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Maths 3 Marks Question

Determinants · Uttarakhand Board (UBSE) STD 12 Science (Uttarakhand - English Medium). 18 questions.

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Question 13 Marks
Evaluate: $\left| {\begin{array}{*{20}{c}} x&y&{x + y} \\ y&{x + y}&x \\ {x+ y}&x&y \end{array}} \right|$
Answer
Let $\Delta = \left| {\begin{array}{*{20}{c}} x&y&{x + y} \\ y&{x + y}&x \\ {x + y}&x&y \end{array}} \right|$$\left[ {{R_1} \to {R_1} + {R_2} + {R_3}} \right]$
$= \left| {\begin{array}{*{20}{c}} {2\left( {x + y} \right)}&{2\left( {x + y} \right)}&{2\left( {x + y} \right)} \\ y&{x + y}&x \\ {x + y}&x&y \end{array}} \right|$
Taking $2(x+y)$ common from first row
$= 2\left( {x + y} \right)\left| {\begin{array}{*{20}{c}} 1&1&1 \\ y&{x + y}&x \\ {x + y}&x&y \end{array}} \right|$
$\left[ {{C_2} \to {C_2} - {C_1}and\,\,{C_3} \to {C_3} - {C_1}} \right]$
$ = 2\left( {x + y} \right)\left| {\begin{array}{*{20}{c}} 1&0&0 \\ y&{x + y - y}&{x - y} \\ {x + y}&{x - x - y}&{y - x - y} \end{array}} \right|$
$= 2\left( {x + y} \right)\left| {\begin{array}{*{20}{c}} 1&0&0 \\ y&x&{x - y} \\ {x + y}&{ - y}&{ - x} \end{array}} \right|$
Expanding along Ist row
$= 2\left( {x + y} \right).1\left| {\begin{array}{*{20}{c}} x&{x - y} \\ { - y}&{ - x} \end{array}} \right|$
$= 2(x + y){ -x^2 + y(x - y)}$
$= 2(x + y)(-x^2 + xy - y^2)$
$= -2(x + y)(x^2 - xy + y^2)$
$= -2(x^3 + y^3)$
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Question 23 Marks
If ${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 3&{ - 1}&1 \\ { - 15}&6&{ - 5} \\ 5&{ - 2}&2 \end{array}} \right]$and $B = \left[ {\begin{array}{*{20}{c}} 1&2&{ - 2} \\ { - 1}&3&0 \\ 0&{ - 2}&1 \end{array}} \right]$find $(AB)^{-1}$
Answer
Given: ${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 3&{ - 1}&1 \\ { - 15}&6&{ - 5} \\ 5&{ - 2}&2 \end{array}} \right]$ and ${B^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 1&2&{ - 2} \\ { - 1}&3&0 \\ 0&{ - 2}&1 \end{array}} \right]$Since, $(AB)^{-1} = B^{-1}A^{-1} $[Reversal law]$ .....(i)$
Now $\left| B \right| = \left| {\begin{array}{*{20}{c}} 1&2&{ - 2} \\ { - 1}&3&0 \\ 0&{ - 2}&1 \end{array}} \right|$
$= 1(3 - 0) - 2(-1 - 0) + (-2)(2 - 0) = 3 + 2 - 4 = 1 = \ne 0$
Therefore, $B^{-1}$ exists.
$\therefore B_{11} = 3, B_{12} = 1, B_{13} = 2$ and $B_{21} = 2, B_{22} = 1, B_{23} = 2$ and $B_{31} = 6, B_{22} = 2, B_{33} = 5$
$\therefore adj.B = \left[ {\begin{array}{*{20}{c}} 3&1&2 \\ 2&1&2 \\ 6&2&5 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3&2&6 \\ 1&1&2 \\ 2&2&5 \end{array}} \right]$
$\therefore {B^{ - 1}} = \frac{1}{{\left| B \right|}}\left( {adj.B} \right) = \frac{1}{1} $ $= \left[ {\begin{array}{*{20}{c}} 3&2&6 \\ 1&1&2 \\ 2&2&5 \end{array}} \right]$
From eq. (i), ${\left( {AB} \right)^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 3&2&6 \\ 1&1&2 \\ 2&2&5 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3&{ - 1}&1 \\ { - 15}&6&{ - 5} \\ 5&{ - 2}&2 \end{array}} \right]$
$\Rightarrow {\left( {AB} \right)^{ - 1}} $ $= \left[ {\begin{array}{*{20}{c}} {9 - 30 + 30}&{ - 3 + 12 - 12}&{3 - 10 + 12} \\ {3 - 15 + 10}&{ - 1 + 6 - 4}&{1 - 5 + 4} \\ {6 - 30 + 25}&{ - 2 + 12 - 10}&{2 - 10 + 10} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} 9&{ - 3}&5 \\ { - 2}&1&0 \\ 1&0&2 \end{array}} \right]$
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Question 33 Marks
Evaluate: $\left| {\begin{array}{*{20}{c}} {\cos \alpha \cos \beta }&{\cos \alpha \sin \beta }&{ - \sin \alpha } \\ { - \sin \beta }&{\cos \beta }&0 \\ {\sin \alpha \cos \beta }&{\sin \alpha \sin \beta }&{\cos \alpha } \end{array}} \right|$
Answer
Let $\Delta = \left| {\begin{array}{*{20}{c}} {\cos \alpha \cos \beta }&{\cos \alpha \sin \beta }&{ - \sin \alpha } \\ { - \sin \beta }&{\cos \beta }&0 \\ {\sin \alpha \cos \beta }&{\sin \alpha \sin \beta }&{\cos \alpha } \end{array}} \right|$

