KC Sinha Mathematics Solution Class 12 Chapter 6 सारणिक (Determinants) Exercise 6.1 (Q26-Q30)
Exercise 6.1
Question 26
दिखाएँ कि [Show that]
$\left|\begin{array}{ccc}(y+z)^{2} & x y & z x \\ x y & (x+z)^{2} & y z \\ x z & z y & (x+y)^{2}\end{array}\right|=2 x y z(x+y+z)^{3}$
Sol: L.H.S
$\left|\begin{array}{ccc}(y+z)^{2} & x y & z x \\ x y & (x+z)^{2} & y z \\ x z & z y & (x+y)^{2}\end{array}\right|$
$R_{1} \rightarrow x R_{1} , R_{2} \rightarrow y R_{2}, R_{3} \rightarrow zR_{3}$
$=\left|\begin{array}{ccc}x(y+z)^{2} & x^{2} y & =x^{2} \\ x y^{2} & y(x+z)^{2} & y^{2} z \\ x z^{2} & z^{2}, y & z(x+y)^{2}\end{array}\right|$
$=\frac{1}{xy z}\left|\begin{array}{lll}x(y+z)^{2} & x^{2} y & z x^{2} \\ x y^{2} & y(x+z)^{2} & y^{2} z \\ x z^{2} & z^{2} y & z(x+y)^{2}\end{array}\right|$
taking out x from C
1 , y from C
2 , z from C
3
$=\frac{1}{x y z} \times x y z\left|\begin{array}{ccc}(y+z)^{2} & x^{2} & x^{2} \\ y^{2} & (x+z)^{2} & y^{2} \\ z^{2} & z^{2} & (x+y)^{2}\end{array}\right|$
C
1→C
1-C
2 , C
2→C
2-C
3
$=\left|\begin{array}{ccc}(y+z)^{2}-x^{2} & 0 & x^{2} \\ y^{2}-(x+z)^{2} & (x+z)^{2}-y^{2} & y^{2} \\ 0 & z^{2}-(x+y)^{2} & (x+y)^{2}\end{array}\right|$
$=\left|\begin{array}{ccc}(y+z+x)(y+z-x) & 0&x^2 \\ (y+x+z)(y-x-z) & (x+z+y)(x+z-y) & y^{2} \\ 0 & (z+x+y)(z-x-y) & (x+y)^{2}\end{array}\right|$
taking out (x+y+z) from C
1 and C
2
$=(x+y+z)^{2}\left|\begin{array}{ccc}y+z-x & 0 & x^{2} \\ y-x-z & x+z-y & y^{2} \\ 0 & z-x-y & (x+y)^{2}\end{array}\right|$
R
3→R
3-(R
1+R
2)
$=(x+y+z)^{2}\left|\begin{array}{ccc}y+z-x & 0 & x^{2} \\ y-x-z & x+z-y & y^{2} \\ 2 x-2 y & -2 x & 2 x y\end{array}\right|$
taking out 2 from R
3
$=2(x+y+z)^{2}\left|\begin{array}{ccc}y+z-x & 0 & x^{2} \\ y-x-z & x+z y & y^{2} \\ x-y & -x & x y\end{array}\right|$
taking out 2 from R
3
$=2(x+y+z)^{2}\left|\begin{array}{ccc}y+z-x & 0 & x^{2} \\ y-x-z & x+z- y & y^{2} \\ x-y & -x & x y\end{array}\right|$
R
2→R
2+R
3
$=2(x+y+x)^{2}\left|\begin{array}{ccc}y+z-x & 0 & x^{2} \\ -z & z-y & y^{2}+xy \\ x-y & -x & x y\end{array}\right|$
Expanding along R
1
$=2(x+y+z)^{2}\left[(y+z-z)\left|\begin{array}{cc}z-y & y^{2}+x y \\ -x & z y\end{array}\right|+x^{2} \bigg| \begin{array}{cc}-z & z-y \\ x -y & -x\end{array}\bigg| \right]$
=2(x+y+z)
2[(y+z-z)(xyz-xy
2+xy
2+x
2y)+x
2(zx-zx+xy+yz-y
2)]
=2(x+y+z)
2[(y+z-z)(xy
2z+x
2y
2+xyz
2+x
2yz-x
2yz-x
3y+x
3y+x
2yz-x
2y
2)]
=2(x+y+z)
2xyz(y+z+x)
=2xyz(x+y+z)
3
R
1→aR
1, R
2→bR
2 , R
3→cR
3
$=\frac{1}{a b c}\left|\begin{array}{ccc}1 & a^{3} & a b c \\ 1 & b^{3} & a b c \\ 1 & c^{3} & a b c\end{array}\right|$
taking out abc from C
3
$=\frac{1}{a b c} \times a b c\left|\begin{array}{ccc}1 & a^3 & 1 \\ 1 & b^{3} & 1 \\ 1 & c^{3} & 1\end{array}\right|$
= 0
<content to be added>
Question 27 किसी चरण में बिना विस्तार किए दिखाएँ कि
[Show without expanding at any stage that:]
(i) $\left|\begin{array}{lll}\frac{1}{a} & a^{2} & b c \\ \frac{1}{b} & b^{2} & c a \\ \frac{1}{c} & c^{2} & a b\end{array}\right|=0$
Sol :
(ii) $\left|\begin{array}{ccc}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{array}\right|=0$
Sol :
L.H.S
$\left|\begin{array}{ccc}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{array}\right|$
C2→C2+C3
$=\left|\begin{array}{ccc}1 & a+b+c & b+c \\ 1 & a+b+c & c+a \\ 1 & a+b+c & c+b\end{array}\right|=0$
<content to be added>
(iii) $\left|\begin{array}{lll}\frac{1}{a} & a & b c \\ \frac{1}{b} & b & c a \\ \frac{1}{c} & c & a b\end{array}\right|=0$
Sol :
(iv) $\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$
Sol :
