 + 4 + 3}&{1 - 6 - 9} \\ {2 - 1 + 6}&{2 - 2 - 3}&{2 + 3 + 9} \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} 4&2&1 \\ { - 3}&8&{ - 14} \\ 7&{ - 3}&{14} \end{array}} \right]$
Now ${A^3} = {A^2}A = \left[ {\begin{array}{*{20}{c}} 4&2&1 \\ { - 3}&8&{ - 14} \\ 7&{ - 3}&{14} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {4 + 2 + 2}&{4 + 4 + 1}&{4 - 6 + 3} \\ { - 3 + 8 - 28}&{ - 3 + 16 + 14}&{ - 3 - 24 - 42} \\ {7 - 3 + 28}&{7 - 6 - 14}&{7 + 9 + 42} \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} 8&7&1 \\ { - 23}&{27}&{ - 69} \\ {32}&{ - 13}&{58} \end{array}} \right]$
L.H.S. $ = {A^3} - 6{A^2} + 5A + 11I$
$= \left[ {\begin{array}{*{20}{c}} 8&7&1 \\ { - 23}&{27}&{ - 69} \\ {32}&{ - 13}&{58} \end{array}} \right] - 6\left[ {\begin{array}{*{20}{c}} 4&2&1 \\ { - 3}&8&{ - 14} \\ 7&{ - 3}&{14} \end{array}} \right] + $$5\left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right] + 11\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} 8&7&1 \\ { - 23}&{27}&{ - 69} \\ {32}&{ - 13}&{58} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} {24}&{12}&6 \\ { - 18}&{48}&{ - 84} \\ {42}&{ - 18}&{84} \end{array}} \right] $ $+ \left[ {\begin{array}{*{20}{c}} 5&5&5 \\ 5&{10}&{ - 15} \\ {10}&{ - 5}&{15} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {11}&0&0 \\ 0&{11}&0 \\ 0&0&{11} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {8 - 24 + 5}&{7 - 12 + 5}&{1 - 6 + 5} \\ { - 23 + 18 + 5}&{27 - 48 + 10}&{ - 69 + 84 - 15} \\ {32 - 42 + 10}&{ - 13 + 18 - 5}&{58 + 84 + 15} \end{array}} \right] $ $+ \left[ {\begin{array}{*{20}{c}} {11}&0&0 \\ 0&{11}&0 \\ 0&0&{11} \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} { - 11}&0&0 \\ 0&{ - 11}&0 \\ 0&0&{ - 11} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {11}&0&0 \\ 0&{11}&0 \\ 0&0&{11} \end{array}} \right] $ $= \left[ {\begin{array}{*{20}{c}} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{array}} \right] = 0$= R.H.S.
Now, to find ${A^{ - 1}}$, multiplying ${A^3} - 6{A^2} + 5A + 11I = 0$ by ${A^{ - 1}}$
$\Rightarrow {A^3}{A^{ - 1}} - 6{A^2}{A^{ - 1}} + 5A{A^{ - 1}} + 11I.{A^{ - 1}} = 0.{A^{ - 1}}$
$ \Rightarrow {A^2} - 6A + 5I + 11{A^{ - 1}} = 0$
$ \Rightarrow 11{A^{ - 1}} = 6A - 5I - {A^2}$
$ \Rightarrow 11{A^{ - 1}} = 6\left[ {\begin{array}{*{20}{c}} 1&1&1 \\ 1&2&{ - 3} \\ 2&{ - 1}&3 \end{array}} \right] $ $- 5\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 4&2&1 \\ { - 3}&8&{ - 14} \\ 7&{ - 3}&{14} \end{array}} \right]$
$ \Rightarrow 11{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 6&6&6 \\ 6&{12}&{ - 18} \\ {12}&{ - 6}&{18} \end{array}} \right] $ $- \left[ {\begin{array}{*{20}{c}} 5&0&0 \\ 0&5&0 \\ 0&0&5 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 4&2&1 \\ { - 3}&8&{ - 14} \\ 7&{ - 3}&{14} \end{array}} \right]$
$\Rightarrow 11{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {6 - 5 - 4}&{6 - 2}&{6 - 1} \\ {6 + 3}&{12 - 5 - 8}&{ - 18 + 14} \\ {12 - 7}&{ - 6 + 3}&{18 - 5 - 14} \end{array}} \right]$
$ \Rightarrow 11{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} { - 3}&4&5 \\ 9&{ - 1}&{ - 4} \\ 5&{ - 3}&{ - 1} \end{array}} \right]$
$ \Rightarrow {A^{ - 1}} = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}} { - 3}&4&5 \\ 9&{ - 1}&{ - 4} \\ 5&{ - 3}&{ - 1} \end{array}} \right]$
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Question 155 Marks
If $ A = \left[ {\begin{array}{*{20}{c}} 3&1 \\ { - 1}&2 \end{array}} \right]$ Show that $A^2 – 5A + 7I = O.$ Hence find $A^{-1}$
Answer
We have, $A^2=A$
$=\begin{bmatrix}3&1\\-1&2\end{bmatrix}\begin{bmatrix}3&1\\-1&2\end{bmatrix}=\begin{bmatrix}8&5\\-5&2\end{bmatrix}$
$|A| = (3)(2) - (1)(-1)$
$= 6 + 1$
$= 7   \neq0$
$\Rightarrow A$ is non singular and hence $A^{-1}$ exists.
$ {A^2} - 5A + 7I$
$= \left[ {\begin{array}{*{20}{c}} 8&5 \\ { - 5}&2 \end{array}} \right] - 5\left[ {\begin{array}{*{20}{c}} 3&1 \\ { - 1}&2 \end{array}} \right] + 7\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {8 - 15 + 7}&{5 - 5 + 0} \\ { - 5 + 50}&{3 - 10 + 7} \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$
$= O$
Now,
$A^2 – 5A + 7I = O \ ($given$)$
$A^2 – 5A = -7I$
Post multiplying by $A^{-1},$ we get,
$A^2A^{-1}-5AA^{-1} = -7IA^{-1}$
$AAA^{-1} – 5AA^{-1} = -7IA^{-1}$
$A – 5I = -7A^{-1} [AA^{-1} = I]$
$7A^{-1} = 5I – A$
$ = 5\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 3&1 \\ { - 1}&2 \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} 2&-1\\ 1&3 \end{array}} \right]$
$ {A^{ - 1}} = \frac{1}{7}\left[ {\begin{array}{*{20}{c}} 2&{ - 1} \\ 1&3 \end{array}} \right]$
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Question 165 Marks
Find the inverse of the matrix (if it exists) given $\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&{\cos \alpha }&{\sin \alpha } \\ 0&{\sin \alpha }&{ - \cos \alpha } \end{array}} \right]$
Answer
Let $A = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&{\cos \alpha }&{\sin \alpha } \\ 0&{\sin \alpha }&{ - \cos \alpha } \end{array}} \right]$