Expanding along first row,

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

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

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

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

$=$ 1

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Question 43 Marks
Solve the system of linear equation, using matrix method 4x - 3y = 3; 3x - 5y = 7
Answer
Matrix form of given equations is AX = B

$\Rightarrow \left[ {\begin{array}{*{20}{c}} 4&{ - 3} \\ 3&{ - 5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3 \\ 7 \end{array}} \right]$

Here $A = \left[ {\begin{array}{*{20}{c}} 4&{ - 3} \\ 3&{ - 5} \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}} 3 \\ 7 \end{array}} \right]$

$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 4&{ - 3} \\ 3&{ - 5} \end{array}} \right| $ = - 20 - (-9) = - 20 + 9 = - 11 $ \ne $ 0

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

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

$= \frac{1}{{ - 11}}\left[ {\begin{array}{*{20}{c}} { - 15 + 21} \\ { - 9 + 28} \end{array}} \right] = \frac{1}{{ - 11}}\left[ {\begin{array}{*{20}{c}} 6 \\ 9 \end{array}} \right]$

$= \left[ {\begin{array}{*{20}{c}} {\frac{{ - 6}}{{11}}} \\ {\frac{{ - 19}}{{11}}} \end{array}} \right]$

Therefore, $x = \frac{{ - 6}}{{11}}$ and $y = \frac{{ - 19}}{{11}}$

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Question 53 Marks
Solve the system of linear equation, using matrix method 2x - y = - 2; 3x + 4y = 3
Answer
Matrix form of given equations is AX = B

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

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

$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 2&{ - 1} \\ 3&4 \end{array}} \right| $ = 8 - (-3) = 8 + 3 $= 11 \ne 0$

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

$\Rightarrow \left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}} 4&1 \\ { - 3}&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 2} \\ 3 \end{array}} \right]$

$= \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}} { - 8 + 3} \\ {6 + 6} \end{array}} \right] = \frac{1}{{11}}\left[ {\begin{array}{*{20}{c}} { - 5} \\ {12} \end{array}} \right]$

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

Therefore, $x = \frac{{ - 5}}{{11}}$ and $y = \frac{{12}}{{11}}$

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Question 63 Marks
Solve the system of linear equation, using matrix method 5x + 2y = 4; 7x + 3y = 5
Answer
Matrix form of given equations is AX = B

$\Rightarrow \left[ {\begin{array}{*{20}{c}} 5&2 \\ 7&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 4 \\ 5 \end{array}} \right]$