L.H.S
$\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|$
C3→C3+C2
$=\left|\begin{array}{ccc}1 & b c & a b+b c+c a \\ 1 & c a & a b+c a+c a \\ 1 & a b & a b+b c+c a\end{array}\right|$
=0
<content to be added>
(v) $\left|\begin{array}{ccc}x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1\end{array}\right|=0$
Sol :
L.H.S
$\left|\begin{array}{ccc}x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1\end{array}\right|$
R1→R1+R2
$=\left|\begin{array}{ccc}x+y+z & x+y+z & x+y+z \\ z & x & y \\ 1 & 1\end{array}\right|$
$=(x+y+z)\left|\begin{array}{ccc}1 & 1 & 1 \\ z & x & y \\ 1 & 1 & 1\end{array}\right|$
=(x+y+z)×0
<content to be added>
(vi) $\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$
Sol :
L.H.S
$\left|\begin{array}{ccc}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|$
C1→C1+C2
$=\left|\begin{array}{ccc}x+a & a & x+a \\ y+b & b & y+b \\ z+c & c & z+c\end{array}\right|=0$
(vii) $\left|\begin{array}{ccc}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$
Sol :
L.H.S
$\left|\begin{array}{ccc}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|$
C1→C1+9C2
$=\left|\begin{array}{ccc}65 & 7 & 65 \\ 75 & 8 & 75 \\ 86 & 9 & 86\end{array}\right|=0$
<to be added>
Question 28
[Show without expanding at any stage that :]
(i) $\left|\begin{array}{lll}a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b & b+c\end{array}\right|=2\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$
Sol :
L.H.S
$\left|\begin{array}{lll}a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b & b+c\end{array}\right|$
R1→R1+R2+R3
$=\left|\begin{array}{ccc}2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\ b+c & c+a & a+b \\ c+a & a+b & b+c\end{array}\right|$
taking out 2 from R
1
$=2 \quad\left|\begin{array}{ccc}a+b+c & a+b+c & a+b+c \\ b+c & c+a & a+b \\ c+a & a+b & b+c\end{array}\right|$
R
1→R
1-R
2
$=2 \quad\left|\begin{array}{ccc}a & b & c \\ b+c & c+a & a+b \\ c+a & a+b & b+c\end{array}\right|$
R
3→R
3-R
1
$=2 \quad\left|\begin{array}{ccc}a & b & c \\ b+c & c+a & a+b \\ c & a & b\end{array}\right|$
R
2→R
2-R
3
$=2 \quad\left|\begin{array}{ccc}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$
(ii) $\begin{vmatrix}0&\sin \alpha &-\cos \alpha \\-\sin \alpha&0& \sin \beta\\ \cos \alpha& -\sin \beta &0\end{vmatrix}$
Sol :
L.H.S
$\left|\begin{array}{ccc}0 & \sin \alpha & -\cos \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\sin \beta & 0\end{array}\right|$
$\left|\begin{array}{ccc}0 & \sin \alpha & -\cos \alpha \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\sin \beta & 0\end{array}\right|=\left|\begin{array}{ccc}0 & -\sin \alpha & \cos \alpha\\\sin \alpha&0& -\sin \beta \\ -\cos \alpha & \sin \beta & 0\end{array}\right|$
$\left|\begin{array}{ccc}0 & \sin \alpha & -\cos \alpha \\ -\operatorname{sin} \alpha & 0 & \sin \beta \\ \cos 2 & -\sin \beta & 0\end{array}\right|=-\left|\begin{array}{ccc}0 & \sin \alpha & -\cos \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\cos \beta & 0\end{array}\right|$
$2\left|\begin{array}{cccc}0 & \sin \alpha & -\cos \alpha \\ -sin \alpha & 0 & sin \beta \\ \cos \alpha & -\sin \beta & 0\end{array}\right|=0$
$\left|\begin{array}{cccc}0 & \sin \alpha & -\cos \alpha \\ -sin \alpha & 0 & sin \beta \\ \cos \alpha & -\sin \beta & 0\end{array}\right|=0$
(iii) $\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|=0$
Sol :
