KC Sinha Mathematics Solution Class 12 Chapter 6 सारणिक (Determinants) Exercise 6.1 (Q1-Q5)

Ex-6.1

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Exercise 6.1

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Question 1

निम्नलिखित का मान निकालें (Evaluate the following):
(i) $\begin{vmatrix}
7 &5 \\
-2 & 3
\end{vmatrix}$
Sol :
= 21-(-10)
= 21+10
= 31

(ii) $\begin{vmatrix}
cos \alpha &sin \alpha \\
sin \alpha & cos \alpha
\end{vmatrix}$

Sol :
= cos²α -sin²α
= cos 2α

(iii) $\begin{vmatrix}
tan \alpha & cosec \alpha \\
sin \alpha & cot \alpha
\end{vmatrix}$

Sol :
= tan α.cotα -sinα.cosecα
= 1-1
= 0

(iv) $\begin{vmatrix}
cos \theta & -sin \theta \\
sin \theta & cos \theta
\end{vmatrix}$

Sol :
=cosθ×cosθ-(-sinθ×sinθ)
=cos²θ+sin²θ [∵sin²θ+cos²θ=1]
=1


(v) $\begin{vmatrix}
\dfrac{1}{3} & \dfrac{1}{5} \\
\dfrac{1}{2} & \dfrac{1}{7}
\end{vmatrix}$

Sol :
$=\dfrac{1}{3} \times \dfrac{1}{7} - \dfrac{1}{5} \times \dfrac{1}{2}$

$=\dfrac{1}{21} - \dfrac{1}{10}$

$=\dfrac{10-21}{210}=\dfrac{-11}{210}$



(vi) $\begin{vmatrix}
x^2-x+1 & x-1  \\
x+1 & x+1
\end{vmatrix}$

Sol :
= (x+1)(x²-x+1)-(x+1)(x-1)
= x³+1³-(x²-1²)
= x³+1-x²+1
=x³-x²+2


(vii) $\begin{vmatrix}
a+ib & c+id \\
-c+id & a-ib
\end{vmatrix}$

Sol :
 =$\begin{vmatrix}
a+ib & c+id \\
-(c-id) & a-ib
\end{vmatrix}$

=(a+ib)(a-ib)+(c+id)(c-id)
=a²-(ib)²+c²-(id)²
=a²-i²b²+c²-i²d²
=a²-(-1)b²+c²-(-1)d²
=a²+b²+c²+d²

(viii) $\begin{vmatrix}
\dfrac{1}{2} & 8 \\
4 & 2
\end{vmatrix}$

Sol :
=$\dfrac{1}{2} \times 2 - 8\times 4$
=1-32
=-31

(ix) यदि (If) $A=\begin{bmatrix} 1&2\\4&2 \end{bmatrix}$ , दिखाएँ कि (show that) $|2A|=4|A|$

Sol :
$|2A|=\begin{vmatrix} 2 \begin{bmatrix} 1&2\\4&2 \end{bmatrix}\end{vmatrix}$

=$\begin{vmatrix}2&4 \\ 8&4 \end{vmatrix}$

=$2 \times 2 \begin{vmatrix}1&2 \\ 4&2 \end{vmatrix}$

= 4|A|

(x) यदि (If) $\mathrm{A}=\left[\begin{array}{cc}2 & 4 \\ -5 & -1\end{array}\right]$, दिखाएँ कि (show that) |3A|=9|A|

Sol :

Question 2
निम्नलिखित सारणिकों में दूसरे स्तम्भ के प्रत्येक अवयव का उपसारणिक तथा सह-खण्ड निकालें साथ ही सारणिक का मान भी निकाले ।

[Write the minors and cofactors of each element of second column in the following determinants and evaluate them]

(i) $\left|\begin{array}{lll}4 & 9 & 7 \\ 3 & 5 & 7 \\ 5 & 4 & 5\end{array}\right|$

Sol :

$\left|\begin{array}{ccc}+-+ \\-+- \\ +-+\end{array}\right|$

$M_{12}=\left|\begin{array}{cc}3 & 7 \\ 5 & 5\end{array}\right|$

=15-35

=-20

$M_{22}=\left|\begin{array}{cc}4 & 7 \\ 5 & 5\end{array}\right|$

=20-35

=-15

$M_{32}=\left|\begin{array}{rr}4 & 7 \\ 3 & 7\end{array}\right|$

=28-21

=7

$A_{12}=-\left|\begin{array}{ll}3 & 7 \\ 5 & 5\end{array}\right|$

=-(15-35)