Here $A = \left[ {\begin{array}{*{20}{c}} 5&2 \\ 7&3 \end{array}} \right],X = \left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right]$and $B = \left[ {\begin{array}{*{20}{c}} 4 \\ 5 \end{array}} \right]$

$\left| A \right| = \left| {\begin{array}{*{20}{c}} 5&2 \\ 7&3 \end{array}} \right| = 15 - 14 = 1 \ne 0$

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

$= \left[ {\begin{array}{*{20}{c}} {12 - 10} \\ { - 28 + 25} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \\ { - 3} \end{array}} \right]$

Therefore, x = 2 and y = - 3

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Question 73 Marks
Examine the consistency of the system of equation 5x - y + 4z = 5; 2x + 3y + 5z = 2; 5x - 2y + 6z = - 1
Answer
Matrix form of given equations is AX = B

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

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

$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 5&{ - 1}&4 \\ 2&3&5 \\ 5&-2&6 \end{array}} \right|$

$\Rightarrow $ |A| = 5(18 + 10) - (-1)(12 - 25) + 4(-4 - 15) = 140 - 13 - 76 = 140 - 89 $ = 51 \ne 0$

Therefore, Unique solution and hence equations are consistent.

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Question 83 Marks
Solve the system of linear equation, using matrix method $5x + 2y = 3; 3x + 2y = 5$
Answer
Matrix form of given equations is $AX = B$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} 5&2 \\ 3&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3 \\ 5 \end{array}} \right]$
Here $A = \left[\begin{array}{ll} 5 & 2 \\ 3 & 2 \end{array}\right]$,
$ X = \left[\begin{array}{l} x \\ y \end{array}\right]$
and $ B = \left[\begin{array}{l} 3 \\ 5 \end{array}\right]$
$\therefore |A| = \left|\begin{array}{ll} 5 & 2 \\ 3 & 2 \end{array}\right|$ 
$= 10 - 6$
$= 4 \ne 0$
Therefore, solution is unique and $X = A^{-1}B = \frac{1}{|A|} (adj\ A) B $
$ = \left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] = \frac{1}{4}\left[ {\begin{array}{*{20}{c}} 2&-2 \\ { - 3}&5 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3 \\ 5 \end{array}} \right]$
$= \frac{1}{4}\left[ {\begin{array}{*{20}{c}} {6 - 10} \\ { - 9 + 25} \end{array}} \right]$
$= \frac{1}{4}\left[ {\begin{array}{*{20}{c}} { - 4} \\ {16} \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} { - 1} \\ 4 \end{array}} \right]$
Therefore, $x = -1$ and $y = 4$
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Question 93 Marks
Find the inverse of the matrix (if it exists) given $\left[ {\begin{array}{*{20}{c}} { - 1}&5 \\ { - 3}&2 \end{array}} \right]$
Answer
Let $A = \left[ {\begin{array}{*{20}{c}} { - 1}&5 \\ { - 3}&2 \end{array}} \right]$$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} { - 1}&5 \\ { - 3}&2 \end{array}} \right| = -2 - (-15) = -2 + 15 = 13 \ne 0$
$\therefore$ Matrix A is non-singular and hence $A^{-1}$ exist.
Now adj. A $= \left[ {\begin{array}{*{20}{c}} 2&{ - 5} \\ 3&{ - 1} \end{array}} \right]$And ${A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A = \frac{1}{{13}}\left[ {\begin{array}{*{20}{c}} 2&{ - 5} \\ 3&{ - 1} \end{array}} \right]$
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Question 103 Marks
Find the inverse of the matrix (if it exists) given $\left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\ 4&3 \end{array}} \right]$
Answer
Let $A = \left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\ 4&3 \end{array}} \right]$

$\therefore \left| A \right| = \left[ {\begin{array}{*{20}{c}} 2&{ - 2} \\ 4&3 \end{array}} \right] = 6 - \left( { - 8} \right) = 6 + 8 = 14 \ne 0$

$\therefore$ Matrix A is non-singular and hence ${A^{ - 1}}$ exist.

