RD sharma solution class 12 chapter determinants

Determinants

Exercise 6.1

Question 1

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case:

(i) A =   \begin{bmatrix}5&~20\\0&-1 \end{bmatrix}

Sol :

In a 2\times 2 matrix, the minor is obtained for a particular element , by deleting that row and column where the element is present .

M_{11}=-1

M_{21}=20

C_{11}=(-1)^{1+1}\times M_{11}      [\because C_{ij}=(-1)^{i+j}\times M_{ij}]

= (+1) (-1)

= -1

C_{21}=(-1)^{2+1}\times M_{21}

=(-1)^3\times 20

= -20

If A = \begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix} then |A|=(a_{11}\times a_{22})-(a_{12}\times a_{21})

Also , |A|=(5\times -1)-(0\times 20)

= -5

 

(ii) A =   \begin{bmatrix}-1&4\\~~~2&3\end{bmatrix}

Sol :

In a 2\times 2 matrix, the minor is obtained for a particular element , by deleting that row and column where the element is present .

M_{11}=3

M_{21}=4

C_{11}=(-1)^{1+1}\times M_{11}     [\because C_{ij}=(-1)^{i+j}\times M_{ij}]

= (+1) (3)

= 3

C_{21}=(-1)^{2+1}\times M_{21}

=(-1)^3\times 4

= -4

If A = \begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix} then |A|=(a_{11}\times a_{22})-(a_{12}\times a_{21})

Also , |A|=(3\times -1)-(4\times 2)

= -3\times -8

= -11

 

(iii) A =   \begin{bmatrix}1&-3&2\\4&-1&2\\3&~~~5&2\end{bmatrix}

Sol :

In 3\times 3 matrix , Mij equals to the determinant of the 2\times 2 sub-matrix obtained by leaving the ith row and jth column of A .

M_{11}=\begin{vmatrix}-1&2\\\phantom{-}5&2\end{vmatrix}

=(-1\times 2)-(5\times 2)

= – 2 – 10

= -12

M_{21}=\begin{vmatrix}-3&2\\\phantom{-}5&2\end{vmatrix}

=(-3\times 2)-(5\times 2)

= – 6 – 10

= -16

M_{31}=\begin{vmatrix}-3&2\\-1&2\end{vmatrix}

=(-3\times 2)-(2\times -1)

= – 6 + 2

= – 4

C_{11}=(-1)^{1+1}\times M_{11}     [\because C_{ij}=(-1)^{i+j}\times M_{ij}]

= (+1) (-12)

= -12

C_{21}=(-1)^{2+1}\times M_{21}

=(-1)^3\times -16

= 16

C_{31}=(-1)^{3+1}\times M_{31}

= – 4

Also , expanding the determinant along the first column .

|A|=a_{11}\{(-1)^{1+1}\times M_{11}\}+a_{21}\{(-1)^{2+1}\times M_{21}\}+a_{31}\{(-1)^{3+1}\times M_{31}\}

=(a_{11}\times C_{11})+(a_{21}\times C_{21})+(a_{31}\times C_{31})

=1\times -12+3(8-6)+2(20+3)

= -12 + 6 + 46

= 40

 

(iv) A =   \begin{bmatrix}1&a&bc\\1&b&ca\\1&c&ab\end{bmatrix}

Sol :

M_{11}=\begin{vmatrix}b&ca\\c&ab\end{vmatrix}

=ab^2-ac^2

M_{21}=\begin{vmatrix}a&bc\\c&ab\end{vmatrix}

=a^2b-c^2b

M_{31}=\begin{vmatrix}a&bc\\b&ca\end{vmatrix}

=a^2c-b^2c

C_{11}=(-1)^{1+1}\times M_{11} \because C_{ij}=(-1)^{i+j}\times M_{ij}

=+(ab^2-ac^2)

C_{21}=(-1)^{2+1}\times M_{21}

=(-1)^3\times (a^2b-c^2b)

C_{31}=(-1)^{3+1}\times M_{31}

=+(a^2c-b^2c)

Also , expanding the determinant along the first column .

