Exercise 6.1
Question 6Evaluate the following :
KC Sinha Mathematics Solution Class 12 Chapter 6 सारणिक (Determinants) Exercise 6.1 (Q6-Q10)
(i) $\begin{vmatrix}a&b+c&a^2\\b&c+a&b^2\\c&a+b&c^2\end{vmatrix}$
Sol :
C2→C2+C1
$\begin{vmatrix}a&a+b+c&a^2\\b&a+b+c&b^2\\c&a+b+c&c^2\end{vmatrix}$
Taking out (a+b+c) from C2
=$(a+b+c)\begin{vmatrix}a&1&a^2\\b&1&b^2\\c&1&c^2\end{vmatrix}$
R1→R1-R2 , R2→R2-R3
$(a+b+c)\begin{vmatrix}a-b&0&a^2-b^2\\b-c&0&b^2-c^2\\c&1&c^2\end{vmatrix}$
$(a+b+c)\begin{vmatrix}a-b&0&(a-b)(a+b)\\b-c&0&(b-c)(b+c)\\c&1&c^2\end{vmatrix}$
taking out (a-b) from R1 and (b-c) from R2
$(a+b+c)(a-b)(b-c)\begin{vmatrix}1&0&(a+b)\\1&0&(b+c)\\c&1&c^2\end{vmatrix}$
Expanding along C2
=$(a+b+c)(a-b)(b-c)(-1)\begin{vmatrix}1&a+b\\1&b+c\end{vmatrix}$
=-(a+b+c)(a-b)(b-c)(b+c-a-b)
=-(a+b+c)(a-b)(b-c)(c-a)
(ii) $\begin{vmatrix}b+c&a-b&a\\c+a&b-c&b\\a+b&c-a&c\end{vmatrix}$
Sol :
C1→C1+C3
$\begin{vmatrix}a+b+c&a-b&a\\a+b+c&b-c&b\\a+b+c&c-a&c\end{vmatrix}$
Taking out (a+b+c) from C1
=$(a+b+c)\begin{vmatrix}1&a-b&a\\1&b-c&b\\1&c-a&c\end{vmatrix} \begin{vmatrix}a-b-b+c\\b-c-c+a\end{vmatrix}$
R1→R1-R2 , R2→R2-R3
=$(a+b+c)\begin{vmatrix}0&a-2b+c&a-b\\0&b-2c+a&b-c\\1&c-a&b-c\end{vmatrix}$
=$(a+b+c).1\begin{vmatrix}a-2b+c&a-b\\b-2c+a&b-c\end{vmatrix}$
=(a+b+c)[(a-2b+c)(b-c)-(b-2c+a)(a-b)]
=(a+b+c)[ab-ac-2b2+2bc+bc-c2-ab+b2+2ac-2bc-a2+ab]
=(a+b+c)[-a2-b2-c2+ab+bc+ac]
=-(a+b+c)[a2+b2+c2-ab-bc-ac]
=-(a3+b3+c3-3abc)
=-a3-b3-c3+3abc
=3abc-a3-b3-c3
(iii) $\begin{vmatrix}a-b&b-c&c-a\\b-c&c-a&a-b\\c-a&a-b&b-c\end{vmatrix}$
Sol :
C1→C1+C2+C3
=$\begin{vmatrix}0&b-c&c-a\\0&c-a&a-b\\0&a-b&b-c\end{vmatrix}$
=$\begin{vmatrix}0&b-c&c-a\\0&c-a&a-b\\0&a-b&b-c\end{vmatrix}$
=0
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(iv) यदि ω इकाई का काल्पनिक मूल है तो निम्नलिखित का मान निकालें।
[If ω is an imaginary cube root of unity m then evaluate the following]
$\begin{vmatrix}1&\omega& {\omega}^2\\ \omega & {\omega}^2&1\\ {\omega}^2&1&\omega\end{vmatrix}$
Sol :
Note: $1+\omega+{\omega}^2=0$
=C1→C1+C2+C3
=$\begin{vmatrix}1+\omega+{\omega}^2&\omega&{\omega}^2\\1+\omega+{\omega}^2&{\omega}^2&1\\1+\omega+{\omega}^2&1&\omega\end{vmatrix}$
=$\begin{vmatrix}0&\omega& {\omega}^2\\ 0& {\omega}^2&1\\0&1&\omega\end{vmatrix}$
=0
(v) यदि a,b,c समानातर श्रेढ़ी में है तो निम्नलिखित का मान निकालें ।
$\begin{vmatrix}1&\omega& {\omega}^2\\ \omega & {\omega}^2&1\\ {\omega}^2&1&\omega\end{vmatrix}$
Sol :
Note: $1+\omega+{\omega}^2=0$
=C1→C1+C2+C3
=$\begin{vmatrix}1+\omega+{\omega}^2&\omega&{\omega}^2\\1+\omega+{\omega}^2&{\omega}^2&1\\1+\omega+{\omega}^2&1&\omega\end{vmatrix}$
=$\begin{vmatrix}0&\omega& {\omega}^2\\ 0& {\omega}^2&1\\0&1&\omega\end{vmatrix}$
=0
(v) यदि a,b,c समानातर श्रेढ़ी में है तो निम्नलिखित का मान निकालें ।
[If a,b,c are in A.P. find the value of the following :]
$\begin{vmatrix}2y+4&5y+7&8y+a\\3y+5&6y+8&9y+b\\4y+6&7y+9&10y+c\end{vmatrix}$
Sol :
b-a=c-b
2b=a+c
∵ a,b,c in A.P.
