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 C1 , y from C, z from C3

$=\frac{1}{x y z} \times x y z\left|\begin{array}{ccc}(y+z)^{2} & x^{2} & x^{2} \\ y^{2} & (x+z)^{2} & y^{2} \\ z^{2} & z^{2} & (x+y)^{2}\end{array}\right|$

C1→C1-C, C2→C2-C3

$=\left|\begin{array}{ccc}(y+z)^{2}-x^{2} & 0 & x^{2} \\ y^{2}-(x+z)^{2} & (x+z)^{2}-y^{2} & y^{2} \\ 0 & z^{2}-(x+y)^{2} & (x+y)^{2}\end{array}\right|$

$=\left|\begin{array}{ccc}(y+z+x)(y+z-x) & 0&x^2 \\ (y+x+z)(y-x-z) & (x+z+y)(x+z-y) & y^{2} \\ 0 & (z+x+y)(z-x-y) & (x+y)^{2}\end{array}\right|$

taking out (x+y+z) from Cand C2

$=(x+y+z)^{2}\left|\begin{array}{ccc}y+z-x & 0 & x^{2} \\ y-x-z & x+z-y & y^{2} \\ 0 & z-x-y & (x+y)^{2}\end{array}\right|$

R3→R3-(R1+R2)

$=(x+y+z)^{2}\left|\begin{array}{ccc}y+z-x & 0 & x^{2} \\ y-x-z & x+z-y & y^{2} \\ 2 x-2 y & -2 x & 2 x y\end{array}\right|$

taking out 2 from R3

$=2(x+y+z)^{2}\left|\begin{array}{ccc}y+z-x & 0 & x^{2} \\ y-x-z & x+z y & y^{2} \\ x-y & -x & x y\end{array}\right|$

taking out 2 from R3

$=2(x+y+z)^{2}\left|\begin{array}{ccc}y+z-x & 0 & x^{2} \\ y-x-z & x+z- y & y^{2} \\ x-y & -x & x y\end{array}\right|$

R2→R2+R3

$=2(x+y+x)^{2}\left|\begin{array}{ccc}y+z-x & 0 & x^{2} \\ -z & z-y & y^{2}+xy \\ x-y & -x & x y\end{array}\right|$

Expanding along R1

$=2(x+y+z)^{2}\left[(y+z-z)\left|\begin{array}{cc}z-y & y^{2}+x y \\ -x & z y\end{array}\right|+x^{2} \bigg| \begin{array}{cc}-z & z-y \\ x -y & -x\end{array}\bigg| \right]$

=2(x+y+z)2[(y+z-z)(xyz-xy2+xy2+x2y)+x2(zx-zx+xy+yz-y2)]

=2(x+y+z)2[(y+z-z)(xy2z+x2y2+xyz2+x2yz-x2yz-x3y+x3y+x2yz-x2y2)]

=2(x+y+z)2xyz(y+z+x)

=2xyz(x+y+z)3

R1→aR1, R2→bR, R3→cR3

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

taking out abc from C3

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

= 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 R1

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

R1→R1-R2

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

R3→R3-R1

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

R2→R2-R3

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


(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

$\left|\begin{array}{ccc}1 & \cos \alpha-\sin \alpha & \cos \alpha+\sin \alpha \\ 1 & \cos \beta-\sin \beta & \cos \beta+\sin \beta \\ 1 & \cos \gamma-\sin \gamma & \cos \gamma+\sin \gamma\end{array}\right|$

C2→C2+C3

$=\left|\begin{array}{lll}1 & 2 \cos \alpha & \cos \alpha+\sin \alpha \\ 1 & 2 \cos \beta & \cos \beta+\sin \beta \\ 2 & 2 \cos \gamma & \cos \gamma+\sin \gamma \end{array}\right|$

taking out 2 from C2

$=2\left|\begin{array}{ccc}1 & \cos \alpha & \cos \alpha+\sin \alpha \\ 1 & \cos \beta & \cos \beta+\sin \beta \\ 1 & \cos \gamma & \cos \gamma+\sin \gamma\end{array}\right|$

C3→C3-C2

$=2\left|\begin{array}{ccc}1 & \operatorname{cos} \alpha & \sin \theta \\ 1 & \cos \beta & \sin \beta \\ 1 & \cos \gamma & \sin \gamma\end{array}\right|$


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
$\begin{vmatrix}1&a&bc\\1&b&ca\\1&c&ab\end{vmatrix}$

R1↔aR1 , R2↔bR2, R3↔cR3

=$\dfrac{1}{abc}\begin{vmatrix}a&a^2&abc\\b&b^2&abc\\c&c^2&abc\end{vmatrix}$

taking out abc from C3

=$\dfrac{1}{abc}\times abc \begin{vmatrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{vmatrix}$

C1↔C3

=$-\begin{vmatrix}1&a^2&a\\1&b^2&b\\1&c^2&c\end{vmatrix}$

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

C2↔C3

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


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

R1↔aR1 , R2↔bR2, R3↔cR3

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

taking out abc from C3

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

C1↔C3

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

C2↔C3

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



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