=20

$A_{22}=\left|\begin{array}{cc}4 & 7 \\ 5 & 5\end{array}\right|$

=20-35

=-15

$A_{32}=-\left|\begin{array}{ll}4 & 7 \\ 3 & 7\end{array}\right|$

=-(28-21)

=-7

$|A|=\left|\begin{array}{ccc}4 & 9 & 7 \\ 3 & 5 & 7 \\ 5 & 4 & 5\end{array}\right|$

=9×20+5×(-15)+4×(-7)

=180-75-28

=180-103

=77

(ii) $\left|\begin{array}{lll}1 & 2 & 4 \\ 1 & 3 & 9 \\ 1 & 4 & 16\end{array}\right|$

Sol :

Question 3

निम्नलिखित सारणिकों में प्रत्येक अवयव का उपसारणिक तथा सह-खण्ड निकालें साथ हो सारणिक का मान भी निकालें ।

[Write the minor and cofactor of each element of the following determinants and also evaluate the determinant in each case.]

(i) $\left|\begin{array}{cc}5 & -10 \\ 0 & 3\end{array}\right|$

Sol :

(ii) $\left|\begin{array}{ccc}1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2\end{array}\right|$

Sol  :

$M_{11}=\left|\begin{array}{rr}5 & -1 \\ 1 & 2\end{array}\right|$

=10-(-1)

=11

$M_{12}=\left|\begin{array}{cc}3 & -1 \\ 0 & 2\end{array}\right|$

=6+0

=6

$M_{13}=\left|\begin{array}{ll}3 & 5 \\ 0 & 1\end{array}\right|$

=3-0

=3

$M_{21}=\left|\begin{array}{ll}0 & 4 \\ 1 & 2\end{array}\right|$

=0-4

=-4

$M_{22}=\begin{vmatrix}1&4\\0&2\end{vmatrix}$
=2-0
=2

$M_{23}=\begin{vmatrix}1&0\\0&1\end{vmatrix}$

=1-0

=1

$M_{31}=\begin{vmatrix}0&4\\5&-1\end{vmatrix}$

=0-20

=-20

$M_{32}=\begin{vmatrix}1&4\\3&-1\end{vmatrix}$

=-1-12

=-13

$M_{33}=\begin{vmatrix}1&0\\3&5\end{vmatrix}$

=5-0

=5

A₁₁=11 , A₁₂=-6 , A₁₃=3

A₂₁=(-4)=4 , A₂₂=2 , A₂₃=-1

A₃₁=-20 , A₃₂=-(-13)=13 , A₃₃=5

Δ$=\begin{vmatrix}1&0&4\\3&5&-1\\0&1&2\end{vmatrix}$

=1×11+0×(-6)+4×3

=11+12

=23

(iii) $\left|\begin{array}{ccc}1 & 3 & -2 \\ 4 & -5 & 6 \\ 3 & 5 & 2\end{array}\right|$

Sol :

(iv) $\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$

Sol :

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Question 4
निम्नलिखित का मान निकालें (Evaluate the following):

(i) $\begin{vmatrix}1 & 5 & 7\\ 6 & 7 & 2\\ 1 & 2 & 3 \end{vmatrix}$

Sol :

Expanding along C₁

=$1\begin{vmatrix}7 & 2 \\2 & 3 \end{vmatrix}-6\begin{vmatrix}5 & 7 \\2 & 3 \end{vmatrix}+1\begin{vmatrix}5 & 7\\7 & 2 \end{vmatrix}$

=1(21-4)-6(15-14)+1(10-49)

=17-6-39

=17-45

=-28

(ii) $\begin{vmatrix}1^2 & 2^2 & 3^2\\2^2 & 3^2 & 4^2\\3^2 & 4^2 & 5^2\end{vmatrix}$

Sol :


(iii) $\begin{vmatrix}a & h & g\\h & b & f\\g & f & c \end{vmatrix}$
Sol :

Expanding along R₁

=$a\begin{vmatrix}b & f \\f & c \end{vmatrix}-h\begin{vmatrix}h & f \\g & c \end{vmatrix}+g\begin{vmatrix}h & b\\g & f \end{vmatrix}$

=a(bc-f²)-h(ch-fg)+g(fh-bg)