Now adj. A $ = \left[ {\begin{array}{*{20}{c}} 3&2 \\ { - 4}&2 \end{array}} \right]$ And ${A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A = \frac{1}{{14}}\left[ {\begin{array}{*{20}{c}} 3&2 \\ { - 4}&2 \end{array}} \right]$

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Question 113 Marks
Verify A (adj. A) = (adj. A) A = |A|I:
$\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right]$
Answer
Let $A = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right]$
$\Rightarrow \left| A \right| = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right]$
$\therefore {A_{11}} =+ \left| {\begin{array}{*{20}{c}} 0&{ - 2} \\ 0&3 \end{array}} \right| = + 0 + 0 = 0,$${A_{12}} = - \left| {\begin{array}{*{20}{c}} 3&{ - 2} \\ 1&3 \end{array}} \right| = - \left( {9 + 2} \right) = - 11$
${A_{13}} = + \left| {\begin{array}{*{20}{c}} 3&0 \\ 1&0 \end{array}} \right| = + \left( {0 - 0} \right) = 0,$${A_{21}} = - \left| {\begin{array}{*{20}{c}} { - 1}&2 \\ 0&3 \end{array}} \right| - \left( { - 3 - 0} \right) = 3$
${A_{22}} = + \left| {\begin{array}{*{20}{c}} 1&2 \\ 1&3 \end{array}} \right| = 3 - 2 = 1,$${A_{23}} = - \left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ 1&0 \end{array}} \right| = - \left( {0 + 1} \right) = - 1$
${A_{31}} = + \left| {\begin{array}{*{20}{c}} { - 1}&2 \\ 0&{ - 2} \end{array}} \right| = 2 - 0 = 2,$ ${A_{32}} = - \left| {\begin{array}{*{20}{c}} 1&2 \\ 3&{ - 2} \end{array}} \right| = - \left( { - 2 - 6} \right) = 8$
${A_{33}} = + \left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ 3&0 \end{array}} \right| = 3 + 0 = 3$
$\therefore adj.A = \left| {\begin{array}{*{20}{c}} 0&{ - 11}&0 \\ 3&1&{ - 1} \\ 2&8&3 \end{array}} \right|$
$= \left| {\begin{array}{*{20}{c}} 0&3&2 \\ { - 11}&1&8 \\ 0&{ - 1}&3 \end{array}} \right|$
$\therefore A.\left( {adj.A} \right) = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0&3&2 \\ { - 11}&1&8 \\ 0&{ - 1}&3 \end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}} {0 + 11 + 0}&{3 - 1 - 2}&{2 - 8 + 6} \\ {0 - 0 - 0}&{9 + 0 + 2}&{6 + 0 - 6} \\ {0 + 0 + 0}&{3 + 0 - 3}&{2 + 0 + 9} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {11}&0&0 \\ 0&{11}&0 \\ 0&0&{11} \end{array}} \right]$...(i)
Again (adj. A). A $= \left[ {\begin{array}{*{20}{c}} 0&3&2 \\ { - 11}&1&8 \\ 0&{ - 1}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {0 + 9 + 2}&{0 + 0 + 0}&{0 - 6 + 6} \\ { - 11 + 3 + 8}&{11 + 0 + 0}&{ - 22 - 2 + 24} \\ {0 - 3 + 3}&{0 - 0 + 0}&{0 + 2 + 9} \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {11}&0&0 \\ 0&{11}&0 \\ 0&0&{11} \end{array}} \right]$….(ii)
And $\left| A \right| = \left| {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 3&0&{ - 2} \\ 1&0&3 \end{array}} \right|$
$= 1\left( {0 - 0} \right) - \left( { - 1} \right)\left( {9 + 2} \right) + 2\left( {0 - 0} \right) = 0 + 11 + 0 = 11$
Also $\left| A \right|I = \left| A \right|{I_3} = 11\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {11}&0&0 \\ 0&{11}&0 \\ 0&0&{11} \end{array}} \right]$...(iii)
$\therefore$ From eq. (i), (ii) and (iii) A. (adj. A) = (adj. A). A = |A|
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Question 123 Marks
Verify A (adj. A) = (adj. A) A= |A|I:
$\left[ {\begin{array}{*{20}{c}} 2&3 \\ { - 4}&{ - 6} \end{array}} \right]$
Answer
Let $A = \left[ {\begin{array}{*{20}{c}} 2&3 \\ { - 4}&{ - 6} \end{array}} \right]$
$\Rightarrow adj.A = \left[ {\begin{array}{*{20}{c}} { - 6}&{ - 3} \\ 4&2 \end{array}} \right]$
$ \Rightarrow A.\left( {adj.A} \right) = \left[ {\begin{array}{*{20}{c}} 2&3 \\ { - 4}&{ - 6} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 6}&{ - 3} \\ 4&2 \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} { - 12 + 12}&{ - 6 + 6} \\ {24 - 24}&{12 - 12} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$ ...(i)
Again (adj. A). A $ = \left[ {\begin{array}{*{20}{c}} { - 6}&{ - 3} \\ 4&2 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2&3 \\ { - 4}&{ - 6} \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} { - 12 + 12}&{ - 18 + 18} \\ {8 + 8}&{12 - 12} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$…..(ii)
And $\left| A \right| = \left| {\begin{array}{*{20}{c}} 2&3 \\ { - 4}&{ - 6} \end{array}} \right| = 2\left( { - 6} \right) - 3\left( { - 4} \right) = - 12 + 12 = 0$
Again $\left| A \right|I = \left| A \right|{I_2} = \left( 0 \right)\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$....(iii)
From eq. (i), (ii) and (iii) A. (adj. A) = (adj. A). A = |A|I
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Question 133 Marks
Find adjoint of the matrix $\left| {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 2&3&5 \\ { - 2}&0&1 \end{array}} \right|$
Answer
Here $A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\ 2&3&5 \\ { - 2}&0&1 \end{array}} \right]$

