KC Sinha Mathematics Solution Class 12 Chapter 6 सारणिक (Determinants) Exercise 6.1 (Q11-Q15)
साबित करे कि (Prove that:)
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$
Sol :
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
R1→R1+R2+R3
$=\left|\begin{array}{ccc}a +b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
taking out (a+b+c) from R1
$=(a+b+c)\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
$=[a+b+c]\left|\begin{array}{ccc}0 & 0 & 1 \\ a+b+c & -(a+b+c) & 2 b \\ 0 & a+b+c & c-a-b\end{array}\right|$
taking out (a+b+c) from C1 and C2
$=(a+b+c)^{3}\left|\begin{array}{cccc}0 & 0 & 1 \\ 1 & -1 & 2 b \\ 0 & 1 & c-a-b\end{array}\right|$
Expanding along R1
$=(a+b+c)^{3} \cdot 1\left|\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right|$
=(a+b+c)3
Question 12
Sol :
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
R1→R1+R2+R3
$=\left|\begin{array}{ccc}a +b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
taking out (a+b+c) from R1
$=(a+b+c)\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
$=[a+b+c]\left|\begin{array}{ccc}0 & 0 & 1 \\ a+b+c & -(a+b+c) & 2 b \\ 0 & a+b+c & c-a-b\end{array}\right|$
taking out (a+b+c) from C1 and C2
$=(a+b+c)^{3}\left|\begin{array}{cccc}0 & 0 & 1 \\ 1 & -1 & 2 b \\ 0 & 1 & c-a-b\end{array}\right|$
Expanding along R1
$=(a+b+c)^{3} \cdot 1\left|\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right|$
=(a+b+c)3
Question 12
साबित करे कि (Prove that:)
$\left|\begin{array}{ccc}a+b+c & -c & -b \\ -c & a+b+c & -a \\ -b & -a & a+b+c\end{array}\right|=2(a+b)(b+c)(c+a)$
Sol :
L.H.S
$\left|\begin{array}{ccc}a+b+c & -c & -b \\ -c & a+b+c & -a \\ -b & -a & a+b+c\end{array}\right|$
C1→C1+C2 , C2→C2+C3
$=\left|\begin{array}{ccc}a+b & -(b+c) & -b \\ a+b & b+c & -a \\ -(a+b) & b+c & a+b+c\end{array}\right|$
taking out (a+b) from C1 and (b+c) from C2
$=(a+b)(b+c)\left|\begin{array}{ccc}1 & -1 & -b \\ 1 & 1 & -a \\ -1 & 1 & a+b+c\end{array}\right|$
C1→C1+C2
$=(a+b)(b+c)\left|\begin{array}{ccc}0 & -1 & -b \\ 2 & 1 & -a \\ 0 & 1 & a+b+c\end{array}\right|$
Expanding along C1
$=(a+b)(b+c)(-2)\left|\begin{array}{cc}-1 & -b \\ 1 & a+b+c\end{array}\right|$
=(a+b)(b+c)(-2)(-a-b-c+b)
=2(a+b)(b+c)(c+a)
Question 13
(i) साबित करे कि (prove that) $\left|\begin{array}{lll}y+z & x & y \\ z+x & z & x \\ x+y & y & z\end{array}\right|=(x+y+z)(x-z)^{2}$
Sol :
L.H.S
$\left|\begin{array}{ccc}y+z & x & y \\ z+x & z & x \\ x+y & y & z\end{array}\right|$
R1→R1+R2+R3
$=\left|\begin{array}{ccc}2(x+y+z) & x+y+z & x+y+z \\ z+x & z & x \\ x+y & y & z\end{array}\right|$
taking out (x+y+z) from R1
