KC Sinha Mathematics Solution Class 12 Chapter 6 सारणिक (Determinants) Exercise 6.1 (Q21-Q25)
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
\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
\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
<content to be added>
Question 24
\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
<content to be added>
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
[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) .
\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) .
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