Simulation · Maths · Class 12

Determinant as an area

From the lesson Second-order determinants in Determinants. Change the values and watch what happens.

The idea behind it

NCERT §4.2.2

For A = [[a, b], [c, d]], |A| = ad − bc: the product down the leading diagonal minus the product down the other diagonal.
Worked example: det [[2, 4], [−1, 2]] = 4 − (−4) = 8. The minus sign in the rule and the negative entry together turn into a plus.
Entries can be expressions. det [[x, x + 1], [x − 1, x]] = x² − (x² − 1) = 1 for every x.
An equation between determinants is an ordinary equation. det [[3, x], [x, 1]] = det [[3, 2], [4, 1]] means 3 − x² = 3 − 8, so x² = 8 and x = ±2√2.
A picture the text does not draw: the columns (a, c) and (b, d) of [[a, b], [c, d]] span a parallelogram whose area is |ad − bc|. The sign of ad − bc says whether going from the first column to the second is an anticlockwise turn (positive) or a clockwise one (negative).
So det = 0 exactly when the two columns lie along one line and the parallelogram is squashed flat. For [[1, 2], [4, 8]] the second column is twice the first and the determinant is 8 − 8 = 0.

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