$=(x+y+z)\left|\begin{array}{ccc}2 & 1 & 1 \\ z+x & z & x \\ x+y & y & z\end{array}\right|$
R1→R1-2R2 , R2→R2-R3
$=(x+y+2)\left|\begin{array}{ccc}0 & 0 & 1 \\ x-z & z-x & x \\ x-y & y-z & z\end{array}\right|$
Expanding along R1
$=(x+y+z)(1)\left|\begin{array}{ll}x-z & z-x \\ x-y & y-z\end{array}\right|$
=(x+y+z)[(xy-zx-yz+z2)-(zx-x2-yz-xy)]
=(x+y+z)(x2-2zx+z2)
=(x+y+z)(x-z)2
(ii) दिखाएँ कि ( Show that )$\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$
Sol :
L.H.S
$\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$
R1→R1+R2+R3
$=\left|\begin{array}{ccc}5 x+4 & 5 x+4 & 5 x+4 \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$
taking out (5x+4) from R1
$=(5 x+4)\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$
C1→C1-C2 , C2→C2-C3
$=(5 x+4)\left|\begin{array}{ccc}0 & 0 & 1 \\ x-4 & 4-x & 2 x \\ 0 & x-4 & x+4\end{array}\right|$
$=(5 x+4)\left|\begin{array}{ccc}0 & 0&1 \\ -(4-x) & 4-x & 2x \\ 0 & -(4-x) & x+4\end{array}\right|$
taking out (4-x) from C1and C2
$=(5 x+4)(4-x)^{2}\left|\begin{array}{ccc}0 & 0 & 1 \\ -1 & 1 & 2 x \\ 0 & -1 & x+4\end{array}\right|$
Expanding along R1
$=(5 x+4)(4-x)^{2} \cdot 1\left|\begin{array}{cc}-1 & 1 \\ 0 & -1\end{array}\right|$
=(5x+4)(4-x)2
Question 14
Sol :
L.H.S
$\left|\begin{array}{ccc}y+z & x & y \\ z+x & z & x \\ x+y & y & z\end{array}\right|$
R1→R1+R2+R3
$=\left|\begin{array}{ccc}2(x+y+z) & x+y+z & x+y+z \\ z+x & z & x \\ x+y & y & z\end{array}\right|$
taking out (x+y+z) from R1
$=(x+y+z)\left|\begin{array}{ccc}2 & 1 & 1 \\ z+x & z & x \\ x+y & y & z\end{array}\right|$
R1→R1-2R2 , R2→R2-R3
$=(x+y+2)\left|\begin{array}{ccc}0 & 0 & 1 \\ x-z & z-x & x \\ x-y & y-z & z\end{array}\right|$
Expanding along R1
$=(x+y+z)(1)\left|\begin{array}{ll}x-z & z-x \\ x-y & y-z\end{array}\right|$
=(x+y+z)[(xy-zx-yz+z2)-(zx-x2-yz-xy)]
=(x+y+z)(x2-2zx+z2)
=(x+y+z)(x-z)2
(ii) दिखाएँ कि ( Show that )$\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2}$
Sol :
L.H.S
$\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$
R1→R1+R2+R3
$=\left|\begin{array}{ccc}5 x+4 & 5 x+4 & 5 x+4 \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$
taking out (5x+4) from R1
$=(5 x+4)\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$
C1→C1-C2 , C2→C2-C3
$=(5 x+4)\left|\begin{array}{ccc}0 & 0 & 1 \\ x-4 & 4-x & 2 x \\ 0 & x-4 & x+4\end{array}\right|$
$=(5 x+4)\left|\begin{array}{ccc}0 & 0&1 \\ -(4-x) & 4-x & 2x \\ 0 & -(4-x) & x+4\end{array}\right|$
taking out (4-x) from C1and C2
$=(5 x+4)(4-x)^{2}\left|\begin{array}{ccc}0 & 0 & 1 \\ -1 & 1 & 2 x \\ 0 & -1 & x+4\end{array}\right|$
Expanding along R1
$=(5 x+4)(4-x)^{2} \cdot 1\left|\begin{array}{cc}-1 & 1 \\ 0 & -1\end{array}\right|$