$\therefore \left| A \right| = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&{\cos \alpha }&{\sin \alpha } \\ 0&{\sin \alpha }&{ - \cos \alpha } \end{array}} \right]$

$ = \left( { - {{\cos }^2}\alpha - {{\sin }^2}\alpha } \right) - 0 + 0 $ $= - \left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right) = - 1 \ne 0$

$\therefore {A^{ - 1}}$ exists.

${A_{11}} = + \left| {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha } \\ {\sin \alpha }&{ - \cos \alpha } \end{array}} \right| = + \left( { - {{\cos }^2}\alpha - {{\sin }^2}\alpha } \right)$

$= - \left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right) = - 1$

${A_{12}} = - \left| {\begin{array}{*{20}{c}} 0&{\sin \alpha } \\ 0&{ - \cos \alpha } \end{array}} \right| = - \left( {0 - 0} \right) = 0,$ ${A_{13}} = + \left| {\begin{array}{*{20}{c}} 0&{\cos \alpha } \\ 0&{\sin \alpha } \end{array}} \right| = + \left( {0 - 0} \right) = 0$

${A_{21}} = - \left| {\begin{array}{*{20}{c}} 0&0 \\ {\sin \alpha }&{-\cos \alpha } \end{array}} \right| = - \left( {0 - 0} \right) = 0,$ ${A_{22}} = + \left| {\begin{array}{*{20}{c}} 1&0 \\ 0&{-\cos \alpha } \end{array}} \right| = + \left( { - \cos \alpha - 0} \right) = - \cos \alpha $

