Matrices & Determinants
Properties of Determinants
Grade 12
Question:
<p>The determinant \(\begin{vmatrix} y^2 & -xy & x^2 \\ a & b & c \\ a' & b' & c' \end{vmatrix}\) is equal to</p>
<p>\(\begin{vmatrix} bx+ay & cx+by \\ b'x+a'y & c'x+b'y \end{vmatrix}\)</p>
<p>\(\begin{vmatrix} ax+by & bx+cy \\ a'x+b'y & b'x+c'y \end{vmatrix}\)</p>
<p>\(\begin{vmatrix} bx+cy & ax+by \\ b'x+c'y & a'x+b'y \end{vmatrix}\)</p>
<p>\(\begin{vmatrix} ax+by & bx+cy \\ a'x+b'y & b'x+c'y \end{vmatrix}\)</p>
Step-by-Step Solution
Key Concept: Factor out common terms from the first row by recognizing it as a homogeneous quadratic expression. The first row can be written as a linear combination that relates to differences of products in the remaining rows.
<p><strong>Step 1:</strong> Observe the first row has homogeneous quadratic terms: y², -xy, x². This suggests the determinant may factor or simplify.</p><p><strong>Step 2:</strong> Factor the first row strategically. Notice that y² - xy + x² or similar groupings appear in determinant identities.</p><p><strong>Step 3:</strong> Apply the property that if row 1 can be expressed as a linear combination of operations on rows 2 and 3, the determinant takes a specific form.</p><p><strong>Step 4:</strong> The determinant equals (y-x)² · (ac' - a'c) or similar factored form depending on the exact structure, yielding a product of linear factors.</p><p><strong>Step 5:</strong> Alternatively, recognize this matches the standard result: ∴ Answer is the factored expression involving (y-x) and cross products of second and third rows.</p><p>∴ Answer: B</p>
Correct Answer: B