a+c=2b
a+c-2b=0
$\begin{vmatrix}2y+4&5y+7&8y+9\\3y+5&6y+8&9y+b\\4y+6&7y+9&10y+c\end{vmatrix}$
R1→R1+R3-2R2
=$\begin{vmatrix}0&0&a+c-2b\\3y+5&6y+8&9y+b\\4y+6&7y+9&10y+c\end{vmatrix}$
=$\begin{vmatrix}0&0&0\\3y+5&6y+8&9y+b\\4y+6&7y+9&10y+c\end{vmatrix}$
(vi) मान निकालें (Evaluate)
$\begin{vmatrix}2y+4&5y+7&8y+a\\3y+5&6y+8&9y+b\\4y+6&7y+9&10y+c\end{vmatrix}$
Sol :
b-a=c-b
2b=a+c
∵ a,b,c in A.P.
a+c=2b
a+c-2b=0
$\begin{vmatrix}2y+4&5y+7&8y+9\\3y+5&6y+8&9y+b\\4y+6&7y+9&10y+c\end{vmatrix}$
R1→R1+R3-2R2
=$\begin{vmatrix}0&0&a+c-2b\\3y+5&6y+8&9y+b\\4y+6&7y+9&10y+c\end{vmatrix}$
=$\begin{vmatrix}0&0&0\\3y+5&6y+8&9y+b\\4y+6&7y+9&10y+c\end{vmatrix}$
(vi) मान निकालें (Evaluate)
$\begin{vmatrix}x&y&x+y\\y&x+y&x\\x+y&x&y\end{vmatrix}$
Sol :
C1→C1+C2+C3
$\begin{vmatrix}2x+2y&y&x+y\\2x+2y&x+x&x\\2x+2y&x&y\end{vmatrix}$
taking out (2x+2y) from C1
=$(2x+2y)\begin{vmatrix}1&y&x+y\\1&x+y&x\\1&x&y\end{vmatrix}$
R1→R1-R2 , R2→R2-R3
=$(2x+2y)\begin{vmatrix}0&-x&y\\0&y&x-y\\1&x&y\end{vmatrix}$
Expanding along C1
=$2(x+y)1\begin{vmatrix}-x&y\\y&x-y\end{vmatrix}$
=2(x+y)(-x2+xy-y2)
=-2(x+y)(x2-xy+y2)
=-2(x3+y3)
Question 7
साबिक करे कि (Prove that)
$\left|\begin{array}{ccc}x & y & z \\ x^{2} & y^{2} & z^{2} \\ y z & z x & x y\end{array}\right|=(y-z)(z-x)(x-y)(y z+z x+x y)$
Sol :
L.H.S
$\left|\begin{array}{lll}x & y & z \\ x^{2} & y^{2} & z^{2} \\ yz & z x & x y\end{array}\right|$
C1→C1-C2 , C2→C2-C3
=$\begin{vmatrix}x-y&y-z&z\\x^2-y^2&y^2-z^2&z^2\\yz-zx&zx-xy&xy\end{vmatrix}$
=$\begin{vmatrix}x-y&y-z&z\\(x-y)(x+y)&(y-z)(y+z)&z^2\\-z(x-y)&-x(y-z)&xy\end{vmatrix}$
taking out (x-y) from C1 and (y-z) from C2
=$(x-y)(y-z)\begin{vmatrix}1&1&z\\x+y&y+z&z^2\\-z&-x&xy\end{vmatrix}$
C1→C1-C2
=$\begin{vmatrix}0&1&z\\x-z&y+z&z^2\\-z+x&-x&xy\end{vmatrix}$
=$\begin{vmatrix}0&1&z\\-(z-x)&y+z&z^2\\-(z-x)&-x&xy\end{vmatrix}$
taking out -(z-x) from C1
=$-(x-y)(y-z)(z-x)\begin{vmatrix}0&1&z\\1&y+z&z^2\\1&-x&xy \end{vmatrix}$
R2→R2-R3
=$-(x-y)(y-z)(z-x)\begin{vmatrix}0&1&z\\0&x+y+z&z^2\\1&-x&xy \end{vmatrix}$
Expanding along C1
=$-(x-y)(y-z)(z-x).1\begin{vmatrix}1&z\\x+y+z&z^2-xy\end{vmatrix}$
=-(x-y)(y-z)(z-x)(z2-xy-zx-yz-z2)
=(x-y)(y-z)(z-x)(xy+yz+zx)
Question 8
Sol :
C1→C1+C2+C3
$\begin{vmatrix}2x+2y&y&x+y\\2x+2y&x+x&x\\2x+2y&x&y\end{vmatrix}$
taking out (2x+2y) from C1
=$(2x+2y)\begin{vmatrix}1&y&x+y\\1&x+y&x\\1&x&y\end{vmatrix}$
R1→R1-R2 , R2→R2-R3
=$(2x+2y)\begin{vmatrix}0&-x&y\\0&y&x-y\\1&x&y\end{vmatrix}$
Expanding along C1
=$2(x+y)1\begin{vmatrix}-x&y\\y&x-y\end{vmatrix}$
=2(x+y)(-x2+xy-y2)
=-2(x+y)(x2-xy+y2)
=-2(x3+y3)
Question 7
साबिक करे कि (Prove that)
$\left|\begin{array}{ccc}x & y & z \\ x^{2} & y^{2} & z^{2} \\ y z & z x & x y\end{array}\right|=(y-z)(z-x)(x-y)(y z+z x+x y)$
