निम्नलिखित का मान निकालें (Evaluate the following):
(i) $\begin{vmatrix}
7 &5 \\
-2 & 3
\end{vmatrix}$
Sol :
= 21-(-10)
= 21+10
= 31
(ii) $\begin{vmatrix}
cos \alpha &sin \alpha \\
sin \alpha & cos \alpha
\end{vmatrix}$
Sol :
= cos2α -sin2α
= cos 2α
(iii) $\begin{vmatrix}
tan \alpha & cosec \alpha \\
sin \alpha & cot \alpha
\end{vmatrix}$
Sol :
= tan α.cotα -sinα.cosecα
= 1-1
= 0
(iv) $\begin{vmatrix}
cos \theta & -sin \theta \\
sin \theta & cos \theta
\end{vmatrix}$
Sol :
=cosθ×cosθ-(-sinθ×sinθ)
=cos2θ+sin2θ [∵sin2θ+cos2θ=1]
=1
(v) $\begin{vmatrix}
\dfrac{1}{3} & \dfrac{1}{5} \\
\dfrac{1}{2} & \dfrac{1}{7}
\end{vmatrix}$
Sol :
$=\dfrac{1}{3} \times \dfrac{1}{7} – \dfrac{1}{5} \times \dfrac{1}{2}$
$=\dfrac{1}{21} – \dfrac{1}{10}$
$=\dfrac{10-21}{210}=\dfrac{-11}{210}$
(vi) $\begin{vmatrix}
x^2-x+1 & x-1 \\
x+1 & x+1
\end{vmatrix}$
Sol :
= (x+1)(x2-x+1)-(x+1)(x-1)
= x3+13-(x2-12)
= x3+1-x2+1
=x3-x2+2
(vii) $\begin{vmatrix}
a+ib & c+id \\
-c+id & a-ib
\end{vmatrix}$
Sol :
=$\begin{vmatrix}
a+ib & c+id \\
-(c-id) & a-ib
\end{vmatrix}$
=(a+ib)(a-ib)+(c+id)(c-id)
=a2-(ib)2+c2-(id)2
=a2-i2b2+c2-i2d2
=a2-(-1)b2+c2-(-1)d2
=a2+b2+c2+d2
(viii) $\begin{vmatrix}
\dfrac{1}{2} & 8 \\
4 & 2
\end{vmatrix}$
Sol :
=$\dfrac{1}{2} \times 2 – 8\times 4$
=1-32
=-31
(ix) यदि (If) $A=\begin{bmatrix} 1&2\\4&2 \end{bmatrix}$ , दिखाएँ कि (show that) $|2A|=4|A|$
Sol :
$|2A|=\begin{vmatrix} 2 \begin{bmatrix} 1&2\\4&2 \end{bmatrix}\end{vmatrix}$
=$\begin{vmatrix}2&4 \\ 8&4 \end{vmatrix}$
=$2 \times 2 \begin{vmatrix}1&2 \\ 4&2 \end{vmatrix}$
= 4|A|
(x) यदि (If) $\mathrm{A}=\left[\begin{array}{cc}2 & 4 \\ -5 & -1\end{array}\right]$, दिखाएँ कि (show that) |3A|=9|A|
Question 2
निम्नलिखित सारणिकों में दूसरे स्तम्भ के प्रत्येक अवयव का उपसारणिक तथा सह-खण्ड निकालें साथ ही सारणिक का मान भी निकाले ।
[Write the minors and cofactors of each element of second column in the following determinants and evaluate them]
(i) $\left|\begin{array}{lll}4 & 9 & 7 \\ 3 & 5 & 7 \\ 5 & 4 & 5\end{array}\right|$
Sol :
$\left|\begin{array}{ccc}+-+ \\-+- \\ +-+\end{array}\right|$
$M_{12}=\left|\begin{array}{cc}3 & 7 \\ 5 & 5\end{array}\right|$
=15-35
=-20
$M_{22}=\left|\begin{array}{cc}4 & 7 \\ 5 & 5\end{array}\right|$
=20-35
=-15
$M_{32}=\left|\begin{array}{rr}4 & 7 \\ 3 & 7\end{array}\right|$
=28-21
=7
$A_{12}=-\left|\begin{array}{ll}3 & 7 \\ 5 & 5\end{array}\right|$
=-(15-35)
=20
$A_{22}=\left|\begin{array}{cc}4 & 7 \\ 5 & 5\end{array}\right|$
=20-35
=-15
$A_{32}=-\left|\begin{array}{ll}4 & 7 \\ 3 & 7\end{array}\right|$
=-(28-21)
=-7
$|A|=\left|\begin{array}{ccc}4 & 9 & 7 \\ 3 & 5 & 7 \\ 5 & 4 & 5\end{array}\right|$
=9×20+5×(-15)+4×(-7)
=180-75-28
=180-103
=77
(ii) $\left|\begin{array}{lll}1 & 2 & 4 \\ 1 & 3 & 9 \\ 1 & 4 & 16\end{array}\right|$
Sol :
Question 3
निम्नलिखित सारणिकों में प्रत्येक अवयव का उपसारणिक तथा सह-खण्ड निकालें साथ हो सारणिक का मान भी निकालें ।
[Write the minor and cofactor of each element of the following determinants and also evaluate the determinant in each case.]