|A|=a_{11}\{(-1)^{1+1}\times M_{11}\}+a_{21}\{(-1)^{2+1}\times M_{21}\}+a_{31}\{(-1)^{3+1}\times M_{31}\}

=(a_{11}\times C_{11})+(a_{21}\times C_{21})+(a_{31}\times C_{31})

=1\times (ab^2-ac^2)+1(c^2b-a^2b)+1(a^2c-b^2c)

=ab^2-ac^2+c^2b-a^2b+a^c-b^2c

 

(v) A =   \begin{bmatrix}0&2&6\\1&5&0\\3&7&1\end{bmatrix}

Sol :

M_{11}=\begin{vmatrix}5&0\\7&1\end{vmatrix}

=(5\times 1)-(0\times 7)

= 5

M_{21}=\begin{vmatrix}2&6\\7&1\end{vmatrix}

=(2\times 1)-(6\times 7)

= 2 – 42

= – 40

M_{31}=\begin{vmatrix}2&6\\5&0\end{vmatrix}

=(2\times 0)-(6\times 5)

= – 30

C_{11}=(-1)^{1+1}\times M_{11} \because C_{ij}=(-1)^{i+j}\times M_{ij}

=(-1)^2\times 5

= 5

C_{21}=(-1)^{2+1}\times M_{21}

=(-1)^3\times (-40)

= 40

C_{31}=(-1)^{3+1}\times M_{31}

=(-1)^4\times (-30)

= – 30

Also , expanding the determinant along the first column .

|A|=a_{11}\{(-1)^{1+1}\times M_{11}\}+a_{21}\{(-1)^{2+1}\times M_{21}\}+a_{31}\{(-1)^{3+1}\times M_{31}\}

=(a_{11}\times C_{11})+(a_{21}\times C_{21})+(a_{31}\times C_{31})

=0\times 5+1\times 40+3\times (-30)

=40-90

= – 50

 

(vi) A =   \begin{bmatrix}a&h&g\\h&b&f\\g&f&c\end{bmatrix}

Sol :

M_{11}=\begin{vmatrix}b&f\\f&c\end{vmatrix}

=(b\times c)-(f\times f)

=bc-f^2

M_{21}=\begin{vmatrix}h&g\\f&c\end{vmatrix}

=(h\times c)-(f\times g)

=hf-bc

M_{31}=\begin{vmatrix}h&g\\b&f\end{vmatrix}

=(h\times f)-(b\times g)

=hf-bg

C_{11}=(-1)^{1+1}\times M_{11} \because C_{ij}=(-1)^{i+j}\times M_{ij}

=(-1)^2\times (bc-f^2)

=bc-f^2

C_{21}=(-1)^{2+1}\times M_{21}

=(-1)^3\times (hc-gf)

=-hc+gf

C_{31}=(-1)^{3+1}\times M_{31}

=(-1)^4\times (hf-bg)

=hf-bg

Also , expanding the determinant along the first column .

|A|=a_{11}\{(-1)^{1+1}\times M_{11}\}+a_{21}\{(-1)^{2+1}\times M_{21}\}+a_{31}\{(-1)^{3+1}\times M_{31}\}

=(a_{11}\times C_{11})+(a_{21}\times C_{21})+(a_{31}\times C_{31})

=a(bc-f^2)+h(-hc+gf)+g(hf-bg)

=abc-af^2+hgf-h^2c+ghf-bg^2

 

(vii) A =   \begin{bmatrix}~~~2&-1&~~~0&~~~1\\-3&~~~0&~~~1&-2\\~~~1&~~~1&-1&~~~1\\~~~2&-1&~~~5&~~~0\end{bmatrix}

Sol :

M_{11}=\begin{vmatrix}\phantom{-}0&\phantom{-}1&-2\\\phantom{-}1&-1&\phantom{-}1\\-1&\phantom{-}5&\phantom{-}0\end{vmatrix}

=0\begin{vmatrix}-1&1\\\phantom{-}5&0\end{vmatrix}-1\begin{vmatrix}\phantom{-}1&1\\-1&0\end{vmatrix}-2\begin{vmatrix}\phantom{-}1&-1\\-1&5\end{vmatrix}