Sol :
L.H.S
$\left|\begin{array}{lll}x & y & z \\ x^{2} & y^{2} & z^{2} \\ yz & z x & x y\end{array}\right|$
C1→C1-C2 , C2→C2-C3
=$\begin{vmatrix}x-y&y-z&z\\x^2-y^2&y^2-z^2&z^2\\yz-zx&zx-xy&xy\end{vmatrix}$
=$\begin{vmatrix}x-y&y-z&z\\(x-y)(x+y)&(y-z)(y+z)&z^2\\-z(x-y)&-x(y-z)&xy\end{vmatrix}$
taking out (x-y) from C1 and (y-z) from C2
=$(x-y)(y-z)\begin{vmatrix}1&1&z\\x+y&y+z&z^2\\-z&-x&xy\end{vmatrix}$
C1→C1-C2
=$\begin{vmatrix}0&1&z\\x-z&y+z&z^2\\-z+x&-x&xy\end{vmatrix}$
=$\begin{vmatrix}0&1&z\\-(z-x)&y+z&z^2\\-(z-x)&-x&xy\end{vmatrix}$
taking out -(z-x) from C1
=$-(x-y)(y-z)(z-x)\begin{vmatrix}0&1&z\\1&y+z&z^2\\1&-x&xy \end{vmatrix}$
R2→R2-R3
=$-(x-y)(y-z)(z-x)\begin{vmatrix}0&1&z\\0&x+y+z&z^2\\1&-x&xy \end{vmatrix}$
Expanding along C1
=$-(x-y)(y-z)(z-x).1\begin{vmatrix}1&z\\x+y+z&z^2-xy\end{vmatrix}$
=-(x-y)(y-z)(z-x)(z2-xy-zx-yz-z2)
=(x-y)(y-z)(z-x)(xy+yz+zx)
Question 8
साबित करें कि ( Prove that )
$\begin{vmatrix}1&a^2+bc&a^3\\1&b^2+ca&b^3\\1&c^2+ab&c^3\end{vmatrix}=-(a-b)(b-c)(c-a)(a^2+b^2+c^2)$
Sol :
L.H.S
$\begin{vmatrix}1&a^2+bc&a^3\\1&b^2+ca&b^3\\1&c^2+ab&c^3\end{vmatrix}$
R1→R1-R2 , R2→R2-R3
=$\begin{vmatrix}0&a^2+bc-b^2-ca&a^3-b^3\\0&b^2+ca-c^2-ab&b^3-c^3\\1&c^2+ab&c^3\end{vmatrix}$
$=\left|\begin{array}{ccc}0 & a^{2}+b c-b^{2}-c a & a^{3}-b^{3} \\ 0 & b^{2}+c a-c^{2} -a b & b^{3}-c^{2} \\ 1 & c^{2}+a b & c^{3}\end{array}\right|$
$=\left|\begin{array}{ccc}0 & (a+b)(a-b)-c(a-b) & (a-b)\left(a^{2}+a b+b^{2}\right) \\ 0 & (b+c)(b-c)-a(b-c) & (b-c)\left(b^{2}+b c+c^{2}\right) \\ 1 & c^{2}+a b & c^{3}\end{array}\right|$
$= \begin{vmatrix} 0 & (a-b)(a+b-c) & (a-b)\left(a^{2}+a b+b^{2}\right) \\ 0 & (b-c)(b+c-a) & (b-c)\left(b^{2}+b c+c^{2}\right) \\1 & c^{2}+a b & c^{3}\end{vmatrix}$
taking out (a-b) from R1 , (b-c) from R2
$=(a-b)(b-c)\left|\begin{array}{ccc}0 & a+b-c & a^{2}+a b+b^{2} \\ 0 & b+c-a & b^{2}+b c+c^{2} \\ 1 & c^{2}+a b & c^{3}\end{array}\right|$
R1→R1-R2
$=(a-b)(b-c)\left|\begin{array}{ccc}0 & a+b+c-b-c+a & a^{2}+a b-b^{2} -b^{2}-b c-c^{2} \\ 0 & b+c-a & b^{2}+b c+c^{2} \\ 1 & c^{2}+a b & c^3 \end{array}\right|$
$=(a-b)(b-c) \quad \begin{vmatrix}0 & -2(c-a) & -(c-a)(c+a+b) \\ 0 & b+c-a & b^{2}+b c+c^{2} \\ 1 & c^{2}+a b & c^{3}\end{vmatrix}$
taking out -(c-a) from R1
$=-(a-b)(b-c)(c-a)\left|\begin{array}{cc}0 & 2 & a+b+c \\ 0 & b+c-a & b^{2}+b c+c^{2} \\ 1 & c^{2}+a b & c^{3}\end{array}\right|$
Expanding along C1
$=-(a-b)(b-c)(c-a) 1 \left|\begin{array}{cc}2 & a+b+c \\ b+c-a & b^{2}+b c+c^{2}\end{array}\right|$
=-(a-b)(b-c)(c-a)(2b2+2bc+2c2-ab-b2-bc-ca-bc-c2+a2+ab+ca)
=-(a-b)(b-c)(c-a)(a2+b2+c2+2bc-2bc)
=-(a-b)(b-c)(c-a)(a2+b2+c2)
Question 9
Using properties of determinants , prove that:
R1→R1-R2 , R2→R2-R3