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

${A_{11}} = + \left| {\begin{array}{*{20}{c}} 3&5 \\ 0&1 \end{array}} \right| = 3$

${A_{12}} = - \left| {\begin{array}{*{20}{c}} 2&5 \\ { - 2}&1 \end{array}} \right| = - \left( {2 + 10} \right) = - 12$

${A_{13}} = + \left| {\begin{array}{*{20}{c}} 2&3 \\ { - 2}&0 \end{array}} \right| = 6$

${A_{21}} = - \left| {\begin{array}{*{20}{c}} { - 1}&2 \\ 0&1 \end{array}} \right| - \left( { - 1} \right) = 1$

${A_{22}} = + \left| {\begin{array}{*{20}{c}} 1&2 \\ { - 2}&1 \end{array}} \right| = 1 + 4 = 5$

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

${A_{31}} = + \left| {\begin{array}{*{20}{c}} { - 1}&2 \\ 3&5 \end{array}} \right| = - 5 - 6 = - 11$

${A_{32}} = + \left| {\begin{array}{*{20}{c}} 1&2 \\ 2&5 \end{array}} \right| = - \left( {5 - 4} \right) = - 1$

${A_{33}} = + \left| {\begin{array}{*{20}{c}} 1&{ - 1} \\ 2&3 \end{array}} \right| = 3 + 2 = 5$

$\therefore Adj.A = \left[ {\begin{array}{*{20}{c}} 3&{ - 12}&6 \\ 1&5&2 \\ { - 11}&{ - 1}&5 \end{array}} \right]'$

$= \left[ {\begin{array}{*{20}{c}} 3&1&{ - 11} \\ { - 12}&5&{ - 1} \\ 6&2&5 \end{array}} \right]$