(i) $\left|\begin{array}{cc}5 & -10 \\ 0 & 3\end{array}\right|$
Sol :
(ii) $\left|\begin{array}{ccc}1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2\end{array}\right|$
Sol :
$M_{11}=\left|\begin{array}{rr}5 & -1 \\ 1 & 2\end{array}\right|$
=10-(-1)
=11
$M_{12}=\left|\begin{array}{cc}3 & -1 \\ 0 & 2\end{array}\right|$
=6+0
=6
$M_{13}=\left|\begin{array}{ll}3 & 5 \\ 0 & 1\end{array}\right|$
=3-0
=3
$M_{21}=\left|\begin{array}{ll}0 & 4 \\ 1 & 2\end{array}\right|$
=0-4
=-4
$M_{22}=\begin{vmatrix}1&4\\0&2\end{vmatrix}$
=2-0
=2
$M_{23}=\begin{vmatrix}1&0\\0&1\end{vmatrix}$
=1-0
=1
$M_{31}=\begin{vmatrix}0&4\\5&-1\end{vmatrix}$
=0-20
=-20
$M_{32}=\begin{vmatrix}1&4\\3&-1\end{vmatrix}$
=-1-12
=-13
$M_{33}=\begin{vmatrix}1&0\\3&5\end{vmatrix}$
=5-0
=5
A11=11 , A12=-6 , A13=3
A21=(-4)=4 , A22=2 , A23=-1
A31=-20 , A32=-(-13)=13 , A33=5
Δ$=\begin{vmatrix}1&0&4\\3&5&-1\\0&1&2\end{vmatrix}$
=1×11+0×(-6)+4×3
=11+12
=23
(iii) $\left|\begin{array}{ccc}1 & 3 & -2 \\ 4 & -5 & 6 \\ 3 & 5 & 2\end{array}\right|$
Sol :
(iv) $\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$
Sol :
Question 4
निम्नलिखित का मान निकालें (Evaluate the following):
(i) $\begin{vmatrix}1 & 5 & 7\\ 6 & 7 & 2\\ 1 & 2 & 3 \end{vmatrix}$
Sol :
Expanding along C1
=$1\begin{vmatrix}7 & 2 \\2 & 3 \end{vmatrix}-6\begin{vmatrix}5 & 7 \\2 & 3 \end{vmatrix}+1\begin{vmatrix}5 & 7\\7 & 2 \end{vmatrix}$
=1(21-4)-6(15-14)+1(10-49)
=17-6-39
=17-45
=-28
(ii) $\begin{vmatrix}1^2 & 2^2 & 3^2\\2^2 & 3^2 & 4^2\\3^2 & 4^2 & 5^2\end{vmatrix}$
Sol :
(iii) $\begin{vmatrix}a & h & g\\h & b & f\\g & f & c \end{vmatrix}$
Sol :
Expanding along R1
=$a\begin{vmatrix}b & f \\f & c \end{vmatrix}-h\begin{vmatrix}h & f \\g & c \end{vmatrix}+g\begin{vmatrix}h & b\\g & f \end{vmatrix}$
=a(bc-f2)-h(ch-fg)+g(fh-bg)
=abc-af2-ch2+fgh+fgh-bg2
=abc+2fgh-af2-bg2-ch2
(iv) $\begin{vmatrix}
43 & 3 & 6\\
35 & 21 & 4\\
17 & 9 & 2
\end{vmatrix}$
Sol :
(v) $\begin{vmatrix}
9 & 9 & 12\\
1 & 3 & -4\\
1 & 9 & 12
\end{vmatrix}$
Sol :
(vi) $\begin{vmatrix}
42 & 1 & 6\\
28 & 7 & 4\\
14 & 3 & 2
\end{vmatrix}$
Sol :
R1→R2-2R3
$=\left|\begin{array}{ccc}42 & 1 & 6 \\ 0 & 1 & 0 \\ 14 & 3 & 2\end{array}\right|$
R2 के सापेक्ष विस्तार करने पर
$=1\left|\begin{array}{cc}42 & 6 \\ 14 & 2\end{array}\right|$
=84-84
=0
(vii) $\begin{vmatrix}
3 & -4 & 5\\
1 & 1 & -2\\
2 & 3 & 1
\end{vmatrix}$
Sol :
(viii) $\begin{vmatrix}
2 & -1 & -2\\
0 & 2 & -1\\
3 & -5 & 0
\end{vmatrix}$
Sol :
(ix) $\begin{vmatrix}
3 & -1 & -2\\
0 & 0 & -1\\
3 & -5 & 0
\end{vmatrix}$
Sol :
R2 के सापेक्ष विस्तार करने पर
$=-(-1)\left|\begin{array}{rr}3 & -1 \\ 3 & -5\end{array}\right|$
={-15-(-3)}
=-15+3
=-12
(x) $\begin{vmatrix}
0 & 1 & 2\\
-1 & 0 & -3\\
-2 & 3 & 0