=$\begin{vmatrix}0&a^2+bc-b^2-ca&a^3-b^3\\0&b^2+ca-c^2-ab&b^3-c^3\\1&c^2+ab&c^3\end{vmatrix}$
$=\left|\begin{array}{ccc}0 & a^{2}+b c-b^{2}-c a & a^{3}-b^{3} \\ 0 & b^{2}+c a-c^{2} -a b & b^{3}-c^{2} \\ 1 & c^{2}+a b & c^{3}\end{array}\right|$
$=\left|\begin{array}{ccc}0 & (a+b)(a-b)-c(a-b) & (a-b)\left(a^{2}+a b+b^{2}\right) \\ 0 & (b+c)(b-c)-a(b-c) & (b-c)\left(b^{2}+b c+c^{2}\right) \\ 1 & c^{2}+a b & c^{3}\end{array}\right|$
$= \begin{vmatrix} 0 & (a-b)(a+b-c) & (a-b)\left(a^{2}+a b+b^{2}\right) \\ 0 & (b-c)(b+c-a) & (b-c)\left(b^{2}+b c+c^{2}\right) \\1 & c^{2}+a b & c^{3}\end{vmatrix}$
taking out (a-b) from R1 , (b-c) from R2
$=(a-b)(b-c)\left|\begin{array}{ccc}0 & a+b-c & a^{2}+a b+b^{2} \\ 0 & b+c-a & b^{2}+b c+c^{2} \\ 1 & c^{2}+a b & c^{3}\end{array}\right|$
R1→R1-R2
$=(a-b)(b-c)\left|\begin{array}{ccc}0 & a+b+c-b-c+a & a^{2}+a b-b^{2} -b^{2}-b c-c^{2} \\ 0 & b+c-a & b^{2}+b c+c^{2} \\ 1 & c^{2}+a b & c^3 \end{array}\right|$
$=(a-b)(b-c) \quad \begin{vmatrix}0 & -2(c-a) & -(c-a)(c+a+b) \\ 0 & b+c-a & b^{2}+b c+c^{2} \\ 1 & c^{2}+a b & c^{3}\end{vmatrix}$
taking out -(c-a) from R1
$=-(a-b)(b-c)(c-a)\left|\begin{array}{cc}0 & 2 & a+b+c \\ 0 & b+c-a & b^{2}+b c+c^{2} \\ 1 & c^{2}+a b & c^{3}\end{array}\right|$
Expanding along C1
$=-(a-b)(b-c)(c-a) 1 \left|\begin{array}{cc}2 & a+b+c \\ b+c-a & b^{2}+b c+c^{2}\end{array}\right|$
=-(a-b)(b-c)(c-a)(2b2+2bc+2c2-ab-b2-bc-ca-bc-c2+a2+ab+ca)
=-(a-b)(b-c)(c-a)(a2+b2+c2+2bc-2bc)
=-(a-b)(b-c)(c-a)(a2+b2+c2)
Question 9
Using properties of determinants , prove that:
(i) $\left|\begin{array}{ccc}x+y & x & x \\ 5 x+4 y & 4 x & 2 x \\ 10 x+8 y & 8 x & 3 x\end{array}\right|=x^{3}$
Sol :
L.H.S
$\left|\begin{array}{ccc}x+y & x & x \\ 5 x+4 y & 4 x & 2 x \\ 10 x+8 y & 8 x & 3 x\end{array}\right|=x^{3}$
$=\left|\begin{array}{ccc}x & x & x \\ 5 x & 4 x & 2 x \\ 10 x & 8 x & 3 x\end{array}\right|+\left|\begin{array}{ccc}y & x & x \\ 4 y & 4 x & 2 x \\ 8 y & 8 x & 3 x\end{array}\right|$
$=x^{3}\left|\begin{array}{ccc}1 & 1 & 1 \\ 5 & 4 & 2 \\ 10 & 8 & 3\end{array}\right|+x^{2} y\left|\begin{array}{ccc}1 & 1 & 1 \\ 4 & 4 & 2 \\ 8 & 8 & 3\end{array}\right|$
$=x^{3}\left|\begin{array}{ccc}1 & 1 & 1 \\ 5 & 4 & 2 \\ 10 & 8 & 3\end{array}\right|+x^{2} y \times 0$
[C1 and C1 are identical]
C1→C1-C2 , C2→C2-C3
$=x^{3}\left|\begin{array}{lll}0 & 0 & 1 \\ 1 & 2 & 2 \\ 2 & 5 & 3\end{array}\right|+0$
Expanding along R1
$=x^{3} \cdot 1\left|\begin{array}{ll}1 & 2 \\ 2 & 5\end{array}\right|$
=x3(5-4)
=x3
(ii) $\left|\begin{array}{ccc}b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{array}\right|=2(a+b+c)\left(a b+b c+c a-a^{2}-b^{2}-c^{2}\right)$
इससे दिखाएँ कि या तो a+b+c=0 or a=b=c