=0-1(0+1)-2(5-1)

= – 1 – 8

= – 9

M_{21}=\begin{vmatrix}-1&\phantom{-}0&\phantom{-}1\\\phantom{-}1&-1&\phantom{-}1\\-1&\phantom{-}5&\phantom{-}0\end{vmatrix}

=-1\begin{vmatrix}-1&1\\\phantom{-}5&0\end{vmatrix}-0\begin{vmatrix}\phantom{-}1&1\\-1&0\end{vmatrix}+1\begin{vmatrix}\phantom{-}1&-1\\-1&\phantom{-}5\end{vmatrix}

=-1(0-5)-0+1(5-1)

= 5 + 4

= 9

M_{31}=\begin{vmatrix}-1&\phantom{-}0&\phantom{-}1\\\phantom{-}0&\phantom{-}1&-2\\-1&\phantom{-}5&\phantom{-}0\end{vmatrix}

=-1\begin{vmatrix}1&-2\\5&\phantom{-}0\end{vmatrix}-0\begin{vmatrix}\phantom{-}0&-2\\-1&\phantom{-}0\end{vmatrix}+1\begin{vmatrix}\phantom{-}0&1\\-1&5\end{vmatrix}

=-1(0+10)-0(0-2)+1(0+1)

= – 10 + 1

= – 9

M_{41}=\begin{vmatrix}-1&\phantom{-}0&\phantom{-}1\\\phantom{-}0&\phantom{-}1&-2\\\phantom{-}1&-1&\phantom{-}1\end{vmatrix}

=-1\begin{vmatrix}\phantom{-}1&-2\\-1&\phantom{-}1\end{vmatrix}-0\begin{vmatrix}0&-2\\1&\phantom{-}1\end{vmatrix}+1\begin{vmatrix}0&\phantom{-}1\\1&-1\end{vmatrix}

=-1(1-2)-0+1(0-1)

= +1 – 1

= 0

C_{11}=(-1)^{1+1}\times M_{11} \because C_{ij}=(-1)^{i+j}\times M_{ij}

= – 9

C_{21}=(-1)^{2+1}\times M_{21}

= – 9

C_{31}=(-1)^{3+1}\times M_{31}

= – 9

C_{41}=(-1)^{4+1}\times M_{41}

= 0

Also , expanding the determinant along the first column .

|A|=a_{11}\{(-1)^{1+1}\times M_{11}\}+a_{21}\{(-1)^{2+1}\times M_{21}\}+a_{31}\{(-1)^{3+1}\times M_{31}\}

=(a_{11}\times C_{11})+(a_{21}\times C_{21})+(a_{31}\times C_{31})

=(2\times -9)+(-3\times -9)+(1\times -9)+(2\times 0)

=-18+27-9

= 0

 

Question 2

(i) A =   \begin{vmatrix}x&-7\\x&5x+1\end{vmatrix}

Sol :

|A|=x(5x+1)-(-7x)

=5x^2+x+7x

=5x^2+8x

 

(ii) A =   \begin{vmatrix}cos~\theta&-sin~\theta\\sin~\theta&~cos~\theta\end{vmatrix}

|A|=(cos~\theta\times cos~\theta)-(-sin~\theta\times sin~\theta)

=cos^2~\theta+sin^2~\theta    [\because sin^2~\theta+cos^2~\theta]

= 1

 

(iii) A =   \begin{vmatrix}cos~15° &sin~15° \\sin~75° &~cos~75° \end{vmatrix}

Sol :

|A|=(cos~15°\times cos~75°)-(sin~15°\times sin~75°)

=cos(15+75)    [\because sin^2~\theta+cos^2~\theta]       [\because cos~A~cos~B-sin~A~sin~B=cos(A+B)]

= cos 90°

= 0

 

(iv) A =   \begin{vmatrix}~~~a+ib&c+id\\-c+id&a-ib\end{vmatrix}

Sol :

|A|=(a+ib)\times(a-ib)-(c+id)(-c+id)

=(a^2-(ib)^2-(id+c)(id-c)

=(a^2-i^2b^2)-[(id)^2-c^2]

