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Vaishali Harpreet Lal
Asked: 2022-11-10T11:03:59+05:30 In:

Let \(\Delta = \begin{vmatrix} Ax & x^2 & 1 \\\ By & y^2 & 1 \\\ Cz & z^2 & 1 \end{vmatrix}\) and \(\Delta_1 = \begin{vmatrix} A & B & C \\\ x & y & z \\\ zy & zx & xy \end{vmatrix}\), then
1. Δ1 = -Δ 
2. Δ1 ≠ Δ
3. Δ1 – Δ = 0
4. Δ1 = Δ = 0

Let \(\Delta = \begin{vmatrix} Ax & x^2 & 1 \\\ By & y^2 & 1 \\\ Cz & z^2 & 1 \end{vmatrix}\) and \(\Delta_1 = \begin{vmatrix} A & B & C \\\ x & y & z \\\ zy & zx & xy \end{vmatrix}\), then
1. Δ1 = -Δ 
2. Δ1 ≠ Δ
3. Δ1 – Δ = 0
4. Δ1 = Δ = 0
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1 Answer

  1. Correct Answer – Option 3 : Δ1 – Δ = 0

    Explanation:

    \(Δ = \begin{vmatrix} Ax & x^2 & 1 \\\ By & y^2 & 1 \\\ Cz & z^2 & 1 \end{vmatrix}\)  and  \(Δ_1 = \begin{vmatrix} A & B & C \\\ x & y & z \\\ zy & zx & xy \end{vmatrix}\)

    Calculating the determinant of Δ1 :

    \(Δ_1 = \begin{vmatrix} A & B & C \\\ x & y & z \\\ zy & zx & xy \end{vmatrix}\)

    Δ1 = A × (x × y2 – x × z2) – B × (x2 × y – z2 × y) + C × (x2 × z – y2 × z)

         = A × x (y2 – z2) – B × y (x2 – z2) + C × z (x2 – y2)………….(1)

    Calculating the determinant of Δ about first column:

    \(Δ = \begin{vmatrix} Ax & x^2 & 1 \\\ By & y^2 & 1 \\\ Cz & z^2 & 1 \end{vmatrix}\)

    Δ = A × x (y2 – z2) – B × y (x2 – z2) + C × z (x2 – y2)…………(2)

    If we subtract (1) and (2) 

    Δ1 – Δ = 0

    Hence option 3 is the correct answer.

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