Question 54 of 58intermediate🔧 ApplyNumerical4 marks

Evaluate ∣xyx+yyx+yxx+yxy∣\begin{vmatrix} x & y & x + y \\ y & x + y & x \\ x + y & x & y \end{vmatrix}

Correct Answer

-2(x3x^{3} + y3y^{3})

Exercise: Miscellaneous Exercises on Chapter 4 | Q: 5 | (Page 24)
For More Understanding

Explanation

This question involves evaluating a 3×33 \times 3 determinant using expansion along a row or column. The context mentions that determinants of order 3 can be determined by expressing them in terms of second order determinants through expansion along a row or column. We apply row operations to simplify the determinant, then expand to find the final expression.

Solution Steps

  1. Step 1: Let Δ = |x y x+y; y x+y x; x+y x y|. Apply R₁ → R₁ + R₂ + R₃ to get Δ = |2x+2y 2x+2y 2x+2y; y x+y x; x+y x y|

  2. Step 2: Take out common factor (2x+2y) from R₁: Δ = (2x+2y) |1 1 1; y x+y x; x+y x y|

  3. Step 3: Apply C₂ → C₂ - C₁ and C₃ → C₃ - C₁: Δ = (2x+2y) |1 0 0; y x x-y; x+y -y -x|

  4. Step 4: Expand along R₁: Δ = (2x+2y) × 1 × |x x-y; -y -x|

  5. Step 5: Evaluate the 2×22 \times 2 determinant: |x x-y; -y -x| = x(-x) - (x-y)(-y) = -x² + xy - y² = -(x² - xy + y²)

  6. Step 6: Simplify: Δ = (2x+2y)[-(x² - xy + y²)] = -2(x+y)(x² - xy + y²) = -2(x³ + y³)

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