KC Sinha Mathematics Solution Class 12 Chapter 6 सारणिक (Determinants) Exercise 6.1 (Q16-Q20)

Ex-6.1

---

Exercise 6.1

Question 16

साबित करे कि (prove that)
$\left|\begin{array}{lll}(b+c)^{2} & a^{2} & b c \\ (c+a)^{2} & b^{2} & c a \\ (a+b)^{2} & c^{2} & a b\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$
Sol :
L.H.S
$\left|\begin{array}{ccc}(b+c)^{2} & a^{2} & b c \\ (c+a)^{2} & b^{2} & c a \\ (a+b)^{2} & c^{2} & a b\end{array}\right|$

C₁→C₁-C₂+C₃

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

taking out (a²+b²+c²) from C₁

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

R₁→R₁-R₂ , R₂→R₂-R₃

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

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

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

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

R₁→R₁-R₂

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

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

taking out (c-a) from R₁

Expanding along C₁

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

=(a-b)(b-c)(c-a)(a+b+c)(a²+b²+c²)



Question 17

साबित करे कि (prove that)
$\left|\begin{array}{lll}a^{2} & a^{2}-(b-c)^{2} & b c \\ b^{2} & b^{2}-(c-a)^{2} & c a \\ c^{2} & c^{2}-(a-b)^{2} & a b\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$
Sol :
L.H.S
$\left|\begin{array}{lll}a^{2} & a^{2}-(b-c)^{2} & b c \\ b^{2} & b^{2}-(c-a)^{2} & c a \\ c^{2} & c^{2}-(a-b)^{2} & a b\end{array}\right|$

C₂→C₂-C₁

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

C₁↔C₂

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

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

Question 18

साबित करे कि (prove that)
$\left|\begin{array}{ccc}b^{2}+c^{2} & a^{2} & a^{2} \\ b^{2} & c^{2}+a^{2} & b^{2} \\ c^{2} & c^{2} & a^{2}+b^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$
Sol :
L.H.S

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

R₁→R₁+R₂+R₃

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

taking out 2 from R₁

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

R₂→R₂-R₁, R₃→R₃-R₁

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

R₁→R₁+R₂+R₃

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

Expanding along R₁

$=2\left.\bigg[-c^{2} \right.\bigg| \begin{array}{cc}-c^{2} & -a^{2} \\ -b^{2} & 0\end{array}\left|+b^{2} \bigg| \begin{array}{cc}-c^{2} & 0 \\ -b^{2} & -a^{2}\end{array}\bigg|\right]$

$=2\left[-c^{2}\left(0-a^{2} b^{2}\right)+b^{2}\left(c^{2} a^{2}+0\right)\right]$

=2(a²b²c²+a²b²c²)

=2(2a²b²c²)

=4a²b²c²


Question 19

साबित करे कि (prove that)
$\left|\begin{array}{ccc}a & b & a x+b y \\ b & c & b x+c y \\ a x+b y & b x+c y & 0\end{array}\right|=\left(b^{2}-a c\right)\left(a x^{2}+2 b x y+c y^{2}\right)$
Sol :
L.H.S

$\left|\begin{array}{ccc}a & b & a x+b y \\ b & c & b x+c y \\ a x+b y & b x+c y & 0\end{array}\right|$

C₁→C₃-xC₁-yC₂

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

Expanding along C₃

$=-\left(a x^{2}+2 b xy+c y^{2}\right)\left|\begin{array}{cc}a & b \\ b & c\end{array}\right|$

=-(ax²+2bxy+cy²)(ac-b²)

=(b²-ac)(ax²+2bxy+cy²)


Question 20

साबित करे कि (prove that)
$\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|=\left(1+a^{2}+b^{2}\right)^{3}$
Sol :
L.H.S

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

C₁→C₁-bC_{3 ,}C₂→C₂+aC₃

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

taking out (1+a²+b²) from C₁ and C₂

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

Expanding along C₁

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

$=\left(1+a^{2}+b^{2}\right)^{2}\left[1-a^{2}-b^{2}+2 a^{2}+b(0+2 b)\right]$

=(1+a²+b²)²(1+a²+b²)

=(1+a²+b²)