InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Let \(\Delta = \begin{vmatrix} Ax & x^2 & 1 \\\ By & y^2 & 1 \\\ Cz & z^2 & 1 \end{vmatrix}\) and \(\Delta_1 = \begin{vmatrix} A & B & C \\\ x & y & z \\\ zy & zx & xy \end{vmatrix}\), then1. Δ1 = -Δ 2. Δ1 ≠ Δ3. Δ1 - Δ = 04. Δ1 = Δ = 0 |
|
Answer» 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. |
|