\end{vmatrix}$
Sol :
(xi) $\begin{vmatrix}
1 & 2 & 4\\
-1 & 3 & 0\\
4 & 1 & 0
\end{vmatrix}$
Sol :
(xii) यदि (If) A=$\begin{bmatrix}1&1&-2\\2&1&-3\\5&4&-9\end{bmatrix}$ , तो |A| का मान निकालें (Find |A|)
Sol :
|A|=$\begin{vmatrix}1&1&-2\\2&1&-3\\5&4&-9\end{vmatrix}$
C1→C1+C3
=$\begin{vmatrix}-1&1&-2\\-1&1&-3\\-4&4&-9\end{vmatrix}$
=$-\begin{vmatrix}1&1&-2\\1&1&-3\\4&4&-9\end{vmatrix}=0$
<content to be added>
(xiii) यदि (If) A=$\begin{bmatrix}1&0&1\\0&1&2\\0&0&4\end{bmatrix}$ , दिखाएँ कि (show that)
|3A|=27|A|
Sol :
L.H.S
|3A|=$\lvert 3\begin{bmatrix}1&0&1\\0&1&2\\0&0&4\end{bmatrix} \rvert$
=$\begin{vmatrix}3&0&3\\0&3&6\\0&0&12\end{vmatrix}$
=$3\times 3\times 3 \begin{vmatrix}1&0&1\\0&1&2\\0&0&4\end{vmatrix}$
=27|A| = R.H.S
Question 5
निम्नलिखित का मान निकालें (Evaluate the following) :
(i) $\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|$
Sol :
$\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|$
C1→C1-C2 , C2→C2-C3
$=\left|\begin{array}{ccc}0 & 0 & 1 \\ -x & x & 1 \\ 0 & -y & 1+y\end{array}\right|$
Expanding along R1→
$=1\left|\begin{array}{cc}-x & x \\ 0 & -y\end{array}\right|$
=xy-0
=xy
(ii) $\left|\begin{array}{lll}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|$
Sol :
$\left|\begin{array}{ccc}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|$
R1→R1-R2, R2→R2-R3
$=\left|\begin{array}{ccc}0 & a-b & b c-c a \\ 0 & b-c & c a-a b \\ 1 & c & a b\end{array}\right|$
taking out (a-b) from R1 and (b-c) from R2
$=(a-b)(b-c)\left|\begin{array}{ccc}0 & 1 & -c \\ 0 & 1 & -a \\ 1 & c & a b\end{array}\right|$
Expanding along C1
$=(a-b)(b-c) 1\left|\begin{array}{cc}1 & -c \\ 1 & -a\end{array}\right|$
=(a-b)(b-c)(-a+c)
=(a-b)(b-c)(c-a)
(iii) $\left|\begin{array}{ccc}x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda\end{array}\right|$
Sol :
$\left|\begin{array}{cccc}x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda\end{array}\right|$
C1→C1+C2+C3
$=\left|\begin{array}{ccc}3 x+\lambda & 1 & x \\ 3 x+\lambda & x+\lambda & x \\ 3 x+\lambda & x & x+\lambda\end{array}\right|$
taking out 3x+λ from C1
$=\left(\begin{array}{ll}3 x+\lambda\end{array}\right)\left|\begin{array}{ccc}1 & x & x \\ 1 & x+\lambda & x \\ 1 & x & x+\lambda\end{array}\right|$
R1→R1-R2, R2→R2-R1
$=(3 x+\lambda)\left|\begin{array}{ccc}0 & -\lambda & 0 \\ 0 & \lambda & -\lambda \\ 1 & x & x+\lambda\end{array}\right|$
Expanding along C1
$=(3 x+\lambda) \cdot 1\left|\begin{array}{cc}-\lambda & 0 \\ x & -\lambda\end{array}\right|$
=(3x+λ)(λ2-0)
=λ2(3x+λ)
(iv) $\left|\begin{array}{ccc}1 & a & a^{2}-b c \\ 1 & b & b^{2}-a c \\ 1 & c & c^{2}-a b\end{array}\right|$
Sol :
$\left|\begin{array}{ccc}1 & a & a^{2}-b c \\ 1 & b & b^{2}-a c \\ 1 & c & c^{2}-a b\end{array}\right|$
