Matrices & Determinants
Differentiation of Determinants
Grade 12
Question:
<p>If \(\Delta_1 = \begin{vmatrix} x & b & b \\ a & x & b \\ a & a & x \end{vmatrix}\) and \(\Delta_2 = \begin{vmatrix} x & b \\ a & x \end{vmatrix}\) are the given determinants, then</p>
<p>(1) \(\Delta_1 = 3(\Delta_2)^2\)</p>
<p>(2) \(\dfrac{d}{dx}(\Delta_1) = 3\Delta_2\)</p>
<p>(3) \(\dfrac{d}{dx}(\Delta_1) = 3(\Delta_2)^2\)</p>
<p>(4) \(\Delta_1 = 3\Delta_2^{3/2}\)</p>
Step-by-Step Solution
Key Concept: Factor out common terms strategically by performing row/column operations, then recognize that Δ₁ can be expressed as a product involving Δ₂. The key is to subtract rows appropriately to reveal the structure: Δ₁ = (x - a)²(x + 2b - a).
<p><strong>Step 1:</strong> Perform row operations on Δ₁. Subtract R₁ from R₂ and R₁ from R₃:</p><p>R₂ → R₂ - R₁ and R₃ → R₃ - R₁</p><p>$$\Delta_1 = \begin{vmatrix} x & b & b \\ a-x & x-b & 0 \\ a-x & a-b & x-b \end{vmatrix}$$</p><p><strong>Step 2:</strong> Factor out (x - a) from rows 2 and 3:</p><p>$$\Delta_1 = (x-a)\begin{vmatrix} x & b & b \\ 1 & -1 & 0 \\ 1 & \frac{a-b}{x-a} & \frac{x-b}{x-a} \end{vmatrix}$$</p><p><strong>Step 3:</strong> After careful expansion and simplification (subtract C₁ from C₂ and C₃), the determinant reduces to:</p><p>$$\Delta_1 = (x-a)^2(x+2b-a)$$</p><p><strong>Step 4:</strong> Note that Δ₂ = x² - ab, and the relationship shows that Δ₁ contains factors of (x - a), revealing the connection between the two determinants through substitution or the expression Δ₁ = (x - a)²(x + 2b - a).</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B