Matrix Determinant

Understanding Matrix Determinants

The determinant of a matrix is a scalar value (a single number) that can be calculated from the elements of a square matrix. The concept of a determinant is very important in linear algebra, one of its uses being to help solve systems of linear equations. Every square matrix has a unique determinant value.

Calculating the Determinant of a 2x2 Order Matrix

A 2x2 order matrix is a matrix that has two rows and two columns. Suppose we have matrix A as follows:

The determinant of matrix A, usually written as $\det(A)$ or $|A|$, is calculated by subtracting the product of the elements of the main diagonal from the product of the elements of the second diagonal.

The formula is:

Note that the notation $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ uses straight lines, which denote the determinant, as opposed to square brackets $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ which denote the matrix itself.

Calculation of a 2x2 Matrix Determinant

Suppose we have matrix B:

To calculate its determinant, we identify $a = -1$, $b = 3$, $c = -7$, and $d = -5$.

Then, the determinant of matrix B is:

So, the determinant value of matrix B is 26.

Solving Systems of Linear Equations with Two Variables (SPLDV) using Determinants

One important application of determinants is to solve systems of linear equations. This method is often called Cramer's Rule.

Consider the following system of linear equations with two variables (SPLDV):

In this system, $x$ and $y$ are the variables whose values we want to find. The coefficients $a_{11}, a_{12}, a_{21}, a_{22}$ and constants $b_1, b_2$ are known numbers.

This system of equations can be converted into matrix multiplication form:

The first step is to calculate the determinant of the coefficient matrix, which we call $D$:

A system of linear equations will have a unique solution if and only if $D \neq 0$.

Next, we calculate two other determinants:

1. $D_x$, which is the determinant of the coefficient matrix where the first column (coefficients of $x$) is replaced by the constant column ($b_1, b_2$ ):

   $$D_x = \begin{vmatrix} b_1 & a_{12} \\ b_2 & a_{22} \end{vmatrix} = b_1a_{22} - a_{12}b_2$$

2. $D_y$, which is the determinant of the coefficient matrix where the second column (coefficients of $y$) is replaced by the constant column ($b_1, b_2$ ):

   $$D_y = \begin{vmatrix} a_{11} & b_1 \\ a_{21} & b_2 \end{vmatrix} = a_{11}b_2 - b_1a_{21}$$

After obtaining the values of $D$, $D_x$, and $D_y$, we can find the values of $x$ and $y$ using the formulas:

These formulas are only valid if $D \neq 0$.

Solving SPLDV with Determinants

Determine the solution of the following system of linear equations:

From the system above, we get:

$a_{11} = 2$, $a_{12} = -1$, $b_1 = 7$

$a_{21} = 1$, $a_{22} = -4$, $b_2 = 14$

Step 1: Calculate the determinant $D$.

Since $D = -7 \neq 0$, this system has a unique solution.

Step 2: Calculate the determinant $D_x$.

Step 3: Calculate the determinant $D_y$.

Step 4: Calculate the values of $x$ and $y$.

So, the solution set of the system of linear equations is $x=2$ and $y=-3$, or can be written as the ordered pair $(2, -3)$.

Exercises

1. Given matrix $M = \begin{bmatrix} 9 & x \\ 8 & -7 \end{bmatrix}$ and $\det(M) = 9$. Determine the value of $x$.

2. Determine the solution of the following system of linear equations:

   $$\begin{cases} 2x - y = 8 \\ x + 3y = -10 \end{cases}$$

Answer Key

1. For matrix $M = \begin{bmatrix} 9 & x \\ 8 & -7 \end{bmatrix}$, its determinant is:

   $$\det(M) = (9 \times -7) - (x \times 8) = -63 - 8x$$

   Given $\det(M) = 9$, then:

   $$-63 - 8x = 9$$

   $$-8x = 9 + 63$$

   $$-8x = 72$$

   $$x = \frac{72}{-8}$$

   $$x = -9$$

   So, the value of $x$ is -9.

2. System of linear equations:

   $$2x - y = 8$$

   $$x + 3y = -10$$

   We determine $D, D_x,$ and $D_y$.

   $$D = \begin{vmatrix} 2 & -1 \\ 1 & 3 \end{vmatrix} = (2 \times 3) - (-1 \times 1) = 6 - (-1) = 6 + 1 = 7$$

   $$D_x = \begin{vmatrix} 8 & -1 \\ -10 & 3 \end{vmatrix} = (8 \times 3) - (-1 \times -10) = 24 - 10 = 14$$

   $$D_y = \begin{vmatrix} 2 & 8 \\ 1 & -10 \end{vmatrix} = (2 \times -10) - (8 \times 1) = -20 - 8 = -28$$

   Then, the values of $x$ and $y$ are:

   $$x = \frac{D_x}{D} = \frac{14}{7} = 2$$

   $$y = \frac{D_y}{D} = \frac{-28}{7} = -4$$

   So, the solution of the system of linear equations is $x=2$ and $y=-4$, or the ordered pair $(2, -4)$.