Matrices & Determinants
System of Linear Equations
Grade 12
Question:
<p>Let \(a, b, c\) be any real numbers. Suppose that there are real numbers \(x, y, z\) not all zero such that \(x = cy + bz\), \(y = az + cx\) and \(z = bx + ay\). Then \(a^2 + b^2 + c^2 + 2abc\) is equal to</p>
<p>\(2\)</p>
<p>\(-1\)</p>
<p>\(0\)</p>
<p>\(1\)</p>
Step-by-Step Solution
Key Concept: The given system has non-trivial solutions iff the coefficient matrix is singular (determinant = 0). Rewrite the system in matrix form and set det = 0 to find the constraint on a, b, c.
<p><strong>Step 1:</strong> Rewrite the system in standard form:</p><p>$x - cy - bz = 0$</p><p>$-cx + y - az = 0$</p><p>$-bx - ay + z = 0$</p><p><strong>Step 2:</strong> For non-trivial solutions ($x, y, z$ not all zero), the coefficient matrix must be singular:</p><p>$$\det\begin{bmatrix}1 & -c & -b\\-c & 1 & -a\\-b & -a & 1\end{bmatrix} = 0$$</p><p><strong>Step 3:</strong> Expand the determinant along the first row:</p><p>$1(1 - a^2) + c(-c + ab) - b(ac + b) = 0$</p><p>$1 - a^2 - c^2 + abc - abc - b^2 = 0$</p><p>$1 - a^2 - b^2 - c^2 = 0$</p><p><strong>Step 4:</strong> Therefore: $a^2 + b^2 + c^2 = 1$</p><p>Alternatively, with proper determinant expansion:</p><p>$1(1 - a^2) - (-c)(-c - ab) - b(ac + b) = 0$</p><p>$1 - a^2 - c^2 - abc - abc - b^2 = 0$</p><p>This gives: $a^2 + b^2 + c^2 + 2abc = 1$</p><p>∴ Answer: $\boxed{1}$</p>
Correct Answer: D