Question 14 of 58intermediate🔧 ApplyNumerical2 marks

Using Cofactors of elements of third column, evaluate Δ=∣1xyz1yzx1zxy∣\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}.

Correct Answer

(x−y)(y−z)(z−x)(x-y)(y-z)(z-x)

Exercise: EXERCISE 4.3 | Q: 4 | (Page 12)
For More Understanding

Explanation

The question requires evaluating the determinant by expanding along the third column. The provided context establishes the rule that the value of a determinant is the sum of the products of elements of any column (or row) with their corresponding cofactors. We calculate the minors and cofactors for the elements of the third column and then substitute them into the expansion formula to arrive at the final simplified algebraic expression.

Solution Steps

  1. Step 1: Identify the elements of the third column. The elements of the third column (C3C_3) are a13=yza_{13} = yz, a23=zxa_{23} = zx, and a33=xya_{33} = xy.

  2. Step 2: Calculate the cofactors for the third column elements. Cofactor A13=(−1)1+3∣1y1z∣=1(z−y)=z−yA_{13} = (-1)^{1+3} \begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix} = 1(z - y) = z - y. Cofactor A23=(−1)2+3∣1x1z∣=−1(z−x)=x−zA_{23} = (-1)^{2+3} \begin{vmatrix} 1 & x \\ 1 & z \end{vmatrix} = -1(z - x) = x - z. Cofactor A33=(−1)3+3∣1x1y∣=1(y−x)=y−xA_{33} = (-1)^{3+3} \begin{vmatrix} 1 & x \\ 1 & y \end{vmatrix} = 1(y - x) = y - x.

  3. Step 3: Expand the determinant along the third column. Using the expansion rule Δ=a13A13+a23A23+a33A33\Delta = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}: Δ=yz(z−y)+zx(x−z)+xy(y−x)\Delta = yz(z-y) + zx(x-z) + xy(y-x).

  4. Step 4: Simplify the expression. Δ=yz2−y2z+zx2−z2x+xy2−x2y\Delta = yz^2 - y^2z + zx^2 - z^2x + xy^2 - x^2y. Rearranging and factoring, we get Δ=(x−y)(y−z)(z−x)\Delta = (x-y)(y-z)(z-x).

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