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Mom's Basement University

Solving Systems Using Matrices

What You’ll Learn

Section titled “What You’ll Learn”

In this lesson you’ll learn how to represent a system of linear equations as a matrix equation and solve it using the inverse of a 2×2 matrix.

The Concept

Section titled “The Concept”

A system of two linear equations can be written in matrix form as AX = B:

[a1b1a2b2][xy]=[c1c2]\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}

Where A is the coefficient matrix, X is the variable column, and B is the constant column.

If A has an inverse, the solution is:

X=A−1BX = A^{-1}B

For a 2×2 matrix, the inverse exists when the determinant ad − bc ≠ 0:

A−1=1ad−bc[d−b−ca]A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

We will explore determinants and matrix inverses in more depth in the future Linear Algebra section. Here we focus on using them to solve small systems.

Worked Example

Section titled “Worked Example”

Solve using matrices:

{3x+4y=102x+5y=11\begin{cases} 3x + 4y = 10 \\ 2x + 5y = 11 \end{cases}

1. Write in matrix form

[3425][xy]=[1011]\begin{bmatrix} 3 & 4 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 11 \end{bmatrix}

2. Find the determinant

det⁡(A)=(3)(5)−(4)(2)=15−8=7≠0\det(A) = (3)(5) - (4)(2) = 15 - 8 = 7 \neq 0

The determinant is not zero, so the inverse exists.

3. Find the inverse of A

A−1=17[5−4−23]A^{-1} = \frac{1}{7} \begin{bmatrix} 5 & -4 \\ -2 & 3 \end{bmatrix}

4. Multiply A⁻¹ by B to find X

To multiply a 2×2 matrix by a 2×1 column, take each row of the matrix and compute the dot product with the column. The dot product means: multiply corresponding entries and add them up.

Row 1 of A⁻¹ is [5, −4]. Dot product with [10, 11]: (5)(10) + (−4)(11) = 50 − 44 = 6.

Row 2 of A⁻¹ is [−2, 3]. Dot product with [10, 11]: (−2)(10) + (3)(11) = −20 + 33 = 13.

Then divide each result by 7 (the determinant):

[xy]=17[5−4−23][1011]=17[613]=[6/713/7]\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 5 & -4 \\ -2 & 3 \end{bmatrix} \begin{bmatrix} 10 \\ 11 \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 6 \\ 13 \end{bmatrix} = \begin{bmatrix} 6/7 \\ 13/7 \end{bmatrix}

Solution: x = 6/7, y = 13/7.

Matrix multiplication, determinants, and inverses are topics we’re just touching on here in Algebra 2 to give you a taste of how powerful matrices are. We’ll revisit all of these concepts in much greater depth in the Linear Algebra section, where you’ll work with larger matrices, learn more multiplication techniques, and see how matrices connect to transformations and higher-dimensional systems.

Real-World Application

Section titled “Real-World Application”

Using matrices to solve systems is efficient when dealing with:

  • Multiple budget categories and constraints
  • Mixing ingredients with several components
  • Balancing electrical circuits with multiple loops
  • Production planning with several products and resources

Example: a nutritionist balancing nutrients across food types can set up the system as a matrix equation and solve it efficiently.

Quiz

Section titled “Quiz”
The matrix equation AX = B represents:
To solve $AX = B$ for $X$ when $A$ is invertible, you compute:
A 2×2 matrix has an inverse only if its determinant is:
The main advantage of using matrices to solve systems is:
Find the determinant of $\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$.