Problem 57 - Differential Equations and Linear Algebra

Question

Show that \(\left|\begin{array}{lll}x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1\end{array}\right|=0\) represents the equation of the straight line through the distinct points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\)

Step-by-Step Solution

Answer

By evaluating the given determinant, we obtain the equation \[x(y_1 - y_2) - y(x_1 - x_2) + x_1 y_2 - x_2 y_1 = 0\]. Upon plugging the points \((x_1, y_1)\) and \((x_2, y_2)\) into the equation and observing that it is true for both points, we conclude that the derived equation represents the straight line passing through the points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\).

Step 1: Evaluating the determinant

To find the value of the given determinant, we will expand it. The given determinant is: \[ \begin{vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{vmatrix} \] Expanding along the first row: \[ (x)*(y_1 - y_2) - (y)*(x_1 - x_2) + (1)*(x_1 y_2 - x_2 y_1) \] Simplifying the expression, we get: \[ x(y_1 - y_2) - y(x_1 - x_2) + x_1 y_2 - x_2 y_1 = 0 \] Now, we have an equation with the terms \(x, y, x_1, y_1, x_2,\) and \(y_2\). We will now prove that this equation represents a line passing through the points \((x_1, y_1)\) and \((x_2, y_2)\).

Step 2: Showing it represents a line through given points

To show that the equation we derived in step 1 represents a line through \((x_1, y_1)\) and \((x_2, y_2)\), we will plug these points into the equation and check if the equation is true. Plugging point \((x_1, y_1)\): \[ x_1(y_1 - y_2) - y_1(x_1 - x_2) + x_1 y_2 - x_2 y_1 = 0 \] We can see that the first term and the third term cancel each other, and the second term also becomes zero. Thus, the equation is true. Now, plugging point \((x_2, y_2)\): \[ x_2(y_1 - y_2) - y_2(x_1 - x_2) + x_1 y_2 - x_2 y_1 = 0 \] Again, we can observe that the terms cancel each other out, and the equation is true. Since the equation is true for both points \((x_1, y_1)\) and \((x_2, y_2)\), we conclude that \[ x(y_1 - y_2) - y(x_1 - x_2) + x_1 y_2 - x_2 y_1 = 0 \] represents the equation of the straight line through the distinct points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\).

Key Concepts

Equation of a Line

An equation of a line in its most basic form expresses a relationship between the x-coordinate and y-coordinate in a 2-dimensional plane. The line can be represented using the standard form, point-slope form, or the slope-intercept form. For the given problem involving a determinant, we are essentially deriving the equation of a line using a set of points.

Using the determinant approach, we notice that the equation you encounter is derived by a determinant matrix representing points in a 2-dimensional space. How does this represent a line? Let's explore.

• The determinant gives a condition for collinearity, using coordinates of points.
• When expanded, the derived equation in the problem shows a linear equation in terms of x and y.
• The derived expression represents the familiar form of a line: if a point (x, y) satisfies the equation, it lies on the line.

Thus, the determinant condition reveals that any point (x, y) forming this equation lies on the line passing through the points \((x_1, y_1)\) and \((x_2, y_2)\). By plugging these values, the terms cancel out, proving collinearity.

Linear Algebra

Linear algebra is the branch of mathematics concerning vector spaces and linear mappings between such spaces. This includes the study of lines, planes, and subspaces. Determinants are a key component of this, useful for multiple purposes such as solving linear equations and checking the linear independence of vectors.

• Determinants: A determinant is a scalar value that can be computed from the elements of a square matrix.
• Collinearity: Determinants help check if points are collinear, meaning they all lie on a single straight line.
• Application in geometry: As in this exercise, determinants can express geometric problems, like verifying if points form a line.

The exercise shows how linear algebra, particularly the use of determinants, helps express an equation of a line purely through algebraic manipulation, which solidifies the practical, computational utility of algebra concepts.

Matrix Expansion

Matrix expansion, or cofactor expansion, is a method of evaluating determinants of larger matrices. The basic idea is to take a determinant down to smaller determinants, which carry the same essential properties as the original.

For a matrix such as \(\begin{vmatrix} x & y & 1 \ x_1 & y_1 & 1 \ x_2 & y_2 & 1 \end{vmatrix}\), expansion involves focusing on a specific row or column.

• Choice of row/column: Each element in the chosen row/column is multiplied by the determinant of its minor.
• Sign pattern: The alternate signs (+ or -) play a key role during expansion to ensure the correct calculation.
• Minor determinant: This continues the process until a size of 2x2 is reached, directly computable.

In conclusion, matrix expansion allows us to handle complex matrix calculations by reducing them to more manageable ones, demonstrating that every non-zero determinant confirms a unique solution—in this case, a line equation emerging from a determinant composed of point values.