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Question

Prove that Δ=∣ ∣axbyczay+bxcx+azay+bxbyczaxbz+cycx+azbz+cyczaxby∣ ∣
=(x2+y2+z2)(a2+b2+c2)(ax+by+cz).

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Solution

Operations are suggested by the form of factors required.
Multiply C1 by x and hence divide by x and then apply
C1+(yC2+zC3), we get
Δ=1x⎢ ⎢a(x2+y2+z2)ay+bxcx+azb(x2+y2+z2)byczaxbz+cyc(x2+y2+z2)bz+cyczaxby⎥ ⎥
Take x2 from C1 and then again multiply R1 by a and divide Δ by a Now operate
R1+(bR2+cR3)
Δ=1ax(x2+y2+z2)
⎢ ⎢ ⎢a2+b2+c2y(a2+b2+c2)z(a2+b2+c2)bbyczaxbz+cycbz+cyczaxby⎥ ⎥ ⎥
=1ax(x2+y2+z2)(a2+b2+c2)
×1yzbbyczaxbz+cycbz+cyczaxby
Now make two zeros by applying R2bR1 and R3CR1
Δ=1yz0(czax)cy0bz(axby)
=(cz+ax)(ax+by)bcyz
=ax(ax+by+cz) Δ=x2.a2.ax

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