Application of Determinants to Coordinate Geometry

Determinants and matrices are useful tools for solving systems of linear equations in two or three variables. This topic explains how a system of equations can be written in matrix form and solved by using the inverse of a matrix when it exists. It also helps in deciding whether a system is consistent or inconsistent.

• In coordinate geometry, the area of a triangle can be calculated if the coordinates of its vertices are known.

• Determinants provide a compact and systematic method to calculate this area using a 3 × 3 matrix formed from the coordinates of the vertices.

Consistent Solution: A system is consistent if it has at least one solution.

Inconsistent Solution: A system is inconsistent if it has no solution.

The same area can be expressed in determinant form using a 3 × 3 determinant:

Area of △ABC = \[ \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\]

• Area is always non-negative:

  The value of the determinant may be positive or negative, but the area is taken as its absolute value.

• Sign and $\pm$ in exam problems:

  When the area is given (for example, Area = 12 sq. units), the determinant value can be +24 or -24 because the formula uses \[\frac{1}{2}|\det|\].

• Condition for collinearity:

  If three points are collinear, the area of the triangle formed by them is zero, so

This is frequently used to check whether three given points lie on the same straight line.

Solve the system of equations

Solution: The system of equations can be written in the form

Now,

Note that

Therefore

i.e.

Hence

• Area of triangle using determinant:

• Expanded coordinate form:

  \[\text{Area} = \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\].

• Area is always taken as positive; use absolute value.

• For collinear points, determinant = 0, so area = 0.

The sum of three numbers is 6. If we multiply the third number by 3 and add the second number to it, we get 11. By adding the first and third numbers, we get twice the second number. Represent it algebraically and find the numbers using the matrix method.

Solution: Let the first, second and third numbers be denoted by\[ x, y\] and\[ z \], respectively.

Then, according to the given conditions, we have

This system can be written as

Here

Hence \[ adj \text{ A} = \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix} \]