R1→R1-R2, R2→R2-R3
$=\left|\begin{array}{ccc}0 & a-b & a^{2}-b c-b^{2}+a c \\ 0 & b-c & b^{2}-a c-c^{2}+a b \\ 1 & c & c^{2}-a b\end{array}\right|$
taking out (a-b) from R1 and (b-c) from R2
$=(a-b)(b-c)\left|\begin{array}{ccc}0 & 1 & a+b+c \\ 0 & 1 & a+b+c \\ 1 & c & c^{2}-a b\end{array}\right|$
=(a-b)(b-c)×0
=0
(v) $\left|\begin{array}{ccc}1 & 1 & 1 \\ \alpha & \beta & \gamma \\ \beta \gamma & \gamma \alpha & \alpha \beta\end{array}\right|$
Sol :
$\left|\begin{array}{ccc}1 & 1 & 1 \\ \alpha & \beta & \gamma \\ \beta \gamma & \gamma \alpha & \alpha \beta\end{array}\right|$
C1→C1-C2, C2→C2-C3
$=\left|\begin{array}{ccc}0 & 0 & 1 \\ \alpha-\beta & \beta-\gamma & \gamma \\ \beta \gamma-\gamma \alpha & \gamma \alpha-\alpha \beta & \alpha \beta\end{array}\right|$
$=\left| \begin{array}{ccc}0 & 0 & 1 \\ \alpha-\beta & \beta-\gamma & \gamma \\ -\gamma(\alpha-\beta) & -\alpha(\beta-\gamma) & \alpha \beta\end{array}\right|$
taking out (α-ꞵ) from C1 and (ꞵ-𝛾) from C2
$=(\alpha-\beta)(\beta-\gamma)\left|\begin{array}{ccc}0 & 0 & 1 \\ 1 & 1 & \gamma \\ -\gamma & -\alpha & \alpha \beta\end{array}\right|$
Expanding along R1
$=(\alpha-\beta)(\beta-\gamma)1 \begin{vmatrix}1&1\\-\gamma&-\alpha \end{vmatrix}$
=(α-β)(β-𝛾)(-α-𝛾)
=(α-β)(β-𝛾)(𝛾-α)
(vi) $\left|\begin{array}{ccc}a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|$
Sol :
$\left|\begin{array}{ccc}a^{n}+2 a & 2 a-1 &1\\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|$
R1→R1-R2, R2→R2-R3
$=\left|\begin{array}{ccc}a^{2}-1 & a-1 & 0 \\ 2 a-2 & a-1 & 0 \\ 3 & 3 & 1\end{array}\right|$
$=\left|\begin{array}{ccc}(a-1)(a+1) & a-1 & 0 \\ 2(a-1) & a-1 & 0 \\ 3 & 3 & 1\end{array}\right|$
taking out (a-1) from R1 and R2
$=(a-1)^{2}\left|\begin{array}{rrr}a+1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1\end{array}\right|$
Expanding along C3
$=(a-1)^{2} \cdot 1 \quad\left|\begin{array}{cc}a+1 & 1 \\ 2 & 1\end{array}\right|$
=(a-1)2(a+1-2)
=(a-1)2(a-1)
=(a-1)3
(vii) $\left|\begin{array}{ccc}1 & 1 & 1 \\ a^{2} & b^{2} & c^{2} \\ a^{3} & b^{3} & c^{3}\end{array}\right|$
Sol :
$\left|\begin{array}{ccc}1 & 1 & 1 \\ a^{2} & b^{2} & c^{2} \\ a^{3} & b^{3} & c^{3}\end{array}\right|$
C1→C1-C2, C2→C2-C3
$=\left|\begin{array}{ccc}0 & 0 & 1 \\ a^{2}-b^{2} & b^{2}-c^{2} & c^{2} \\ a^{3}-b^{3} & b^{3}-c^{3} & c^{3}\end{array}\right|$
taking out (a-b) from C1 and (b-c) from C2
$=(a-b)(b-c)\left|\begin{array}{ccc}0 & 0 \\ a+b & b+c & c^{2} \\ a^{2}+a b+b^{2} & b^{2}+b c+c^{2} & c^{3}\end{array}\right|$
C1→C1-C3
$=(a-b)(b-c) \quad\left|\begin{array}{ccc}0 & 0 & 1 \\ a-c & b+c & c^{2} \\ a^{2}+a b-b c-c^{2} & b^{2}+b c+c^{2} & c^{3}\end{array}\right|$