[Hence show that either a+b+c=0 or a=b=c]
Sol :
If $\left|\begin{array}{ccc}b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{array}\right|=0$
then a+b+c=0 or a=b=c
L.H.S
$\left|\begin{array}{ccc}b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{array}\right|$
C1→C1-C2
$=\left|\begin{array}{ccc}2(a+b+c) & c+a & a+b \\ 2(a+b+c) & a+b & b+c \\ 2(a+b+c) & b+c & c+a\end{array}\right|$
taking out 2(a+b+c) from C1
$=2(a+b+c)\left|\begin{array}{ccc}1 & c+a & a+b \\ 1 & a+b & b+c \\ 1 & b+c & c+a\end{array}\right|$
R1→R1-R2 , R2→R2-R3
$=2(a+b+c)\left|\begin{array}{ccc}0 & c-b & a-c \\ 0 & a-c & b-a \\ 1 & b+c & c+a\end{array}\right|$
Expanding along C1
$=2(a+b+c) \cdot 1\left|\begin{array}{cc}c-b & a-c \\ a-c & b-a\end{array}\right|$
=2(a+b+c)[(bc-ca-b2+ab)-(a2-ca-ca+c2)]
=2(a+b+c)(bc-ca-b2+ab-a2+ca+ca-c2)
=2(a+b+c)(ab+bc+ca-a2-b2-c2)
$\left|\begin{array}{cccc}b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{array}\right|=0$
=2(a+b+c)(ab+bc+ca-a2-b2-c2)=0
=-2(a+b+c)(a2+b2+c2-ab-bc-ca=0
=-(a+b+c)(2a2+2b2+2c2-2ab-2bc-2ca)=0
=(a+b+c)(a2+b2-2ab+b2+c2-2bc+c2+a2-2ca)=0
=(a+b+c)[(a-b)2+(b-c)2+(c-a)2]=0
$\begin{array}{r|l}a+b+c=0&(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0\\&a-b=0,b-c=0,c-a=0\\&a=b,b=c,c=a\\&\therefore a=b=c\end{array}$
(iii) $\left|\begin{array}{ccc}1 & x+y & x^{2}+y^{2} \\ 1 & y+z & y^{2}+z^{2} \\ 1 & z+x & z^{2}+x^{2}\end{array}\right|=(x-y)(y-z)(z-x)$
Sol :
L.H.S
$\left|\begin{array}{ccc}1 & x+y & x^{2}+y^{2} \\ 1 & y+z & y^{2}+z^{2} \\ 1 & z+x & z^{2}+x^{2}\end{array}\right|$
R1→R1-R2 , R2→R2-R1
$=\left|\begin{array}{lll}0 & x-2 & x^{2}-z^{2} \\ 0 & y-x & y^{2}-x^{2} \\ 1 & z+x & z^{2}+x^{2}\end{array}\right|$
$=\left|\begin{array}{ccc}0 & x-z & (x-z)(x+z) \\ 0 & y-x & (y-x)(y+x) \\ 1 & z+x & z^{2}+x^{2}\end{array}\right|$
taking out (x-z) from R1 and (y-x) from R2
$=(x-z)(y-x)\left|\begin{array}{ccc}0 & 1 & x+z \\ 0 & 1 & y+x \\ 1 & z+x & z^{2}+x^{2}\end{array}\right|$
$=(z-x)(x-y)\left|\begin{array}{ccc}0 & 1 & x+z \\ 0 & 1 & y+x \\ 1 & z+x & z^{2}+x^{2}\end{array}\right|$
Expanding along C1
$=(z-x)(x-y) \cdot 1 \cdot\left|\begin{array}{cc}1 & x+z \\ 1 & y+x\end{array}\right|$
=(z-x)(x-y)(y+x-x-z)
=(x-y)(y-z)(z-x)
Question 10
(i) यदि x,y,z असमान हों तथा [If x,y,z are different and]
$\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3}\end{array}\right|=0$
तो दिखाएँ कि (then show that) 1+xyz=0
Sol :
$\left|\begin{array}{ccc}x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3}\end{array}\right|=0$
$\left|\begin{array}{ccc}x & x^{2} & 1 \\ y & y^{2} & 1 \\ 2 & z^{2} & 1\end{array}\right|+\left|\begin{array}{ccc}x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3}\end{array}\right|=0$
C1↔C3
