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Question

Mathematics Question on Applications of Determinants and Matrices

The parameter on which the value of the determinant Δ=1aa2 cos(pd)xcos pxcos(p+d)x sin(pd)xsin pxsin(p+d)x \Rightarrow \Delta=\begin{vmatrix}1& a & a^2\\\ \cos(p-d)x& \cos\ px & \cos(p+d)x\\\ \sin(p-d)x& \sin\ px& \sin(p+d)x\\\ \end{vmatrix} does not depend upon, is

A

a

B

p

C

d

D

x

Answer

p

Explanation

Solution

=1aa2 cos(pd)xcos pxcos(p+d)x sin(pd)sin pxsin(p+d)x=\begin{vmatrix}1 & a & a^2\\\ \cos (p-d)x& \cos\ px & \cos(p+d)x\\\ \sin (p-d)& \sin\ px& \sin(p+d)x\\\\\end{vmatrix}
Applying C1Cx+C3C_1 \rightarrow C x + C_3
=1+a2aa2 cos(pd)x+cos(p+d)xcos pxcos(p+d)x sin(pd)+sin(p+d)xsin pxsin(p+d)x=\begin{vmatrix}1+a^2 & a & a^2\\\ \cos (p-d)x+\cos(p+d)x& \cos\ px & \cos(p+d)x\\\ \sin (p-d)+ \sin(p+d)x& \sin\ px& \sin(p+d)x\\\\\end{vmatrix}
Δ=1+a2aa2\2cos px cos dxcos pxcos(p+d)x\2sin px sin dxsin pxsin(p+d)x\Rightarrow \Delta=\begin{vmatrix}1+a^2 & a & a^2\\\2 \cos\ px\ \cos\ dx& \cos\ px & \cos(p+d)x\\\2 \sin\ px\ \sin\ dx& \sin\ px& \sin(p+d)x\\\\\end{vmatrix}

Applying C1C12cosdxC2C_1 \rightarrow C_1-2 \cos dx C_2
Δ=1+a22a cos dxaa2\0cos pxcos(p+d)x\0sin pxsin(p+d)x\Rightarrow \Delta=\begin{vmatrix}1+a^2-2a\ \cos\ dx & a & a^2\\\0& \cos\ px & \cos(p+d)x\\\0& \sin\ px& \sin(p+d)x\\\\\end{vmatrix}
Δ=(1+a22acosdx)[sin(p+d)xcospxsinpxcos(p+d)x]\Rightarrow \Delta= (1 + a^2 - 2a \cos dx) [\sin (p + d) x \cos px - \sin p x \cos (p + d ) x]
Δ=(1+a22acosdx)sindx\Rightarrow \Delta= (1 + a^2 - 2a \cos dx) \sin d x
which is independent of p.