=abc-af²-ch²+fgh+fgh-bg²

=abc+2fgh-af²-bg²-ch²

(iv) $\begin{vmatrix}
43 & 3 & 6\\
35 & 21 & 4\\
17 & 9 & 2
\end{vmatrix}$
Sol :


(v) $\begin{vmatrix}
9 & 9 & 12\\
1 & 3 & -4\\
1 & 9 & 12
\end{vmatrix}$
Sol :

(vi) $\begin{vmatrix}
42 & 1 & 6\\
28 & 7 & 4\\
14 & 3 & 2
\end{vmatrix}$
Sol :
R₁→R₂-2R₃

$=\left|\begin{array}{ccc}42 & 1 & 6 \\ 0 & 1 & 0 \\ 14 & 3 & 2\end{array}\right|$

R₂ के सापेक्ष विस्तार करने पर

$=1\left|\begin{array}{cc}42 & 6 \\ 14 & 2\end{array}\right|$

=84-84

=0

(vii) $\begin{vmatrix}
3 & -4 & 5\\
1 & 1 & -2\\
2 & 3 & 1
\end{vmatrix}$
Sol :

(viii) $\begin{vmatrix}
2 & -1 & -2\\
0 & 2 & -1\\
3 & -5 & 0
\end{vmatrix}$
Sol :


(ix) $\begin{vmatrix}
3 & -1 & -2\\
0 & 0 & -1\\
3 & -5 & 0
\end{vmatrix}$
Sol :

R₂ के सापेक्ष विस्तार करने पर

$=-(-1)\left|\begin{array}{rr}3 & -1 \\ 3 & -5\end{array}\right|$

={-15-(-3)}
=-15+3
=-12

(x) $\begin{vmatrix}
0 & 1 & 2\\
-1 & 0 & -3\\
-2 & 3 & 0
\end{vmatrix}$
Sol :

(xi) $\begin{vmatrix}
1 & 2 & 4\\
-1 & 3 & 0\\
4 & 1 & 0
\end{vmatrix}$
Sol :

(xii) यदि (If) A=$\begin{bmatrix}1&1&-2\\2&1&-3\\5&4&-9\end{bmatrix}$ , तो |A| का मान निकालें (Find |A|)
Sol :
|A|=$\begin{vmatrix}1&1&-2\\2&1&-3\\5&4&-9\end{vmatrix}$

C₁→C₁+C₃

=$\begin{vmatrix}-1&1&-2\\-1&1&-3\\-4&4&-9\end{vmatrix}$

=$-\begin{vmatrix}1&1&-2\\1&1&-3\\4&4&-9\end{vmatrix}=0$

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(xiii) यदि (If) A=$\begin{bmatrix}1&0&1\\0&1&2\\0&0&4\end{bmatrix}$ , दिखाएँ कि (show that)
|3A|=27|A|
Sol :
L.H.S
|3A|=$\lvert 3\begin{bmatrix}1&0&1\\0&1&2\\0&0&4\end{bmatrix} \rvert$

=$\begin{vmatrix}3&0&3\\0&3&6\\0&0&12\end{vmatrix}$

=$3\times 3\times 3 \begin{vmatrix}1&0&1\\0&1&2\\0&0&4\end{vmatrix}$

=27|A| = R.H.S


Question 5
निम्नलिखित का मान निकालें (Evaluate the following) :

(i) $\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|$
Sol :
$\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|$

C₁→C₁-C₂ , C₂→C₂-C₃

$=\left|\begin{array}{ccc}0 & 0 & 1 \\ -x & x & 1 \\ 0 & -y & 1+y\end{array}\right|$

Expanding along R₁→

$=1\left|\begin{array}{cc}-x & x \\ 0 & -y\end{array}\right|$

=xy-0
=xy

(ii) $\left|\begin{array}{lll}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|$
Sol :
$\left|\begin{array}{ccc}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|$

R₁→R₁-R₂ , R₂→R₂-R₃

$=\left|\begin{array}{ccc}0 & a-b & b c-c a \\ 0 & b-c & c a-a b \\ 1 & c & a b\end{array}\right|$

taking out (a-b) from R₁ and (b-c) from R₂

$=(a-b)(b-c)\left|\begin{array}{ccc}0 & 1 & -c \\ 0 & 1 & -a \\ 1 & c & a b\end{array}\right|$

Expanding along C₁

$=(a-b)(b-c) 1\left|\begin{array}{cc}1 & -c \\ 1 & -a\end{array}\right|$

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

(iii) $\left|\begin{array}{ccc}x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda\end{array}\right|$
Sol :