$=(a-b)(b-c)\left|\begin{array}{ccc}0 & 0 \\ -(c-a) & b+c & c^{2} \\ -(c-a)(a+b+c) & b^{2}+b c+c^{2} & c^{3}\end{array}\right|$
taking out (c-a) from C1
$=(a-b)(b-c)(c-a)\left|\begin{array}{ccc}0 & 0 & 1 \\ -1 & b+c & c^{2} \\ -(a+1+c) & b^{2}+b c+c^{2} & c^{3}\end{array}\right|$
Expanding along C1
$=(a-b)(b-c)(c-a) \cdot\left|\begin{array}{cc}-1 & b+c \\ -(a+b+c) & b^{2}+bc +c^2\end{array} \right|$
=(a-b)(b-c)(c-a)(-b2-bc-c2+ab+ca+b2+bc+bc+c2)
=(a-b)(b-c)(c-a)(ab+bc+ca)
(viii) $\left|\begin{array}{lll}a & b & c \\ a^{2} & b^{2} & c^{2} \\ a^{3} & b^{3} & c^{3}\end{array}\right|$
Sol :
taking out a from C1 , b from C2 and c from C3
$=a bc\left|\begin{array}{lll}1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2}\end{array}\right|$
C1→C1-C2, C2→C2-C3
(ix) $\left|\begin{array}{lll}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{array}\right|$
Sol :
$\left|\begin{array}{lll}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{array}\right|$
C2→C2+C3
$=\left|\begin{array}{ccc}1 & a+b+c & b+c \\ 1 & a+b+c & c+a \\ 1 & a-1 b+c & a- b\end{array}\right|=0$
taking out a2 from C1, b2 from C2, c2 from C3
$=a^{2} b^{2} c^{2}\left|\begin{array}{ccc}0 & a & a \\ b & 0 & b \\ c & c & 0\end{array}\right|$
taking out a from R1 , b from R2 and c from R3
$=a^{3} b^{3} c^{3}\left|\begin{array}{ccc}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right|$
(x) $\left|\begin{array}{ccc}0 & a b^{2} & a c^{2} \\ a^{2} b & 0 & b c^{2} \\ a^{2} c & c b^{2} & 0\end{array}\right|$
Sol :
(xi) $\left|\begin{array}{ccc}1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y\end{array}\right|$
Sol :
$\left|\begin{array}{ccc}1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y\end{array}\right|$
R1→R1-R2, R2→R2-R3
$=\left|\begin{array}{ccc}0 & -y & 0 \\ 0 & y & -x \\ 1 & x & x+y\end{array}\right|$
Expanding along C1
$=1\left|\begin{array}{cc}-y & 0 \\ y & -x\end{array}\right|$
=xy-0
=xy
Question 6
Evaluate the following :
(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
=$\begin{vmatrix}0&b-c&c-a\\0&c-a&a-b\\0&a-b&b-c\end{vmatrix}$
=0
<content to be added>
(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 ।
[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)
$\begin{vmatrix}x&y&x+y\\y&x+y&x\\x+y&x&y\end{vmatrix}$
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
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:
(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
[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
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
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
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
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)
