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3.5. Applications of determinant: curves and surfaces

Everything in this section is based on the following fact:

A homogeneous linear system of n equations in n unknowns has nontrivial solutions if and only if the determinant of the coefficient matrix is zero.

A = 0

1. If det A 0 => A-1 exists <=> A-1 A = A-1 = 0 <=> =
2. If det A = 0, then the rank of A is not full; it is n - r < n and there are n - r unknowns which can be chosen arbitrarily.
=> the system has nontrivial ( 0) solutions.
For example, for a line passing through the two distinct points (x1 , y1 ) and (x2 , y2 ) we have

The equation of the line: c1 x + c2 y + c3 = 0
The equation for the first point: c1 x1 + c2 y1 + c3 = 0
The equation for the second point: c1 x2 + c2 y2 + c3 = 0

The three equations above form a linear system of equations, which we write in the matrix form:
x y 1
x1 y1 1
x2 y2 1
c1
c2
c3
= 0
0
0

From this we obtain:

A. The equation of a line passing through the two distinct points (x1 , y1 ) and (x2 , y2 ) is

xy1 = 0.
x1 y1 1
x2 y2 1

Example 1: The equation of a line

B. The equation of a circle passing through the three distinct points (x1 , y1 ), (x2 , y2 ) and (x3 , y3 ) is

x2 + y2xy1 = 0.
x21 + y21 x1 y1 1
x22 + y22 x2 y2 1
x23 + y23 x3 y3 1

Example 2: The equation of a circle

C. The equation of a conic section(parabola, hyperbola or an ellipse) passing through the five distinct points (xi , yi ),  i = 1, ..., 5 is

x2 xy y2 x y 1 = 0.
x21 x1 y1 y21 x1 y1 1
x22 x2 y2 y22 x2 y2 1
x23 x3 y3 y23 x3 y3 1
x24 x4 y4 y24 x4 y4 1
x25 x5 y5 y25 x5 y5 1

D. The equation of a plane passing through the three distinct points (xi , yi , zi ),  i = 1, 2, 3, which are not collinear, i.e., they do not lie on a same line, is

xyz1 = 0.
x1 y1 z1 1
x2 y2 z2 1
x3 y3 z3 1

E. The equation of a spherical surface passing through the four distinct points (xi , yi , zi ),  i = 1, ..., 4, which are not coplanar, i.e., they do not lie on a same plane, is

x2 + y2 + z2xyz1 = 0.
x21 + y21 + z21 x1 y1 z1 1
x22 + y22 + z22 x2 y2 z2 1
x23 + y23 + z23 x3 y3 z3 1
x24 + y24 + z24 x4 y4 z4 1


Exercises: E31, E32
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