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Question 143 Marks
For the matrix $A = \left[ {\begin{array}{*{20}{c}} 3&2 \\ 1&1 \end{array}} \right]$, find the numbers a and b such that $A^2 + aA + bI = 0.$
Answer
Given: $A = \left[ {\begin{array}{*{20}{c}} 3&2 \\ 1&1 \end{array}} \right]$$\therefore {A^2} = A.A = \left[ {\begin{array}{*{20}{c}} 3&2 \\ 1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3&2 \\ 1&1 \end{array}} \right]$
$= \left[ {\begin{array}{*{20}{c}} {9 + 2}&{6 + 2} \\ {3 + 1}&{2 + 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {11}&8 \\ 4&3 \end{array}} \right]$
$\therefore {A^2} + aA + b{I_2} = 0$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} {11}&8 \\ 4&3 \end{array}} \right] + a\left[ {\begin{array}{*{20}{c}} 3&2 \\ 1&1 \end{array}} \right] + b\left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] = 0$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} {11}&8 \\ 4&3 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {3a}&{2a} \\ a&a \end{array}} \right] $ $+ \left[ {\begin{array}{*{20}{c}} b&0 \\ 0&b \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} {11 + 3a + b}&{8 + 2a + 0} \\ {4 + a + 0}&{3 + a + b} \end{array}} \right] $ $= \left[ {\begin{array}{*{20}{c}} 0&0 \\ 0&0 \end{array}} \right]$
$\therefore$ We have $11 + 3a + b = 0….(i)$
$8 + 2a + 0 = 0 ...(ii)$
$\Rightarrow 2a = - 8$
$\Rightarrow a = - 4$
Here $a = -4$ satisfies $4 + a + 0 = 0$ also, therefore $a = -4$
Putting $a = -4$ in eq. (i), $11 - 12 + b = 0$ $ \Rightarrow b - 1 = 0 \Rightarrow b = 1$
Here also $b = 1$ satisfies $3 + a + b = 0,$ therefore $b = 1$
Therefore, $a = - 4$ and $b = 1$
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Question 153 Marks
Let $A = \left[ {\begin{array}{*{20}{c}} 3&7 \\ 2&5 \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}} 6&8 \\ 7&9 \end{array}} \right]$verify that $(AB)^{-1} = B^{-1}A^{-1}$
Answer
Given: Matrix $A = \left[ {\begin{array}{*{20}{c}} 3&7 \\ 2&5 \end{array}} \right]$$\therefore \left| A \right| = \left| {\begin{array}{*{20}{c}} 3&7 \\ 2&5 \end{array}} \right| = 15 - 14 = 1 \ne 0$
$\therefore {A^{ - 1}} = \frac{1}{{\left| A \right|}}adj.A $ $= \frac{1}{1}\left[ {\begin{array}{*{20}{c}} 5&{ - 7} \\ { - 2}&3 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5&{ - 7} \\ { - 2}&3 \end{array}} \right]$
Matrix $B = \left[ {\begin{array}{*{20}{c}} 6&8 \\ 7&9 \end{array}} \right]$
$\therefore \left| B \right| = \left| {\begin{array}{*{20}{c}} 6&8 \\ 7&9 \end{array}} \right| = 54 - 56 = - 2 \ne 0$
$\therefore {B^{ - 1}} = \frac{1}{{\left| B \right|}}adj.B = \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} 9&{ - 8} \\ { - 7}&6 \end{array}} \right]$
Now $AB = \left[ {\begin{array}{*{20}{c}} 3&7 \\ 2&5 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 6&8 \\ 7&9 \end{array}} \right] $ $= \left[ {\begin{array}{*{20}{c}} {18 + 49}&{24 + 63} \\ {12 + 35}&{16 + 45} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {67}&{87} \\ {47}&{61} \end{array}} \right]$