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-bC3 ,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)
साबित करे कि (prove that)
$\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{array}\right|=1$
Sol :
L.H.S
$\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{array}\right|$
R1→cos𝛼 R1 , R3→sin𝛼 R3
$=\frac{1}{\cos \alpha \sin \alpha}\left|\begin{array}{ccc}\cos ^{2} \alpha \cos \beta & \cos ^{2} \alpha \sin \beta & -\cos \alpha \sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin ^{2} \alpha \cos \beta & \sin ^{2} \alpha \sin \beta & \cos \alpha \sin \alpha\end{array}\right|$
R1→R1+R3
$=\frac{1}{\cos \alpha \sin \alpha}\left|\begin{array}{ccc}\cos \left(\cos ^{2} \alpha+\sin ^{2} \alpha\right) & \sin \beta\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right) & 0 \\ -\sin \beta & \cos \beta & 0 \\ \sin ^{2} \alpha \cos \beta & \sin ^{2} \alpha \operatorname{sin} \beta & \cos \alpha \sin \alpha\end{array}\right|$
$=\frac{1}{\cos \alpha \sin\alpha}\left|\begin{array}{ccc}\cos \beta & \sin \beta & 0 \\ -\sin \beta & \cos \beta & 0 \\ \sin ^{2} \alpha \cos \beta & \sin ^{2} \alpha \sin \beta & \cos \alpha \sin \alpha\end{array}\right|$
Expanding along C3
$=\frac{1}{\cos \alpha \sin \alpha} \times \cos \alpha \sin \alpha \begin{vmatrix}\cos \beta & \sin \beta\\-\sin \beta & \cos \beta\end{vmatrix}$
=cos2β+sin2β
Question 22
साबित करे कि (prove that)
$\left|\begin{array}{ccc}a+b x & c+d x & p+q x \\ a x+b & c x+d & p x+q \\ u & v & w\end{array}\right|=\left(1-x^{2}\right)\left|\begin{array}{lll}a & c & p \\ b & d & q \\ u & v & w\end{array}\right|$
Sol :
L.H.S
$\left|\begin{array}{ccc}a+b x & c+d x & p+q x \\ a x+b & c x+d & p x+q \\ u & v & w\end{array}\right|$
R1→R1-xR2
$=\left|\begin{array}{ccc}a\left(1-x^{2}\right) & c\left(1-x^{2}\right) & p\left(1-x^{2}\right) \\ a x+b & (x+d) & p x+q \\ u & v & w\end{array}\right|$
taking out (1-x2) from R1
$=\left(1-x^{2}\right)\left|\begin{array}{ccc}a & c & p \\ a x+b & c x+d & p x+q \\ u & v & w\end{array}\right|$
R2→R2-xR1
$=\left(1-x^{2}\right)\left|\begin{array}{ccc}a & c & \rho \\ b & d & q \\ u & v & w\end{array}\right|$
Question 23
दिखाएँ कि (Show that 🙂
$\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma+\delta)\end{array}\right|=0$
Sol :
L.H.S
$\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma+\delta)\end{array}\right|$
C3→C3-cosẟC2+sinẟC1
$=\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & 0 \\ \sin \beta & \cos \beta & 0 \\ \sin \gamma & \cos \gamma & 0\end{array}\right|$
=0
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Question 24
(i) यदि (If) $\Delta=\left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin (\alpha+\delta) \\ \sin \beta & \cos \beta & \sin (\beta+\delta) \\ \sin \gamma & \cos \gamma & \sin (\gamma+\delta)\end{array}\right|$ तो साबित करे कि Δ,𝛼,β,ɣ तथा ẟ]
