Matrix Determinant

What a Determinant Tells Us

The determinant assigns one scalar value to each square matrix. That value tells us, among other things, whether the matrix is invertible and whether a linear system has a unique solution.

Calculating the Determinant of a Two by Two Matrix

A $2 \times 2$ 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 the product along the main diagonal minus the product along the other diagonal.

The formula is:

The vertical bars in the formula denote the determinant $\det(A)$. The square brackets in the definition of $A$ enclose the matrix entries.

Worked Determinant of a Two by Two Matrix

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 Two Linear Equations Using Determinants

Determinants can be used to solve systems of linear equations through Cramer's rule.

Consider the following system of two linear equations:

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 $D \neq 0$.

If $D=0$, the fractions in Cramer's Rule are undefined. That does not by itself distinguish between no solution and infinitely many solutions, so use elimination or row reduction to classify the system.

Solving a Linear System 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

Each problem below uses the determinant for a different purpose. The first asks for an unknown entry inside a matrix with a given determinant, the second asks for the solution of a system of linear equations.

1. The following matrix satisfies $\det(M) = 9$. Determine $x$.

   $$M = \begin{bmatrix} 9 & x \\ 8 & -7 \end{bmatrix}$$

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

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

Worked Solutions

1. The matrix in the first exercise is:

   $$M = \begin{bmatrix} 9 & x \\ 8 & -7 \end{bmatrix}$$

   Its determinant is:

   $$\begin{aligned} \det(M) &= (9 \times -7) - (x \times 8) \\ &= -63 - 8x \end{aligned}$$

   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 $D_y$.

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

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

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

   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 $(2, -4)$.