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

Step 1: Put everything in homogeneous form.
Bring all terms to one side: \[x-cy-bz=0,\quad -cx+y-az=0,\quad -bx-ay+z=0.\]
Step 2: Write the condition for a non-zero solution.
A homogeneous system has a non-trivial solution only when the coefficient determinant is zero: \[\begin{vmatrix}1&-c&-b\\-c&1&-a\\-b&-a&1\end{vmatrix}=0.\]
Step 3: Expand the determinant.
Expanding along the first row: \[1(1-a^2)+c(-c-ab)-b(ca+b).\]
Step 4: Simplify the terms.
This gives \(1-a^2-c^2-abc-abc-b^2 = 1-a^2-b^2-c^2-2abc\).

Step 5: Set it equal to zero.
\[1-a^2-b^2-c^2-2abc=0\;\Rightarrow\; a^2+b^2+c^2+2abc=1.\]
Step 6: Build \((a+b+c)^2\) from this.
We know \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\). Replace \(a^2+b^2+c^2=1-2abc\): \[(a+b+c)^2=1-2abc+2(ab+bc+ca)=1+2(ab+bc+ca-abc).\] This matches option 4.
\[ \boxed{(a+b+c)^2=1+2(ab+bc+ca-abc)} \]