=(a^2-i^2b^2)-[i^2d^2-c^2]       i^2=-1

=(a^2+b^2)-[-d^2-c^2]

=a^2+b^2+d^2+c^2]

 

Question 3

Evaluate :

(iv) A =   \begin{vmatrix}2&3&7\\13&17&5\\15&20&12\end{vmatrix}^2

Sol :

Let |A|=2\begin{vmatrix}17&5\\20&12\end{vmatrix}-3\begin{vmatrix}13&5\\15&12\end{vmatrix}+7\begin{vmatrix}13&17\\15&20\end{vmatrix}

=2(204-100)-3(156-75)+7(260-255)

=2(104)-3(81)+7(5)

= 208 – 243 + 35

= 243 -243

= 0

\because \begin{vmatrix}2&3&7\\13&17&5\\15&20&12\end{vmatrix}=0

\Rightarrow \begin{vmatrix}2&3&7\\13&17&5\\15&20&12\end{vmatrix}^2=0^2 = 0

\because |A^2|=(~|A|~)^2

 

Question 4

Show that

  \begin{vmatrix}sin~10^\circ &-cos~10^\circ \\sin~80^\circ &~cos~80^\circ \end{vmatrix} =1

Sol :

=(sin~10°)\times(cos~80°)-(-cos~10°)\times(sin~80°)

=(sin~10°)\times(cos~80°)+(cos~10°)\times(sin~80°)

=sin(10°+80°)     [\because sin~A~cos~B+cos~A~sin~B]

= sin 90°

= 1

Alternate method

=(sin~10°)\times(cos~80°)+(cos~10°)\times(sin~80°)

=(sin~10°)\times[cos~(90°-10°)]+(cos~10°)\times[sin~(90°-10°)]     \because cos~\theta=sin(90°-\theta) and also \because sin~\theta=cos(90°-\theta)

=(sin~10°)\times(sin~10°)+(cos~10°)\times(cos~10°)

=sin^2~10°+cos^2~10°

= 1  \because sin^2~\theta+cos^2~theta=1

 

Question 5 (working)

Evaluate :

(iv) A =   \begin{vmatrix}~~~2&3&-5\\~~~7&1&-2\\-3&4&~~~1\end{vmatrix} by two methods.

Sol :

First method

 

 

 

Question 6

Evaluate  \Delta =   \begin{vmatrix}0&sin~\alpha&-cos~\alpha\\-sin~\alpha&0&sin~\beta\\cos~\alpha&-sin~\beta&0\end{vmatrix}

 

Question 7

Evaluate  \Delta =   \begin{vmatrix}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{vmatrix}

 

Question 8

If A =   \begin{bmatrix}2&5\\2&1\end{bmatrix} and B =   \begin{bmatrix}4&-3\\2&~~~5\end{bmatrix} , verify that   \begin{vmatrix}AB\end{vmatrix} =   \lvert{A}\rvert~\lvert{B}\rvert

 

Question 9

If A =   \begin{bmatrix}1&0&1\\0&1&2\\0&0&4\end{bmatrix} ,then show that  \lvert{3A}\rvert = 27    \lvert{A}\rvert

 

 

Question 10

Find the value of x, if

(i)   \begin{vmatrix}2&4\\5&1\end{vmatrix}=\begin{vmatrix}2x&4\\6&x\end{vmatrix}

(ii)   \begin{vmatrix}2&3\\4&5\end{vmatrix}=\begin{vmatrix}x&3\\2x&5\end{vmatrix}

(iii)   \begin{vmatrix}3&x\\x&1\end{vmatrix}=\begin{vmatrix}3&2\\4&1\end{vmatrix}

(iv) If   \begin{vmatrix}3x&7\\2&4\end{vmatrix} = 10 , find the value of x . 

 

Question 11

Find the integral value of x, if   \begin{bmatrix}x^2&x&1\\0&2&1\\3&1&4\end{bmatrix}=28

 

Question 12 

For what value of x the matrix A =   \begin{bmatrix}x-1&1&1\\1&x-1&1\\1&1&x-1\end{bmatrix} is singular ?

Sol :

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