Matrices & Determinants
Properties of Determinants
Grade 12
Question:
<p>Which of the following is/are true for \(\Delta = \begin{vmatrix} a^2 & 1 & a+c \\ 0 & b^2+1 & b+c \\ 0 & b+c & c^2+1 \end{vmatrix}\)?</p>
<p>\(\Delta \geq 0\) for real values of \(a, b, c\).</p>
<p>\(\Delta \leq 0\) for real values of \(a, b, c\).</p>
<p>\(\Delta = \begin{vmatrix} bc-1 & 0 & 0 \\ 1 & ac & -a \\ -b & -a & ab \end{vmatrix}\)</p>
<p>\(\Delta = 0\) if \(bc = 1\), where \(a, b, c\) are non-zero.</p>
Step-by-Step Solution
Key Concept: Expand along the first column to reduce the 3×3 determinant to a 2×2 determinant, then recognize the resulting expression as a perfect square or factorable form.
<p><strong>Step 1:</strong> Expand along the first column (which has two zeros in positions (2,1) and (3,1)):</p><p>Δ = a² · |b²+1 b+c| - 0 + 0</p><p> |b+c c²+1|</p><p><strong>Step 2:</strong> Calculate the 2×2 determinant:</p><p>= a² [(b²+1)(c²+1) - (b+c)²]</p><p><strong>Step 3:</strong> Expand and simplify the inner expression:</p><p>(b²+1)(c²+1) - (b+c)² = b²c² + b² + c² + 1 - b² - 2bc - c²</p><p>= b²c² + 1 - 2bc = (bc - 1)² + (b-c)²</p><p><strong>Step 4:</strong> Therefore:</p><p>Δ = a²[(bc-1)² + (b-c)²]</p><p>This shows Δ ≥ 0 for all real a, b, c (always non-negative).</p><p>Δ = 0 if and only if a = 0, b = c, and bc = 1 (simultaneous conditions required).</p><p>∴ Answer: A,D</p>
Correct Answer: A,D