[then prove that Δ is independent of 𝛼,β,ɣ and ẟ]
(ii) साबित करे कि[ Prove that ]
$\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|$ θ से स्वतन्त्र है।
[is independent of θ]
Sol :
$\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right|$
Expanding along R1
$\left.=x\left|\begin{array}{cc}-x & 1 \\ 1 & x\end{array}\right|-\operatorname{sin} \theta\left|\begin{array}{cc}-\sin \theta & 1 \\ \cos \theta & x\end{array}\right| + \cos \theta \left| \begin{array}{cc}-\sin \theta & -x \\ \cos \theta & 1\end{array}\right| \right.$
=x(-x2-1)-sinθ(-x sinθ-cosθ)+cosθ(-sinθ+xcosθ)
=-x3-x+xsin2θ+sinθcosθ-sinθcosθ+xcos2θ
=-x3-x+x(sin2θ+cos2θ)
=-x3-x+x
=-x3 is independent of θ
Question 25
दिखाएँ कि [Show that]
$\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x)$
Sol :
L.H.S
$\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=$
$=\left|\begin{array}{ccc}x & x^{2} & 1 \\ y & y^{2} & 1 \\ 2 & z^{2} & 1\end{array}\right|+\left|\begin{array}{ccc}x & x^{2} & p x^{3} \\ y & y^{2} & p y^{3} \\ z & z^{2} & p z^{3}\end{array}\right|$
C1↔C3
Taking out p from C3
$=-\left|\begin{array}{ccc}1 & x^{2} & x \\ 1 & y^{2} & y \\ 1 & z^{2} & z\end{array}\right|+p\left|\begin{array}{ccc}x & x^{2} & x^{3} \\ y & y^{2} & y^{3} \\ z & z^{2} & z^{3}\end{array}\right|$
C2↔C3
Taking out x from R1 , y from R2 and z from R3
C2↔C3
$\left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y_{2} \\ 1 & z & z^{2}\end{array}\right|+p xy_{2}\left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & 2 & z^{2}\end{array}\right|$
$=(1+p x yz)\left|\begin{array}{ccc}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|$
R1→R1-R2 ,R2→R2-R3
$=(1+p x y z) \left| \begin{array}{ccc}0 & x-y & x^{2}-y^{2} \\ 0 & y-z & y^{2}-z^{2} \\ 1 & z & z^{2}\end{array}\right|$
$=(1+p x y z)\left|\begin{array}{ccc}0 & x-y & (x-y)(x+y) \\ 0 & y-z & (y-z)(y+z) \\ 1 & z & z^{2}\end{array}\right|$
Taking out (x-y) from R1 and (y-z) from R2
$=(1+p x y z)(x-y)(y-z)\left|\begin{array}{ccc}0 & 1 & x+y \\ 0 & 1 & y+z \\1 & z & z^{2}\end{array}\right|$
Expanding along C1 ,
$=(1+p x y z)(x-y](y-z) \cdot 1\left|\begin{array}{cc}1 & x+y \\ 1 & y+z\end{array}\right|$
=(1+p x y z)(x-y)(y-z)(y+z-x-y)
=(1+p x yz)(x-y)(y-z)(z-x) .
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 C2 , 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-C2 , 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 C1 and 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→bR2 , 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
$=\left|\begin{array}{ccc}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$