Evaluation of Determinants and Area of Triangles using Determinants

1. Introduction to Determinants

A determinant is a scalar value that can be computed from the elements of a square matrix. It is a fundamental concept in linear algebra with wide-ranging applications in solving systems of linear equations, finding eigenvalues, and calculating areas and volumes. Determinants are only defined for square matrices (matrices with the same number of rows and columns).

2. Determinants of Order 2

For a 2x2 matrix, say $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is calculated as the product of the elements on the main diagonal minus the product of the elements on the off-diagonal.

The determinant of matrix A, denoted as $|A|$ or $\det(A)$, is given by: $|A| = ad - bc$

Example: If $A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$, then $|A| = (2 \times 5) - (3 \times 4) = 10 - 12 = -2$.

3. Determinants of Order 3

For a 3x3 matrix, say $A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$, the determinant can be calculated using the cofactor expansion method. We can expand along any row or any column. The most common method is to expand along the first row.

$|A| = a \begin{vmatrix} e & f \\ h & i \end{vmatrix} - b \begin{vmatrix} d & f \\ g & i \end{vmatrix} + c \begin{vmatrix} d & e \\ g & h \end{vmatrix}$

Here, $\begin{vmatrix} e & f \\ h & i \end{vmatrix}$, $\begin{vmatrix} d & f \\ g & i \end{vmatrix}$, and $\begin{vmatrix} d & e \\ g & h \end{vmatrix}$ are determinants of 2x2 matrices, which we can evaluate using the formula from Section 2.

So, $|A| = a(ei - fh) - b(di - fg) + c(dh - eg)$

$|A| = aei + bfg + cdh - ceg - bdi - afh$

3.1. Sarrus' Rule for 3x3 Determinants (Shortcut)

Sarrus' rule provides a visual shortcut for calculating 3x3 determinants. 1. Write down the 3x3 matrix. 2. Rewrite the first two columns to the right of the matrix. 3. Sum the products of the elements along the three downward diagonals. 4. Sum the products of the elements along the three upward diagonals. 5. Subtract the sum of the upward diagonal products from the sum of the downward diagonal products.

For matrix $A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$:

$\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}$ $\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix} \begin{matrix} a & b \\ d & e \\ g & h \end{matrix}$

Sum of downward diagonals: $(a \times e \times i) + (b \times f \times g) + (c \times d \times h)$

Sum of upward diagonals: $(c \times e \times g) + (a \times f \times h) + (b \times d \times i)$

Determinant = (Sum of downward diagonals) - (Sum of upward diagonals)

This yields the same formula: $aei + bfg + cdh - ceg - afh - bdi$.

Example: If $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$, using Sarrus' rule:

Downward diagonals: $(1 \times 5 \times 9) + (2 \times 6 \times 7) + (3 \times 4 \times 8) = 45 + 84 + 96 = 225$ Upward diagonals: $(3 \times 5 \times 7) + (1 \times 6 \times 8) + (2 \times 4 \times 9) = 105 + 48 + 72 = 225$ Determinant = $225 - 225 = 0$.

4. Properties of Determinants

Understanding the properties of determinants can significantly simplify calculations and is crucial for solving problems efficiently.

• Property 1: The determinant of a matrix is equal to the determinant of its transpose. $|A| = |A^T|$.
• Property 2: If any two rows (or columns) of a determinant are interchanged, the sign of the determinant changes.
• Property 3: If any two rows (or columns) of a determinant are identical, the value of the determinant is zero.
• Property 4: If a row (or column) of a determinant consists entirely of zeros, the value of the determinant is zero.
• Property 5: If a row (or column) of a determinant is multiplied by a scalar k, the determinant is multiplied by k. This implies that a common factor from any row or column can be taken out.
• Property 6: If a row (or column) of a determinant is the sum of two or more rows (or columns), the determinant can be expressed as the sum of determinants. For example, if the first row is $(a_1+a_2, b_1+b_2, c_1+c_2)$, then the determinant is the sum of two determinants with $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ as the first row, respectively.
• Property 7: If a multiple of one row (or column) is added to another row (or column), the value of the determinant does not change. This is the most useful property for simplifying determinants to a form where evaluation is easy (e.g., upper or lower triangular form).
• Property 8: The determinant of a triangular matrix (upper or lower) is the product of the elements on its main diagonal. For $A = \begin{bmatrix} a & b & c \\ 0 & e & f \\ 0 & 0 & i \end{bmatrix}$, $|A| = aei$. Similarly for a lower triangular matrix.
• Property 9: The determinant of the product of two square matrices is the product of their determinants. $|AB| = |A||B|$.
• Property 10: If a matrix has linearly dependent rows or columns, its determinant is zero.