$=-\left|\begin{array}{ccc}1 & x^{2} & x \\ 1 & y^{2} & y \\ 1 & z^{2} & z\end{array}\right|+x yz \begin{vmatrix}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{vmatrix}$
C2↔C3
$=1\left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|+x yz\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|=0$
$(1+x y z)\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^2 \\ 1 & z & z^{2}\end{array}\right|=0$
R1→R1-R2 , R2→R2-R3
$(1+x y z)\left|\begin{array}{ccc}0 & x-y & x^{2}-y^{2} \\ 0 & y-z & y^{2}-z^{2} \\ 1 & z & z^{2}\end{array}\right|=0$
taking out (x-y) from R1 and (y-z) from R2
$(1+x y z)(x-y)(y-z)\left|\begin{array}{rrr}0 & 1 & x+y \\ 0 & 1 & y+z \\ 1 & z & z^{2}\end{array}\right|=0$
Expanding along C1
$(1+x y z)(x-y)(y-z) 1\left|\begin{array}{ll}1 & x+y \\ 1 & y+z\end{array}\right|=0$
(1+xyz)(x-y)(y-z)(y+z-x-y)=0
(1+xyz)(x-y)(y-z)(z-x)=0
1+xyz=0
(ii)
यदि x,y,z असमान हों तथा
$\left|\begin{array}{ccc}x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3}\end{array}\right|=0$
$\left|\begin{array}{ccc}x & x^{2} & 1 \\ y & y^{2} & 1 \\ 2 & z^{2} & 1\end{array}\right|+\left|\begin{array}{ccc}x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3}\end{array}\right|=0$
C1↔C3
$=-\left|\begin{array}{ccc}1 & x^{2} & x \\ 1 & y^{2} & y \\ 1 & z^{2} & z\end{array}\right|+x yz \begin{vmatrix}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{vmatrix}$
C2↔C3
$=1\left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|+x yz\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|=0$
$(1+x y z)\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^2 \\ 1 & z & z^{2}\end{array}\right|=0$
R1→R1-R2 , R2→R2-R3
$(1+x y z)\left|\begin{array}{ccc}0 & x-y & x^{2}-y^{2} \\ 0 & y-z & y^{2}-z^{2} \\ 1 & z & z^{2}\end{array}\right|=0$
taking out (x-y) from R1 and (y-z) from R2
$(1+x y z)(x-y)(y-z)\left|\begin{array}{rrr}0 & 1 & x+y \\ 0 & 1 & y+z \\ 1 & z & z^{2}\end{array}\right|=0$
Expanding along C1
$(1+x y z)(x-y)(y-z) 1\left|\begin{array}{ll}1 & x+y \\ 1 & y+z\end{array}\right|=0$
(1+xyz)(x-y)(y-z)(y+z-x-y)=0
(1+xyz)(x-y)(y-z)(z-x)=0
1+xyz=0
(ii)
यदि x,y,z असमान हों तथा
[If x,y,z are distinct and]
$\left|\begin{array}{lll}x & x^{3} & x^{4}-1 \\ y & y^{3} & y^{4}-1 \\ z & z^{3} & z^{4}-1\end{array}\right|=0$ , तो साबित करें कि (then prove that)
(xyz)(xy+yz+zx)=(x+y+z)
Sol :
$\left|\begin{array}{lll}x & x^{3} & x^{4}-1 \\ y & y^{3} & y^{4}-1 \\ z & z^{3} & z^{4}-1\end{array}\right|=0$
$x y z\left|\begin{array}{ccc}1 & x^{2} & x^{3} \\ 1 & y^{2} & y^{3} \\ 1 & z^{2} & z^{3}\end{array}\right|=\left|\begin{array}{ccc}1 & x^{3} & x \\ 1 & y^{3} & y \\ 1 & z^{3} & z\end{array}\right|$