=(5x+4)(4-x)2
Question 14
साबित करे कि (prove that) $\left|\begin{array}{ccc}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|=\left(1-x^{3}\right)^{2}$
Sol :
LH.S
$\left|\begin{array}{ccc}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|$
R1→R1+R2+R3
$=\left| \begin{array}{ccc}1+x+x^{2} & 1+x+x^{2} & 1+x+x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|$
taking out (1+x+x2) from R1
$=\left(1+x+x^{2}\right)\left|\begin{array}{ccc}1 & 1 & 1 \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|$
C1→C1-C2 , C2→C2-C3
$=\left(1+x+x^{2}\right)\left|\begin{array}{ccc}0 & 0 & 1 \\ x^{2}-1 & 1-x & x \\ x-x^{2} & x^{2}-1 & 1\end{array}\right|$
$=\left(1+x+x^{2}\right)\left|\begin{array}{ccc}0 & 0 & 1 \\ -\left(1-x^{2}\right) & 1-x & x \\ x(1-x) & -\left(1-x^{2}\right) & 1\end{array}\right|$
$=\left(1+x+x^{2}\right)\left|\begin{array}{ccc}0 & 0 & 1 \\ -(1-x)(1+x) & 1-x & x \\ x(1-x) & -(1-x)(1+x) & 1\end{array}\right|$
taking out (1-x) from C1 and C2
$=\left(1+x+x^{2}\right)(1-x)^{2} \left| \begin{array}{ccc}0 & 0 & 1 \\ -(1+x) & 1 & x \\ x & -(1+x) & 1\end{array}\right|$
Expanding along R1
$=\left(1+x+z^{2}\right)(1-x)^{2} (1)\left|\begin{array}{cc}-(1+x) & 1 \\ x & -(1+x)\end{array}\right|$
=(1+x+x2)(1-x)2(1+2x+x2-x)
=(1+x+x2)(1-x)2(1+x+x2)
=(1+x+x2)2(1-x)2
=[(1-x)(12+1.x+x2)]2
=[13-x3]2
=(1-x3)2
Question 15
Sol :
LH.S
$\left|\begin{array}{ccc}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|$
R1→R1+R2+R3
$=\left| \begin{array}{ccc}1+x+x^{2} & 1+x+x^{2} & 1+x+x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|$
taking out (1+x+x2) from R1
$=\left(1+x+x^{2}\right)\left|\begin{array}{ccc}1 & 1 & 1 \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|$
C1→C1-C2 , C2→C2-C3
$=\left(1+x+x^{2}\right)\left|\begin{array}{ccc}0 & 0 & 1 \\ x^{2}-1 & 1-x & x \\ x-x^{2} & x^{2}-1 & 1\end{array}\right|$
$=\left(1+x+x^{2}\right)\left|\begin{array}{ccc}0 & 0 & 1 \\ -\left(1-x^{2}\right) & 1-x & x \\ x(1-x) & -\left(1-x^{2}\right) & 1\end{array}\right|$
$=\left(1+x+x^{2}\right)\left|\begin{array}{ccc}0 & 0 & 1 \\ -(1-x)(1+x) & 1-x & x \\ x(1-x) & -(1-x)(1+x) & 1\end{array}\right|$
taking out (1-x) from C1 and C2
$=\left(1+x+x^{2}\right)(1-x)^{2} \left| \begin{array}{ccc}0 & 0 & 1 \\ -(1+x) & 1 & x \\ x & -(1+x) & 1\end{array}\right|$
Expanding along R1
$=\left(1+x+z^{2}\right)(1-x)^{2} (1)\left|\begin{array}{cc}-(1+x) & 1 \\ x & -(1+x)\end{array}\right|$
=(1+x+x2)(1-x)2(1+2x+x2-x)
=(1+x+x2)(1-x)2(1+x+x2)
=(1+x+x2)2(1-x)2
=[(1-x)(12+1.x+x2)]2
=[13-x3]2
=(1-x3)2
Question 15
साबित करे कि (prove that)