$\left|\begin{array}{cccc}x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda\end{array}\right|$

C₁→C₁+C₂+C₃

$=\left|\begin{array}{ccc}3 x+\lambda & 1 & x \\ 3 x+\lambda & x+\lambda & x \\ 3 x+\lambda & x & x+\lambda\end{array}\right|$

taking out 3x+λ from C₁

$=\left(\begin{array}{ll}3 x+\lambda\end{array}\right)\left|\begin{array}{ccc}1 & x & x \\ 1 & x+\lambda & x \\ 1 & x & x+\lambda\end{array}\right|$

R₁→R₁-R₂, R₂→R₂-R₁

$=(3 x+\lambda)\left|\begin{array}{ccc}0 & -\lambda & 0 \\ 0 & \lambda & -\lambda \\ 1 & x & x+\lambda\end{array}\right|$

Expanding along C₁

$=(3 x+\lambda) \cdot 1\left|\begin{array}{cc}-\lambda & 0 \\ x & -\lambda\end{array}\right|$

=(3x+λ)(λ²-0)
=λ²(3x+λ)

(iv) $\left|\begin{array}{ccc}1 & a & a^{2}-b c \\ 1 & b & b^{2}-a c \\ 1 & c & c^{2}-a b\end{array}\right|$
Sol :
$\left|\begin{array}{ccc}1 & a & a^{2}-b c \\ 1 & b & b^{2}-a c \\ 1 & c & c^{2}-a b\end{array}\right|$

R₁→R₁-R₂, R₂→R₂-R₃

$=\left|\begin{array}{ccc}0 & a-b & a^{2}-b c-b^{2}+a c \\ 0 & b-c & b^{2}-a c-c^{2}+a b \\ 1 & c & c^{2}-a b\end{array}\right|$

taking out (a-b) from R₁ and (b-c) from R₂

$=(a-b)(b-c)\left|\begin{array}{ccc}0 & 1 & a+b+c \\ 0 & 1 & a+b+c \\ 1 & c & c^{2}-a b\end{array}\right|$
=(a-b)(b-c)×0
=0


(v) $\left|\begin{array}{ccc}1 & 1 & 1 \\ \alpha & \beta & \gamma \\ \beta \gamma & \gamma \alpha & \alpha \beta\end{array}\right|$
Sol :
$\left|\begin{array}{ccc}1 & 1 & 1 \\ \alpha & \beta & \gamma \\ \beta \gamma & \gamma \alpha & \alpha \beta\end{array}\right|$

C₁→C₁-C₂, C₂→C₂-C₃

$=\left|\begin{array}{ccc}0 & 0 & 1 \\ \alpha-\beta & \beta-\gamma & \gamma \\ \beta \gamma-\gamma \alpha & \gamma \alpha-\alpha \beta & \alpha \beta\end{array}\right|$

$=\left| \begin{array}{ccc}0 & 0 & 1 \\ \alpha-\beta & \beta-\gamma & \gamma \\ -\gamma(\alpha-\beta) & -\alpha(\beta-\gamma) & \alpha \beta\end{array}\right|$

taking out (α-ꞵ) from C₁ and (ꞵ-𝛾) from C₂

$=(\alpha-\beta)(\beta-\gamma)\left|\begin{array}{ccc}0 & 0 & 1 \\ 1 & 1 & \gamma \\ -\gamma & -\alpha & \alpha \beta\end{array}\right|$

Expanding along R₁

$=(\alpha-\beta)(\beta-\gamma)1 \begin{vmatrix}1&1\\-\gamma&-\alpha \end{vmatrix}$

=(α-β)(β-𝛾)(-α-𝛾)
=(α-β)(β-𝛾)(𝛾-α)


(vi) $\left|\begin{array}{ccc}a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|$
Sol :
$\left|\begin{array}{ccc}a^{n}+2 a & 2 a-1 &1\\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|$

R₁→R₁-R₂ , R₂→R₂-R₃

$=\left|\begin{array}{ccc}a^{2}-1 & a-1 & 0 \\ 2 a-2 & a-1 & 0 \\ 3 & 3 & 1\end{array}\right|$

$=\left|\begin{array}{ccc}(a-1)(a+1) & a-1 & 0 \\ 2(a-1) & a-1 & 0 \\ 3 & 3 & 1\end{array}\right|$

taking out (a-1) from R₁ and R₂

$=(a-1)^{2}\left|\begin{array}{rrr}a+1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1\end{array}\right|$