=$\begin{aligned}x y z\left|\begin{array}{ccc}1 & x^{2} & x^{3} \\ 1 & y^{2} & y^{3} \\ 1 & z^{2} & z^{3}\end{array}\right|=\left|{\begin{array}{ccc}x & x^{3} & 1 \\ y & y^{3} & 1 \\ z & z^{3} & 1\end{array}}\right|\\{C_1\leftrightarrow C_3}\end{aligned}$
$\begin{aligned}x y z\left|\begin{array}{ccc}1 & x^{2} & x^{3} \\ 1 & y^{2} & y^{3} \\ 1 & z^{2} & z^{3}\end{array}\right|=-\left|{\begin{array}{ccc}1 & x^{3} & x \\1 & y^{3} & y \\ 1 & z^{3} & z\end{array}}\right|\\{C_1\leftrightarrow C_3}\end{aligned}$
$x y z\left|\begin{array}{ccc}1 & x^{2} & x^{3} \\ 1 & y^{2} & y^{3} \\ 1 & z^{2} & z^{3}\end{array}\right|=\left|{\begin{array}{ccc}1 & x& x^{3} \\1 & y & y^{3} \\ 1 & z & z^{3}\end{array}}\right|$
R1→R1-R2 , R2→R2-R3
$x yz\left|\begin{array}{ccc}0 & x^{2}-y^{2} & x^{3}-y^{3} \\ 0 & y^{2}-z^{2} & y^{3}-z^{3} \\ 1 & z^{2} & z^{3}\end{array}\right|=\left|\begin{array}{ccc}0 & x-y & x^{3}-y^3 \\ 0 & y-z & y^{3}-z^3 \\ 1 & z & z^{3}\end{array}\right|$
$x yz\left|\begin{array}{ccc}0 & (x-y) (x+y) & (x-y)\left(x^{2}+x y+y\right)^{2} \\ 0 & (y-z)(y+z) & (y-z)\left(y^{2}+y z+z^{2}\right) \\1& z^{2} & z^{3}\end{array}\right|= \begin{vmatrix}0 & x-y & (x-y)(x^2+xy+y^2) \\ 0 & y-z & (y-z)\left(y^{2}+yz+z^2\right) \\ 1 & z & z^3\end{vmatrix}$
taking out (x-y) from R1and (y-z) from R2
$x yz(x - y)(y-z)\left|\begin{array}{ccc}0 & x+y & x^{2}+x y+y^{2} \\ 0 & y+z & y^{2}+y z+z^{2} \\1& z^{2} & z^{3}\end{array}\right|=(x - y)(y-z)\left|\begin{array}{ccc}0 &1 & x^{2}+x y+y^{2} \\ 0 & 1 & y^{2}+y z+z^{2} \\1& z & z^{3}\end{array}\right|$
R1→R1-R2
$x y z\left|\begin{array}{ccc}0 & x-z & x^{2}+x y-y z-z^{2} \\ 0 & y+z & y^{2}+y z+z^{2} \\ 1 & z^{2} & z^{3}\end{array}\right|=\left|\begin{array}{ccc}0 & 0 & x^{2}+x y-y z \\ 0 & 1 & y^{2}+y z+z^{2} \\ 1 & z & z^{3}\end{array}\right|$
$x y=\left|\begin{array}{ccc}0 & x-z & (x-z)(x+y+z) \\ 0 & y+z & y^{2}+y z+z^{2} \\ 1 & z^{2} & z^{3}\end{array}\right|=\left|\begin{array}{ccc}0 & 0 & (x-z)(x+y+z) \\ 0 & 1 & y^{2}+y z+z^{2} \\ 1 & z & z^{3} \end{array}\right|$
taking out (x-y) from R1
$x yz(x-z)\left|\begin{array}{ccc}0 & 1 & x+y+z \\ 0 & y+z & y^{2}+y z+z^{2} \\ 1 & z^{2} & z^{3}\end{array}\right|=$
$=(x-z)\left|\begin{array}{ccc}0 & 0 & x+y+z \\ 0 & 1 & y^{2}+y z+z^2 \\ 1 & z & z^{3}\end{array}\right|$
Expanding along C1
$x y z \cdot 1\left|\begin{array}{cc}1 & x+y+z \\ y+z & y^{2}+y z+z^2\end{array}\right|=1 \cdot\left|\begin{array}{cc}0 & x+y+z \\1& y^2+yz+z^{2}\end{array}\right|$
xyz(y2+yz+z2-xy-y2-yz-zx-yz-z2)=(0-x-y-z)
-xyz(xy+yz+zx)=-(x+y+z)
xyz(xy+yz+zx)=x+y+z
$\left|\begin{array}{lll}x & x^{3} & x^{4}-1 \\ y & y^{3} & y^{4}-1 \\ z & z^{3} & z^{4}-1\end{array}\right|=0$
$x y z\left|\begin{array}{ccc}1 & x^{2} & x^{3} \\ 1 & y^{2} & y^{3} \\ 1 & z^{2} & z^{3}\end{array}\right|=\left|\begin{array}{ccc}1 & x^{3} & x \\ 1 & y^{3} & y \\ 1 & z^{3} & z\end{array}\right|$