$\left|\begin{array}{ccc}a & b-c & c+b \\ a+c & b & c-a \\ a-b & a+b & c\end{array}\right|=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$
Sol :
L.H.S
$\left|\begin{array}{ccc}a & b-c & c+b \\ a+c & b & c-a \\ a-b & a+b & c\end{array}\right|$
C1→aC1 , C2→bC2 , C3→cC3
$=\frac{1}{a b c}\left|\begin{array}{ccc}a^{2} & b^{2}-b c & c^{2}+b c \\ a^{2}+a c & b^{2} & c^{2}-a c \\ a^{2}-a b & a b+b^{2} & c^{2}\end{array}\right|$
C1→C1+C2+C3
$=\frac{1}{a b c} \left| \begin{array}{ccc}a^{2}+b^{2}+c^{2} & b^{2}-b c & c^{2}+b c \\ a^{2}+b^{2}+c^{2} & b^{2} & c^{2}-a c \\ a^{2}+b^{2}+c^{2} & a b+b^{2} & c^{2}\end{array}\right|$
taking out (a2+b2+c2) from C1, b from C2 , from C3
$=\dfrac{1}{a bc} \times b c\left(a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc}1 & b-c & c+b \\ 1 & b & c-a \\ 1 & a+b & c\end{array}\right|$
R1→R1-R2 , R2→R2-R3
$=\frac{1}{a}\left(a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc}0 & -c & b+a \\ 0 & -a & -a \\ 1 & a+b & c\end{array}\right|$
Expanding along C1
$=\frac{1}{a}\left(a^{2}+b^{2}+c^{2}\right) \left| \begin{array}{cc}-c & b+a \\ -a & -a\end{array} \right|$
$=\frac{1}{a}\left(a^{2}+b^{2}+c^{2}\right)\left(c a+a b+a^{2}\right)$
$=\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right) \cdot a(c+b+a)$
=(a+b+c)(a2+b2+c2)
$\left|\begin{array}{ccc}a & b-c & c+b \\ a+c & b & c-a \\ a-b & a+b & c\end{array}\right|=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$
Sol :
L.H.S
$\left|\begin{array}{ccc}a & b-c & c+b \\ a+c & b & c-a \\ a-b & a+b & c\end{array}\right|$
C1→aC1 , C2→bC2 , C3→cC3
$=\frac{1}{a b c}\left|\begin{array}{ccc}a^{2} & b^{2}-b c & c^{2}+b c \\ a^{2}+a c & b^{2} & c^{2}-a c \\ a^{2}-a b & a b+b^{2} & c^{2}\end{array}\right|$
C1→C1+C2+C3
$=\frac{1}{a b c} \left| \begin{array}{ccc}a^{2}+b^{2}+c^{2} & b^{2}-b c & c^{2}+b c \\ a^{2}+b^{2}+c^{2} & b^{2} & c^{2}-a c \\ a^{2}+b^{2}+c^{2} & a b+b^{2} & c^{2}\end{array}\right|$
taking out (a2+b2+c2) from C1, b from C2 , from C3
$=\dfrac{1}{a bc} \times b c\left(a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc}1 & b-c & c+b \\ 1 & b & c-a \\ 1 & a+b & c\end{array}\right|$
R1→R1-R2 , R2→R2-R3
$=\frac{1}{a}\left(a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc}0 & -c & b+a \\ 0 & -a & -a \\ 1 & a+b & c\end{array}\right|$
Expanding along C1
$=\frac{1}{a}\left(a^{2}+b^{2}+c^{2}\right) \left| \begin{array}{cc}-c & b+a \\ -a & -a\end{array} \right|$
$=\frac{1}{a}\left(a^{2}+b^{2}+c^{2}\right)\left(c a+a b+a^{2}\right)$
$=\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right) \cdot a(c+b+a)$
=(a+b+c)(a2+b2+c2)
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