Question

Let \(a,b\) and \(c\) be real numbers. Suppose there exist real numbers \(x,y,z\) which are not all zero such that the system of equations \(x = cy + bz\), \(y = cx + az\) and \(z = bx + ay\) has a non-zero solution then \(\left(a+b+c\right)^2 =\)

• A. \(a^2 + b^2 + c^2 + 2abc - 1\)
• B. \(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - 1\)
• C. \(1 + 2(ab + bc + ca + abc)\)
• D. \(1 + 2(ab + bc + ca - abc)\)

Hint:

A homogeneous system (everything equal to zero form) has a non-zero solution only when the determinant of its coefficient matrix is zero. We write the system in that form, set the determinant to zero, and simplify.

Answer 1

The Correct Option is D

Concept:
A homogeneous system (everything equal to zero form) has a non-zero solution only when the determinant of its coefficient matrix is zero. We write the system in that form, set the determinant to zero, and simplify.

Step 1:
Rewrite each equation with all terms on one side: \[x - cy - bz = 0,\quad -cx + y - az = 0,\quad -bx - ay + z = 0.\]

Step 2:
For a non-zero solution the determinant must vanish: \[\begin{vmatrix} 1 & -c & -b \\ -c & 1 & -a \\ -b & -a & 1 \end{vmatrix} = 0.\]

Step 3:
Expand the determinant. \[1(1 - a^2) - (-c)(-c - ab) + (-b)(ca + b).\] This gives \(1 - a^2 - c(c+ab) - b(ca+b)\) which equals \(1 - a^2 - b^2 - c^2 - 2abc\). Setting it to zero: \[1 - a^2 - b^2 - c^2 - 2abc = 0,\] so \(a^2 + b^2 + c^2 + 2abc = 1\).

Step 4:
Now expand the asked quantity: \[(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca).\] Replace \(a^2+b^2+c^2\) using the result above, namely \(a^2+b^2+c^2 = 1 - 2abc\): \[(a+b+c)^2 = 1 - 2abc + 2(ab+bc+ca) = 1 + 2(ab+bc+ca - abc).\]

Answer: Option (4) — \(1 + 2(ab + bc + ca - abc)\).