$\therefore \left| {AB} \right| = \left| {\begin{array}{*{20}{c}} {67}&{87} \\ {47}&{61} \end{array}} \right| = 67(61) - 87(47) = 4087 - 4089$ $= - 2 \ne 0$
Now L.H.S. = ${\left( {AB} \right)^{ - 1}} = \frac{1}{{\left| {AB} \right|}}adj.\left( {AB} \right) $ $= \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} {61}&{ - 87} \\ { - 47}&{67} \end{array}} \right]$….(i)
R.H.S. $ = B^{-1}A^{-1} = \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} 9&{ - 8} \\ { - 7}&6 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5&{ - 7} \\ { - 2}&3 \end{array}} \right]$
$= \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} {45 + 16}&{ - 63 - 24} \\ { - 35 - 12}&{49 + 18} \end{array}} \right]$
$= \frac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} {61}&{ - 87} \\ { - 47}&{67} \end{array}} \right]$….(ii)
$\therefore$ From eq. (i) and (ii), we get
L.H.S. = R.H.S.
$ \Rightarrow {\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$
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Question 163 Marks
Using cofactors of elements of third column, evaluate $\Delta = \left| {\begin{array}{*{20}{c}} 1&x&{yz} \\ 1&y&{zx} \\ 1&z&{xy} \end{array}} \right|$
Answer
$ \Delta=a_{13} A_{13}+a_{23} A_{23}+a_{33} A_{33} $
$ =y z(z-y)+z x(x-z)+x y(y-x) $
$ =y z^2-y^2 z+z x^2-z^2 x+x y^2-x^2 y $
$ =z x^2-x^2 y+x y^2-z^2 x+y z^2-y^2 z $
$ =x^2(z-y)+x\left(y^2-z^2\right)+y z(z-y) $
$ =(z-y)\left[x^2-x(z+y)+y z\right] $
$ =(z-y)\left[x^2-x z-x y+y z\right] $
$ =(z-y)[x(x-y)-z(x-y)] $
$ =(z-y)[(x-y)(x-z)] $
$ =(z-y)(x-y)(x-z)$
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Question 173 Marks
Write minors and cofactors of the element of $\left| {\begin{array}{*{20}{c}} 1&0&4 \\ 3&5&{ - 1} \\ 0&1&2 \end{array}} \right|$
Answer
Let $\Delta = \left| {\begin{array}{*{20}{c}} 1&0&4 \\ 3&5&{ - 1} \\ 0&1&2 \end{array}} \right|$$M_{11} =$ Minor of ${a_{11}} = \left| {\begin{array}{*{20}{c}} 5&{ - 1} \\ 1&2 \end{array}} \right| = 10 - \left( { - 1} \right) = 11$ and ${A_{11}} = {\left( { - 1} \right)^{1 + 1}}{M_{11}} = {\left( { - 1} \right)^2}\left( {11} \right) = 11$
$M_{12} =$ Minor of ${a_{12}} = \left| {\begin{array}{*{20}{c}} 3&{ - 1} \\ 0&2 \end{array}} \right| = 6 - 0 = 6$ and ${A_{12}} = {\left( { - 1} \right)^{1 + 2}}{M_{12}} = {\left( { - 1} \right)^3}\left( 6 \right) = - 6$
$M_{13} =$ Minor of ${a_{13}} = \left| {\begin{array}{*{20}{c}} 3&5 \\ 0&1 \end{array}} \right| = 3 - 0 = 3$ and ${A_{13}} = {\left( { - 1} \right)^{1 + 2}}{M_{13}} = {\left( { - 1} \right)^4}\left( 3 \right) = 3$
$M_{21} =$ Minor of ${a_{21}} = \left| {\begin{array}{*{20}{c}} 0&4 \\ 1&2 \end{array}} \right| = 0 - 4 = - 4$ and ${A_{21}} = {\left( { - 1} \right)^{2 + 1}}{M_{21}} = {\left( { - 1} \right)^3}\left( { - 4} \right) = 4$
$M_{22} =$ Minor of ${a_{22}} = \left| {\begin{array}{*{20}{c}} 1&4 \\ 0&2 \end{array}} \right| = 2 - 0 = 2$ and ${A_{22}} = {\left( { - 1} \right)^{2 + 2}}{M_{22}} = \left( { - {1^4}} \right)\left( 2 \right) = 2$
$M_{23} =$ Minor of ${a_{23}} = \left| {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right| = 1 - 0 = 1$ and ${A_{23}} = {\left( { - 1} \right)^{2 + 2}}{M_{23}} = {\left( { - 1} \right)^5}\left( 1 \right) = - 1$