${A_{23}} = - \left| {\begin{array}{*{20}{c}} 1&0 \\ 0&{\sin \alpha } \end{array}} \right| = - \left( {\sin \alpha - 0} \right) = \sin \alpha ,$ ${A_{31}} = + \left| {\begin{array}{*{20}{c}} 0&0 \\ {\cos \alpha }&{\sin \alpha } \end{array}} \right| = \left( {0 - 0} \right) = 0$

${A_{32}} = - \left| {\begin{array}{*{20}{c}} 1&0 \\ 0&{\sin \alpha } \end{array}} \right| = - \left( {\sin \alpha - 0} \right) = - \sin \alpha $, ${A_{33}} = + \left| {\begin{array}{*{20}{c}} 1&0 \\ 0&{\cos \alpha } \end{array}} \right| = + \left( {\cos \alpha - 0} \right) = \cos \alpha $

$\therefore adj.A = \left[ {\begin{array}{*{20}{c}} { - 1}&0&0 \\ 0&{ - \cos \alpha }&{ \sin \alpha } \\ 0&{ - \sin \alpha }&{\cos \alpha } \end{array}} \right]' $ $= \left[ {\begin{array}{*{20}{c}} { - 1}&0&0 \\ 0&{ - \cos \alpha }&{ -\sin \alpha } \\ 0&{ \sin \alpha }&{\cos \alpha } \end{array}} \right]$

$\therefore {A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A $ $= - \left[ {\begin{array}{*{20}{c}} { - 1}&0&0 \\ 0&{ - \cos \alpha }&{ - \sin \alpha } \\ 0&{ \sin \alpha }&{\cos \alpha } \end{array}} \right] $ $= \left[ {\begin{array}{*{20}{c}} { 1}&0&0 \\ 0&{ \cos \alpha }&{\sin \alpha } \\ 0&{ - \sin \alpha }&{ - \cos \alpha } \end{array}} \right]$

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Question 175 Marks
Find the inverse of the matrix (if it exists) given $\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 0&2&{ - 3} \\ 3&{ - 2}&4 \end{array}} \right]$
Answer
Let $A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 0&2&{ - 3} \\ 3&{ - 2}&4 \end{array}} \right]$

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

$\therefore {A^{ - 1}}$ exists.

${A_{11}} = + \left| {\begin{array}{*{20}{c}} 2&{ - 3} \\ { - 2}&4 \end{array}} \right| = + \left( {8 - 6} \right) = 2,{A_{12}} = - \left| {\begin{array}{*{20}{c}} 0&{ - 3} \\ 3&4 \end{array}} \right| = - \left( {0 + 9} \right) = - 9$

${A_{13}} = + \left| {\begin{array}{*{20}{c}} 0&2 \\ 3&{ - 2} \end{array}} \right| = + \left( {0 - 6} \right) = 6,{A_{21}} = - \left| {\begin{array}{*{20}{c}} { - 1}&2 \\ { - 2}&4 \end{array}} \right| = - \left( { - 4 + 4} \right) = 0$

${A_{22}} = + \left| {\begin{array}{*{20}{c}} 2&2 \\ 3&4 \end{array}} \right| = + \left( {4 - 6} \right) = - 2,{A_{23}} = - \left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ 3&{ - 2} \end{array}} \right| = - \left( { - 2 + 3} \right) = - 1$

${A_{31}} = + \left| {\begin{array}{*{20}{c}} { - 1}&2 \\ 2&3 \end{array}} \right| = + \left( {3 - 4} \right) = - 1,{A_{32}} = - \left| {\begin{array}{*{20}{c}} 1&2 \\ 0&3 \end{array}} \right| = - \left( { - 3 - 0} \right) = 3$

${A_{33}} = + \left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ 0&2 \end{array}} \right| = + \left( {2 - 0} \right) = 2,$

$\therefore adj.A = \left[ {\begin{array}{*{20}{c}} 2&{ - 9}&{ - 6} \\ 0&{ - 2}&{ - 1} \\ { - 1}&3&2 \end{array}} \right]' = \left[ {\begin{array}{*{20}{c}} 2&0&{ - 1} \\ { - 9}&{ - 2}&3 \\ { - 6}&{ - 1}&2 \end{array}} \right]$

$\therefore {A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A = \frac{{ - 1}}{1} \left[ {\begin{array}{*{20}{c}} 2&0&{ - 1} \\ { - 9}&{ - 2}&3 \\ { - 6}&{ - 1}&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 2}&0&1 \\ 9&2&{ - 3} \\ 6&1&{ - 2} \end{array}} \right]$

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