4.1. Simplification using Properties

Let's evaluate the determinant $D = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$ using properties. We noticed earlier this determinant is 0. Let's see how properties help. Apply $R_2 \rightarrow R_2 - 4R_1$ and $R_3 \rightarrow R_3 - 7R_1$. This operation does not change the determinant's value. $D = \begin{vmatrix} 1 & 2 & 3 \\ 4 - 4(1) & 5 - 4(2) & 6 - 4(3) \\ 7 - 7(1) & 8 - 7(2) & 9 - 7(3) \end{vmatrix}$ $D = \begin{vmatrix} 1 & 2 & 3 \\ 0 & 5 - 8 & 6 - 12 \\ 0 & 8 - 14 & 9 - 21 \end{vmatrix}$ $D = \begin{vmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -12 \end{vmatrix}$ Now, we can see that the third row is exactly twice the second row ($R_3 = 2R_2$). This means the rows are linearly dependent. Alternatively, apply $R_3 \rightarrow R_3 - 2R_2$: $D = \begin{vmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 - 2(0) & -6 - 2(-3) & -12 - 2(-6) \end{vmatrix}$ $D = \begin{vmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 + 6 & -12 + 12 \end{vmatrix}$ $D = \begin{vmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & 0 & 0 \end{vmatrix}$ According to Property 4, if a row consists entirely of zeros, the determinant is zero.

5. Area of a Triangle using Determinants

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ can be calculated using determinants. The formula is derived from the concept of vectors and cross products, but it can be expressed directly using coordinates.

Area $= \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$

The absolute value is taken because the area must be a non-negative quantity. The determinant itself can be positive or negative depending on the order of the vertices (clockwise or counterclockwise).

Let's expand the determinant: $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = x_1(y_2 \cdot 1 - y_3 \cdot 1) - y_1(x_2 \cdot 1 - x_3 \cdot 1) + 1(x_2 y_3 - x_3 y_2)$ $= x_1(y_2 - y_3) - y_1(x_2 - x_3) + x_2 y_3 - x_3 y_2$ $= x_1 y_2 - x_1 y_3 - x_2 y_1 + x_3 y_1 + x_2 y_3 - x_3 y_2$

So, Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Example: Find the area of a triangle with vertices A(1, 2), B(3, 4), and C(5, 6).

Area $= \frac{1}{2} \left| \begin{vmatrix} 1 & 2 & 1 \\ 3 & 4 & 1 \\ 5 & 6 & 1 \end{vmatrix} \right|$

Evaluate the determinant: $1(4 \cdot 1 - 6 \cdot 1) - 2(3 \cdot 1 - 5 \cdot 1) + 1(3 \cdot 6 - 5 \cdot 4)$ $= 1(4 - 6) - 2(3 - 5) + 1(18 - 20)$ $= 1(-2) - 2(-2) + 1(-2)$ $= -2 + 4 - 2$ $= 0$

Area $= \frac{1}{2} |0| = 0$.

Interpretation: An area of zero means that the three points are collinear (lie on the same straight line). In this case, the points (1,2), (3,4), and (5,6) lie on the line $y = x+1$.

5.1. Condition for Collinearity using Determinants

Three points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ are collinear if and only if the area of the triangle formed by them is zero.

Condition for collinearity: $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0$

This is a very common application of determinants in coordinate geometry problems.

5.2. Area of a Polygon using Determinants

The area of a polygon with vertices $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$ taken in order (either clockwise or counterclockwise) can be calculated using the shoelace formula, which is related to determinants.

Area $= \frac{1}{2} |(x_1 y_2 + x_2 y_3 + \dots + x_n y_1) - (y_1 x_2 + y_2 x_3 + \dots + y_n x_1)|$

This can be visualized by writing the coordinates in two columns and drawing "shoelaces" to connect them for multiplication.

Example: Area of a quadrilateral with vertices (1,1), (3,2), (4,5), (2,4).

Write coordinates: 1 1 3 2 4 5 2 4 1 1 (repeat the first point)

Downward products: $(1 \times 2) + (3 \times 5) + (4 \times 4) + (2 \times 1) = 2 + 15 + 16 + 2 = 35$

Upward products: $(1 \times 3) + (2 \times 4) + (5 \times 2) + (4 \times 1) = 3 + 8 + 10 + 4 = 25$

Area $= \frac{1}{2} |35 - 25| = \frac{1}{2} |10| = 5$.

6. Determinants in Solving Systems of Linear Equations (Brief Mention)

Determinants are fundamental to Cramer's Rule, a method for solving systems of linear equations. For a system of $n$ linear equations in $n$ variables, if the determinant of the coefficient matrix is non-zero, a unique solution exists, and Cramer's Rule provides a formula for each variable using determinants.

For a system: $a_1 x + b_1 y + c_1 z = d_1$ $a_2 x + b_2 y + c_2 z = d_2$ $a_3 x + b_3 y + c_3 z = d_3$

Let $D = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$. Let $D_x = \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{vmatrix}$ (replace x-coefficients with constants). Let $D_y = \begin{vmatrix} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{vmatrix}$ (replace y-coefficients with constants). Let $D_z = \begin{vmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{vmatrix}$ (replace z-coefficients with constants).

If $D \neq 0$, then $x = \frac{D_x}{D}$, $y = \frac{D_y}{D}$, $z = \frac{D_z}{D}$.

If $D=0$, the system either has no solution or infinitely many solutions. This is a crucial link between determinants and the consistency of linear systems.