$M_{31} =$ Minor of ${a_{31}} = \left| {\begin{array}{*{20}{c}} 0&4 \\ 5&{ - 1} \end{array}} \right| = 0 - 20 = - 20$ and ${A_{31}} = {\left( { - 1} \right)^{2 + 2}}{M_{31}} = {\left( { - 1} \right)^4}\left( { - 20} \right) = - 20$
$M_{32} =$ Minor of ${a_{32}} = \left| {\begin{array}{*{20}{c}} 1&4 \\ 3&{ - 1} \end{array}} \right| = - 1 - 12 = - 13$ and ${A_{32}} = {\left( { - 1} \right)^{3 + 2}}{M_{32}} = {\left( { - 1} \right)^5}\left( { - 13} \right) = 13$
$M_{33} $= Minor of ${a_{33}} = \left| {\begin{array}{*{20}{c}} 1&0 \\ 3&5 \end{array}} \right| = 5 - 0 = 5$ and ${A_{33}} = {\left( { - 1} \right)^{3 + 3}}{M_{33}} = {\left( { - 1} \right)^6}\left( 5 \right) = 5$
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Question 183 Marks
Write minors and cofactors of the element of $\left| {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right|$
Answer
Let $\Delta = \left| {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right|$$M_{11} =$ Minor of ${a_{11}} = \left| {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right| = 1 - 0 = 1$ and ${A_{11}} = {\left( { - 1} \right)^{1 + 1}}{M_{11}} = {\left( { - 1} \right)^2}\left( 1 \right) = 1$
$M_{12} =$ Minor of ${a_{12}} = \left| {\begin{array}{*{20}{c}} 0&0 \\ 0&1 \end{array}} \right| = 0 - 0 = 0$ and ${A_{12}} = {\left( { - 1} \right)^{1 + 2}}{M_{12}} = {\left( { - 1} \right)^3}\left( 0 \right) = 0$
$M_{13} =$ Minor of ${a_{13}} = \left| {\begin{array}{*{20}{c}} 0&1 \\ 0&0 \end{array}} \right| = 0 - 0 = 0$ and ${A_{13}} = {\left( { - 1} \right)^{1 + 3}}{M_{13}} = {\left( { - 1} \right)^4}\left( 0 \right) = 0$
$M_{21} =$ Minor of ${a_{21}} = \left| {\begin{array}{*{20}{c}} 0&0 \\ 0&1 \end{array}} \right| = 0 - 0 = 0$ and ${A_{21}} = {\left( { - 1} \right)^{2 + 1}}{M_{21}} = {\left( { - 1} \right)^3}\left( 0 \right) = 0$
$M_{22} =$ Minor of ${a_{22}} = \left| {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right| = 1 - 0 = 1$ and ${A_{22}} = {\left( { - 1} \right)^{2 + 3}}{M_{22}} = {\left( { - 1} \right)^4}\left( 1 \right) = 1$
$M_{23} =$ Minor of ${a_{23}} = \left| {\begin{array}{*{20}{c}} 1&0 \\ 0&0 \end{array}} \right| = 0 - 0 = 0$ and ${A_{23}} = {\left( { - 1} \right)^{2 + 3}}{M_{23}} = {\left( { - 1} \right)^5}\left( 0 \right) = 0$
$M_{31} =$ Minor of ${a_{31}} = \left| {\begin{array}{*{20}{c}} 0&0 \\ 1&0 \end{array}} \right| = 0 - 0 = 0$ and ${A_{31}} = {\left( { - 1} \right)^{3 + 1}}{M_{31}} = {\left( { - 1} \right)^4}\left( 0 \right) = 0$
$M_{32} =$ Minor of ${a_{32}} = \left| {\begin{array}{*{20}{c}} 1&0 \\ 0&0 \end{array}} \right| = 0 - 0 = 0$ and ${A_{32}} = {\left( { - 1} \right)^{3 + 2}}{M_{32}} = {\left( { - 1} \right)^5}\left( 0 \right) = 0$
$M_{33} =$ Minor of ${a_{33}} = \left| {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right| = 1 - 0 = 1$ and ${A_{33}} = {\left( { - 1} \right)^{3 + 3}}{M_{33}} = {\left( { - 1} \right)^6}\left( 1 \right) = 1$
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