Mathematics · Ch 4 — Determinants

Applications of Determinants and Matrices

4.6

Applications of Determinants and Matrices

4.6 Applications of Determinants and Matrices

Determinants and matrices are powerful tools for solving systems of linear equations. In this section, we focus on systems with two or three variables, using these tools to determine whether a solution exists and, if so, to find it uniquely.

Consistent and Inconsistent Systems

A system of equations is a collection of two or more equations involving the same set of variables. For such a system, we ask: does a solution exist?

Important

Consistent system: A system of equations is said to be consistent if it has at least one solution (one or more solutions exist).

Inconsistent system: A system of equations is said to be inconsistent if it has no solution.

In this chapter, we restrict our study to systems of linear equations that have unique solutions only. This means we will only consider consistent systems where exactly one set of values satisfies all equations simultaneously.

Note

The term "consistent" does not guarantee a unique solution — a system could have infinitely many solutions and still be consistent. However, our focus here is on systems with exactly one solution.

Representing a System of Linear Equations Using Matrices

Consider a system of $n$ linear equations in $n$ unknowns. For $n = 2$ :

\begin{aligned} a_1 x + b_1 y &= c_1 \\ a_2 x + b_2 y &= c_2 \end{aligned}

For $n = 3$ :

\begin{aligned} 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 \end{aligned}

We can write any such system in matrix form as:

A X = B

where:

$A$ is the coefficient matrix (a square matrix of order $n$ )
$X$ is the variable matrix (a column matrix of order $n \times 1$ )
$B$ is the constant matrix (a column matrix of order $n \times 1$ )

For a $3 \times 3$ system:

A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}

The matrix equation $A X = B$ is equivalent to the original system of equations.

Solving a System Using the Inverse of a Matrix

If $A$ is a square matrix and $|A| \neq 0$ , then $A$ is invertible (its inverse exists). We can solve $A X = B$ by multiplying both sides on the left by $A^{-1}$ :

A^{-1} (A X) = A^{-1} B

Since matrix multiplication is associative and $A^{-1} A = I$ (the identity matrix):

(A^{-1} A) X = I X = X = A^{-1} B

X = A^{-1} B

This is the matrix method for solving a system of linear equations. The solution exists and is unique if and only if $|A| \neq 0$ .

Watch out

A common mistake is to write $X = B A^{-1}$ . Remember that matrix multiplication is not commutative — the order matters. Since $A^{-1}$ multiplies $A$ on the left, it must also multiply $B$ on the left.

Checking Consistency Using Determinants

The determinant of the coefficient matrix $A$ tells us about the nature of the system:

If $|A| \neq 0$ , the system is consistent and has a unique solution.
If $|A| = 0$ , the system may be consistent (with infinitely many solutions) or inconsistent (no solution). Further investigation is needed.

Tip

For the systems we study (unique solutions only), we only need to check that $|A| \neq 0$ . If $|A| = 0$ , the system either has no solution or infinitely many — both cases are outside our scope here.

Solving a $2 \times 2$ System Using the Matrix Method

Example: Solve the system

\begin{aligned} 2x + 3y &= 7 \\ 3x - y &= 5 \end{aligned}

Step 1: Write in matrix form $A X = B$ .

A = \begin{bmatrix} 2 & 3 \\ 3 & -1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad B = \begin{bmatrix} 7 \\ 5 \end{bmatrix}

Step 2: Find $|A|$ .

|A| = (2)(-1) - (3)(3) = -2 - 9 = -11 \neq 0

Since $|A| \neq 0$ , the system is consistent with a unique solution.

Step 3: Find $A^{-1}$ .

For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , the inverse is:

A^{-1} = \frac{1}{|A|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

Here:

A^{-1} = \frac{1}{-11} \begin{bmatrix} -1 & -3 \\ -3 & 2 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 1 & 3 \\ 3 & -2 \end{bmatrix}

Step 4: Compute $X = A^{-1} B$ .

X = \frac{1}{11} \begin{bmatrix} 1 & 3 \\ 3 & -2 \end{bmatrix} \begin{bmatrix} 7 \\ 5 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 1(7) + 3(5) \\ 3(7) + (-2)(5) \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 7 + 15 \\ 21 - 10 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 22 \\ 11 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}

Thus, $x = 2$ , $y = 1$ .

Important

Always verify your solution by substituting back into the original equations. For this example: $2(2) + 3(1) = 4 + 3 = 7$ ✓ and $3(2) - 1 = 6 - 1 = 5$ ✓.

Solving a $3 \times 3$ System Using the Matrix Method

Example: Solve the system

\begin{aligned} x + 2y + z &= 7 \\ 2x - y + z &= 4 \\ 3x + y - 2z &= 1 \end{aligned}

Step 1: Write in matrix form.

A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & -1 & 1 \\ 3 & 1 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 7 \\ 4 \\ 1 \end{bmatrix}

Step 2: Find $|A|$ .

Expanding along the first row:

\begin{aligned} |A| &= 1 \cdot \begin{vmatrix} -1 & 1 \\ 1 & -2 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 1 \\ 3 & -2 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} \\ &= 1[(-1)(-2) - (1)(1)] - 2[(2)(-2) - (1)(3)] + 1[(2)(1) - (-1)(3)] \\ &= 1[2 - 1] - 2[-4 - 3] + 1[2 + 3] \\ &= 1(1) - 2(-7) + 1(5) \\ &= 1 + 14 + 5 = 20 \neq 0 \end{aligned}

Since $|A| \neq 0$ , the system has a unique solution.

Step 3: Find $A^{-1}$ using the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$ .

First, find the cofactors: …

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Applications of Determinants and Matrices