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Question

# Given, x=cy+bz,y=az+cx,z=bx+ay, where x, y, z are not all zero, and a, b, c are real numbers. The value of a2+b2+c2+2abc is

A
\N
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B
4
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C
2
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D
1
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Solution

## The correct option is D 1Given systems of equations can be rewritten as −x+cy+by=0,cx−y+az=0 and bx+ay−z=0 Above system of equations are homogeneous equation. Since, x, y and z are not all zero, so it has non - trivial solution. Therefore, the coefficient of determinant must be zero. ∴∣∣ ∣∣−1cbc−1aba−1∣∣ ∣∣=0 ⇒−1(1−a2)−c(−c−ab)+b(ca+b)=0 ⇒a2+b2+c2+2abc−1=0 ⇒a2+b2+c2+2abc=1

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