Question 49 of 58intermediate🔧 ApplyNumerical4 marks

Evaluate ∣1xy1x+yy1xx+y∣\begin{vmatrix} 1 & x & y \\ 1 & x + y & y \\ 1 & x & x + y \end{vmatrix}

Correct Answer

xy

Exercise: Miscellaneous Exercises on Chapter 4 | Q: 6 | (Chapter: 24)
For More Understanding

Explanation

To evaluate the determinant, we expand along the first row (R₁) using the method demonstrated in the context. This involves multiplying each element of the row by its corresponding cofactor and summing the results. We then simplify the resulting 2x2 determinants and algebraic expressions to arrive at the final value.

Solution Steps

  1. Step 1: Expand along Row 1. We express the determinant Δ\Delta by expanding along the first row: Δ=1∣x+yyxx+y∣−x∣1y1x+y∣+y∣1x+y1x∣\Delta = 1 \begin{vmatrix} x+y & y \\ x & x+y \end{vmatrix} - x \begin{vmatrix} 1 & y \\ 1 & x+y \end{vmatrix} + y \begin{vmatrix} 1 & x+y \\ 1 & x \end{vmatrix}

  2. Step 2: Evaluate the 2x2 determinants. We calculate the value of each minor determinant using the formula ∣abcd∣=ad−bc\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc: ∣x+yyxx+y∣=(x+y)(x+y)−(x)(y)=x2+2xy+y2−xy=x2+xy+y2\begin{vmatrix} x+y & y \\ x & x+y \end{vmatrix} = (x+y)(x+y) - (x)(y) = x^2 + 2xy + y^2 - xy = x^2 + xy + y^2 ∣1y1x+y∣=1(x+y)−1(y)=x+y−y=x\begin{vmatrix} 1 & y \\ 1 & x+y \end{vmatrix} = 1(x+y) - 1(y) = x + y - y = x ∣1x+y1x∣=1(x)−1(x+y)=x−x−y=−y\begin{vmatrix} 1 & x+y \\ 1 & x \end{vmatrix} = 1(x) - 1(x+y) = x - x - y = -y

  3. Step 3: Substitute and simplify. Substitute these values back into the expansion equation: Δ=1(x2+xy+y2)−x(x)+y(−y)\Delta = 1(x^2 + xy + y^2) - x(x) + y(-y) Δ=x2+xy+y2−x2−y2\Delta = x^2 + xy + y^2 - x^2 - y^2 Δ=xy\Delta = xy

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