Expanding along C₃

$=(a-1)^{2} \cdot 1 \quad\left|\begin{array}{cc}a+1 & 1 \\ 2 & 1\end{array}\right|$

=(a-1)²(a+1-2)
=(a-1)²(a-1)
=(a-1)³

(vii) $\left|\begin{array}{ccc}1 & 1 & 1 \\ a^{2} & b^{2} & c^{2} \\ a^{3} & b^{3} & c^{3}\end{array}\right|$
Sol :
$\left|\begin{array}{ccc}1 & 1 & 1 \\ a^{2} & b^{2} & c^{2} \\ a^{3} & b^{3} & c^{3}\end{array}\right|$

C₁→C₁-C₂, C₂→C₂-C₃

$=\left|\begin{array}{ccc}0 & 0 & 1 \\ a^{2}-b^{2} & b^{2}-c^{2} & c^{2} \\ a^{3}-b^{3} & b^{3}-c^{3} & c^{3}\end{array}\right|$

taking out (a-b) from C₁ and (b-c) from C₂

$=(a-b)(b-c)\left|\begin{array}{ccc}0 & 0 \\ a+b & b+c & c^{2} \\ a^{2}+a b+b^{2} & b^{2}+b c+c^{2} & c^{3}\end{array}\right|$

C₁→C₁-C₃

$=(a-b)(b-c) \quad\left|\begin{array}{ccc}0 & 0 & 1 \\ a-c & b+c & c^{2} \\ a^{2}+a b-b c-c^{2} & b^{2}+b c+c^{2} & c^{3}\end{array}\right|$

$=(a-b)(b-c)\left|\begin{array}{ccc}0 & 0 \\ -(c-a) & b+c & c^{2} \\ -(c-a)(a+b+c) & b^{2}+b c+c^{2} & c^{3}\end{array}\right|$

taking out (c-a) from C₁

$=(a-b)(b-c)(c-a)\left|\begin{array}{ccc}0 & 0 & 1 \\ -1 & b+c & c^{2} \\ -(a+1+c) & b^{2}+b c+c^{2} & c^{3}\end{array}\right|$

Expanding along C₁

$=(a-b)(b-c)(c-a) \cdot\left|\begin{array}{cc}-1 & b+c \\ -(a+b+c) & b^{2}+bc +c^2\end{array} \right|$

=(a-b)(b-c)(c-a)(-b²-bc-c²+ab+ca+b²+bc+bc+c²)

=(a-b)(b-c)(c-a)(ab+bc+ca)


(viii) $\left|\begin{array}{lll}a & b & c \\ a^{2} & b^{2} & c^{2} \\ a^{3} & b^{3} & c^{3}\end{array}\right|$
Sol :
taking out a from C₁ , b from C₂ and c from C₃

$=a bc\left|\begin{array}{lll}1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2}\end{array}\right|$

C₁→C₁-C₂, C₂→C₂-C₃


(ix) $\left|\begin{array}{lll}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{array}\right|$
Sol :
$\left|\begin{array}{lll}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{array}\right|$

C₂→C₂+C₃

$=\left|\begin{array}{ccc}1 & a+b+c & b+c \\ 1 & a+b+c & c+a \\ 1 & a-1 b+c & a- b\end{array}\right|=0$

taking out a² from C₁, b² from C₂, c² from C₃

$=a^{2} b^{2} c^{2}\left|\begin{array}{ccc}0 & a & a \\ b & 0 & b \\ c & c & 0\end{array}\right|$

taking out a from R₁ , b from R₂  and c from R₃

$=a^{3} b^{3} c^{3}\left|\begin{array}{ccc}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right|$

(x) $\left|\begin{array}{ccc}0 & a b^{2} & a c^{2} \\ a^{2} b & 0 & b c^{2} \\ a^{2} c & c b^{2} & 0\end{array}\right|$
Sol :

(xi) $\left|\begin{array}{ccc}1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y\end{array}\right|$
Sol :
$\left|\begin{array}{ccc}1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y\end{array}\right|$

R₁→R₁-R₂ , R₂→R₂-R₃

$=\left|\begin{array}{ccc}0 & -y & 0 \\ 0 & y & -x \\ 1 & x & x+y\end{array}\right|$

Expanding along C₁

$=1\left|\begin{array}{cc}-y & 0 \\ y & -x\end{array}\right|$

=xy-0

=xy