(iv) $\left|\begin{array}{ccc}1 & \cos \alpha-\sin \alpha & \cos \alpha+\sin \alpha \\ 1 & \cos \beta-\sin \beta & \cos \beta+\sin \beta \\ 1 & \cos \gamma-\sin \gamma & \cos \gamma+\sin \gamma\end{array}\right|=2\left|\begin{array}{ccc}1 & \cos \alpha & \sin \alpha \\ 1 & \cos \beta & \sin \beta \\ 1 & \cos \gamma & \sin \gamma\end{array}\right|$
Sol :
L.H.S
$\left|\begin{array}{ccc}1 & \cos \alpha-\sin \alpha & \cos \alpha+\sin \alpha \\ 1 & \cos \beta-\sin \beta & \cos \beta+\sin \beta \\ 1 & \cos \gamma-\sin \gamma & \cos \gamma+\sin \gamma\end{array}\right|$
C
2→C
2+C
3
$=\left|\begin{array}{lll}1 & 2 \cos \alpha & \cos \alpha+\sin \alpha \\ 1 & 2 \cos \beta & \cos \beta+\sin \beta \\ 2 & 2 \cos \gamma & \cos \gamma+\sin \gamma \end{array}\right|$
taking out 2 from C
2
$=2\left|\begin{array}{ccc}1 & \cos \alpha & \cos \alpha+\sin \alpha \\ 1 & \cos \beta & \cos \beta+\sin \beta \\ 1 & \cos \gamma & \cos \gamma+\sin \gamma\end{array}\right|$
C
3→C
3-C
2
$=2\left|\begin{array}{ccc}1 & \operatorname{cos} \alpha & \sin \theta \\ 1 & \cos \beta & \sin \beta \\ 1 & \cos \gamma & \sin \gamma\end{array}\right|$
Question 29
(i) $\left|\begin{array}{ccc}(a-1)^{2} & a^{2}+1 & a \\ (b-1)^{2} & b^{2}+1 & b \\ (c-1)^{2} & c^{2}+1 & c\end{array}\right|=0$
Sol :
L.H.S
$\left|\begin{array}{ccc}(a-1)^{2} & a^{2}+1 & a \\ (b-1)^{2} & b^{2}+1 & b \\ (c-1)^{2} & c^{2}+1 & c\end{array}\right|$
C1→C1-C2+2C3
$=\left|\begin{array}{ccc}0 & a^{2}+1 & a \\ 0 & b^{2}+1 & b \\ 0 & c^{2}+1 & c\end{array}\right|=0$
<to be added>
(ii) $\left|\begin{array}{ccc}a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z\end{array}\right|=0$
Sol :
L.H.S
$\left|\begin{array}{ccc}a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z\end{array}\right|$
R2→R2-R1-2R3
$=\left|\begin{array}{ccc}a & b & c \\ 0 & 0 & 0 \\ x & y & z\end{array}\right|=0$
Question 30
(i) $\begin{vmatrix}1&a&bc\\1&b&ca\\1&c&ab\end{vmatrix}=\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}$
Sol :
L.H.S
$\begin{vmatrix}1&a&bc\\1&b&ca\\1&c&ab\end{vmatrix}$
R
1↔aR
1 , R
2↔bR
2, R
3↔cR
3
=$\dfrac{1}{abc}\begin{vmatrix}a&a^2&abc\\b&b^2&abc\\c&c^2&abc\end{vmatrix}$
taking out abc from C
3
=$\dfrac{1}{abc}\times abc \begin{vmatrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{vmatrix}$
C
1↔C
3
=$-\begin{vmatrix}1&a^2&a\\1&b^2&b\\1&c^2&c\end{vmatrix}$
$=-\quad\left|\begin{array}{ccc}1 & a^{2} & a \\ 1 & b^{2} & b \\ 1 & c^{2} & c\end{array}\right|$
C
2↔C
3
$=\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|$
(ii) $\left|\begin{array}{ccc}a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z\end{array}\right|=0$
Sol :
Question 30
किसी चकण मे बिना विस्तार किए दिखाएँ कि
(Show without expanding at any stage that)
(i) $\left|\begin{array}{lll}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|=\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|$
Sol :
(ii) $\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{ccc}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$
Sol :
L.H.S
$\left|\begin{array}{ccc}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|$
R
1↔aR
1 , R
2↔bR
2, R
3↔cR
3
$=\frac{1}{a b c}\left|\begin{array}{ccc}a^{2} & a^{3} & a b c \\ b^{2} & b^{3} & a b c \\ c^{2} & c^{3} & a b c\end{array}\right|$
taking out abc from C
3
$=\frac{1}{a b c} \times abc\left|\begin{array}{ccc}a^{2} & a^{3} & 1 \\ b^{2} & b^{3} & 1 \\ c^{2} & c^{3} & 1\end{array}\right|$
C
1↔C
3
$=-\left|\begin{array}{ccc}1 & a^{3} & a^{2} \\ 1 & b^{3} & b^{2} \\ 1 & c^{3} & c^{2}\end{array}\right|$
C
2↔C
3
$=\left|\begin{array}{ccc}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$
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