=$\begin{aligned}x y z\left|\begin{array}{ccc}1 & x^{2} & x^{3} \\ 1 & y^{2} & y^{3} \\ 1 & z^{2} & z^{3}\end{array}\right|=\left|{\begin{array}{ccc}x & x^{3} & 1 \\ y & y^{3} & 1 \\ z & z^{3} & 1\end{array}}\right|\\{C_1\leftrightarrow C_3}\end{aligned}$
$\begin{aligned}x y z\left|\begin{array}{ccc}1 & x^{2} & x^{3} \\ 1 & y^{2} & y^{3} \\ 1 & z^{2} & z^{3}\end{array}\right|=-\left|{\begin{array}{ccc}1 & x^{3} & x \\1 & y^{3} & y \\ 1 & z^{3} & z\end{array}}\right|\\{C_1\leftrightarrow C_3}\end{aligned}$
$x y z\left|\begin{array}{ccc}1 & x^{2} & x^{3} \\ 1 & y^{2} & y^{3} \\ 1 & z^{2} & z^{3}\end{array}\right|=\left|{\begin{array}{ccc}1 & x& x^{3} \\1 & y & y^{3} \\ 1 & z & z^{3}\end{array}}\right|$
R1→R1-R2 , R2→R2-R3
$x yz\left|\begin{array}{ccc}0 & x^{2}-y^{2} & x^{3}-y^{3} \\ 0 & y^{2}-z^{2} & y^{3}-z^{3} \\ 1 & z^{2} & z^{3}\end{array}\right|=\left|\begin{array}{ccc}0 & x-y & x^{3}-y^3 \\ 0 & y-z & y^{3}-z^3 \\ 1 & z & z^{3}\end{array}\right|$
$x yz\left|\begin{array}{ccc}0 & (x-y) (x+y) & (x-y)\left(x^{2}+x y+y\right)^{2} \\ 0 & (y-z)(y+z) & (y-z)\left(y^{2}+y z+z^{2}\right) \\1& z^{2} & z^{3}\end{array}\right|= \begin{vmatrix}0 & x-y & (x-y)(x^2+xy+y^2) \\ 0 & y-z & (y-z)\left(y^{2}+yz+z^2\right) \\ 1 & z & z^3\end{vmatrix}$
taking out (x-y) from R1and (y-z) from R2
$x yz(x - y)(y-z)\left|\begin{array}{ccc}0 & x+y & x^{2}+x y+y^{2} \\ 0 & y+z & y^{2}+y z+z^{2} \\1& z^{2} & z^{3}\end{array}\right|=(x - y)(y-z)\left|\begin{array}{ccc}0 &1 & x^{2}+x y+y^{2} \\ 0 & 1 & y^{2}+y z+z^{2} \\1& z & z^{3}\end{array}\right|$
R1→R1-R2
$x y z\left|\begin{array}{ccc}0 & x-z & x^{2}+x y-y z-z^{2} \\ 0 & y+z & y^{2}+y z+z^{2} \\ 1 & z^{2} & z^{3}\end{array}\right|=\left|\begin{array}{ccc}0 & 0 & x^{2}+x y-y z \\ 0 & 1 & y^{2}+y z+z^{2} \\ 1 & z & z^{3}\end{array}\right|$
$x y=\left|\begin{array}{ccc}0 & x-z & (x-z)(x+y+z) \\ 0 & y+z & y^{2}+y z+z^{2} \\ 1 & z^{2} & z^{3}\end{array}\right|=\left|\begin{array}{ccc}0 & 0 & (x-z)(x+y+z) \\ 0 & 1 & y^{2}+y z+z^{2} \\ 1 & z & z^{3} \end{array}\right|$
taking out (x-y) from R1
$x yz(x-z)\left|\begin{array}{ccc}0 & 1 & x+y+z \\ 0 & y+z & y^{2}+y z+z^{2} \\ 1 & z^{2} & z^{3}\end{array}\right|=$
$=(x-z)\left|\begin{array}{ccc}0 & 0 & x+y+z \\ 0 & 1 & y^{2}+y z+z^2 \\ 1 & z & z^{3}\end{array}\right|$
Expanding along C1
$x y z \cdot 1\left|\begin{array}{cc}1 & x+y+z \\ y+z & y^{2}+y z+z^2\end{array}\right|=1 \cdot\left|\begin{array}{cc}0 & x+y+z \\1& y^2+yz+z^{2}\end{array}\right|$
xyz(y2+yz+z2-xy-y2-yz-zx-yz-z2)=(0-x-y-z)
-xyz(xy+yz+zx)=-(x+y+z)
xyz(xy+yz+zx)=x+y+z
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