Solving Systems Using
Matrix Inverses

Learn how to write a system as $ AX = B $ and use the inverse matrix $ A^{-1} $ to solve for unknowns efficiently and elegantly.

Algebra 2

What Is a Matrix?

A matrix is a rectangular array of numbers arranged in rows and columns. We use matrices to organize the coefficients of a system of equations into a compact form that enables powerful algebraic operations.

The Matrix Equation $ AX = B $

Any system of linear equations can be written as a single matrix equation. We separate the system into three parts: the coefficient matrix $ A $, the variable matrix $ X $, and the constant matrix $ B $.

What Is the Inverse of a Matrix?

The inverse of a square matrix $ A $, written $ A^{-1} $, is the unique matrix such that:

$$ A \cdot A^{-1} = A^{-1} \cdot A = I $$

Here $ I $ is the identity matrix: the matrix equivalent of the number 1. For a $ 2 \times 2 $ system:

$$ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$

The Determinant: Does the Inverse Exist?

Before finding an inverse, you must check the determinant. For a $ 2 \times 2 $ matrix $ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $:

$$ \det(A) = ad - bc $$

✓ $ \det(A) \neq 0 $

The inverse exists. The system has a unique solution.

✕ $ \det(A) = 0 $

The inverse does not exist. The matrix is singular: no unique solution.

The $ 2 \times 2 $ Inverse Formula

For $ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $ with $ \det(A) \neq 0 $:

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

1

Swap the main diagonal entries: $ a \leftrightarrow d $.

2

Negate the off-diagonal entries: $ b \to -b $ and $ c \to -c $.

3

Multiply the entire matrix by $ \frac{1}{\det(A)} $.

Solving $ AX = B $ with the Inverse

Once you have $ A^{-1} $, the solution to the system is found by multiplying both sides on the left by $ A^{-1} $:

$$ AX = B \implies A^{-1}(AX) = A^{-1}B \implies IX = A^{-1}B \implies X = A^{-1}B $$

Matrix Inverse vs. Other Methods

Feature	Matrix Inverse	Elimination	Cramer's Rule
Setup	Write $ AX = B $	Augmented matrix	Determinants
Best for	Multiple $ B $ vectors	One-time solve	Small systems
Requires	$ \det(A) \neq 0 $	Row reduction	$ \det(A) \neq 0 $
Scalability	Great with tech	Good	Poor for large

How to Apply the Method

1

Write the system in matrix form $ AX = B $. Identify the coefficient matrix $ A $, variable matrix $ X $, and constant matrix $ B $.

2

Compute the determinant $ \det(A) = ad - bc $. If $ \det(A) = 0 $, stop: no unique solution exists.

3

Find the inverse $ A^{-1} $ using the formula (for $ 2 \times 2 $) or technology (for $ 3 \times 3 $ and above).

4

Multiply: $ X = A^{-1}B $. Perform the matrix multiplication carefully.

5

Check your answer by substituting back into the original equations.

Example 1: A $ 2 \times 2 $ System

Solve the system using matrix inverses:

$$ 3x + 4y = 5 $$ $$ 2x + 3y = 4 $$

Example 1: Finding the Solution

Example 2: Fractional Inverse

Solve the system:

$$ x + 2y = 10 $$ $$ 3x + 5y = 26 $$

Example 3: A $ 3 \times 3 $ System (with Technology)

Solve using a calculator or software:

$$ x + y + z = 6,\quad 2x - y + z = 3,\quad x + 2y - z = 2 $$

When the Inverse Does Not Exist

Consider the system $ 2x + 4y = 6 $ and $ x + 2y = 3 $.

$$ A = \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix}, \quad \det(A) = (2)(2)-(4)(1) = 0 $$

Singular Matrix

$ \det(A) = 0 $ means $ A^{-1} $ does not exist. The rows are proportional: row 1 is 2 times row 2.

Geometric Meaning

The two equations represent the same line: infinitely many solutions. (If constants differed, the lines would be parallel: no solution.)

Common Mistakes to Avoid

✕ Multiplying on the Wrong Side

Writing $ X = BA^{-1} $ instead of $ X = A^{-1}B $. Matrix multiplication is not commutative.

✕ Forgetting to Check the Determinant

Attempting to compute $ A^{-1} $ when $ \det(A) = 0 $. Always verify the determinant is nonzero first.

✕ Swapping the Wrong Entries

Negating the main diagonal instead of the off-diagonal. Remember: swap $ a \leftrightarrow d $, negate $ b $ and $ c $.

Summary: The Complete Process

Step	Action	Key Formula
1	Write $ AX = B $	Identify $ A, X, B $
2	Compute determinant	$ ad - bc $
3	Find inverse	$ \frac{1}{\det(A)}\begin{bmatrix}d & -b\\-c & a\end{bmatrix} $
4	Multiply	$ X = A^{-1}B $
5	Verify	Substitute back

Question 1 of 5

What is the determinant of $ A = \begin{bmatrix} 5 & 3 \\ 2 & 4 \end{bmatrix} $?

The determinant is $ ad - bc = (5)(4) - (3)(2) = 20 - 6 = 14 $. Remember: multiply the main diagonal and subtract the product of the off-diagonal.

Question 2 of 5

If $ A = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix} $, what is $ A^{-1} $?

$ \det(A) = (2)(3) - (1)(5) = 1 $. By the formula: swap $ a $ and $ d $ to get 3 and 2, negate $ b $ and $ c $ to get $ -1 $ and $ -5 $. Since $ \det = 1 $, the scalar is just 1. Result: $ \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix} $.

Question 3 of 5

To solve $ AX = B $ using the inverse, you compute:

Multiply both sides on the left by $ A^{-1} $: $ A^{-1}(AX) = A^{-1}B $, so $ IX = A^{-1}B $, giving $ X = A^{-1}B $. Order matters because matrix multiplication is not commutative.

Question 4 of 5

What happens when $ \det(A) = 0 $?

When the determinant is zero, the matrix is singular and has no inverse. The system is either inconsistent (no solution) or dependent (infinitely many solutions). You must use another method to classify it.

Question 5 of 5

Use the inverse to solve: $ 4x + y = 9 $ and $ 3x + y = 7 $. What is $ x $?

$ A = \begin{bmatrix} 4 & 1 \\ 3 & 1 \end{bmatrix} $, $ \det(A) = 4 - 3 = 1 $. So $ A^{-1} = \begin{bmatrix} 1 & -1 \\ -3 & 4 \end{bmatrix} $. Then $ X = A^{-1}B = \begin{bmatrix} 1 & -1 \\ -3 & 4 \end{bmatrix}\begin{bmatrix} 9 \\ 7 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} $. So $ x = 2 $.

Solving Systems Using
Matrix Inverses

What Is a Matrix?

The Matrix Equation \( AX = B \)

What Is the Inverse of a Matrix?

The Determinant: Does the Inverse Exist?

The \( 2 \times 2 \) Inverse Formula

Solving \( AX = B \) with the Inverse

Matrix Inverse vs. Other Methods

How to Apply the Method

Example 1: A \( 2 \times 2 \) System

Example 1: Finding the Solution

Example 2: Fractional Inverse

Example 3: A \( 3 \times 3 \) System (with Technology)

When the Inverse Does Not Exist

Common Mistakes to Avoid

Summary: The Complete Process

Question 1 of 5

Question 2 of 5

Question 3 of 5

Question 4 of 5

Question 5 of 5

Feature	Matrix Inverse	Elimination	Cramer's Rule
Setup	Write \( AX = B \)	Augmented matrix	Determinants
Best for	Multiple \( B \) vectors	One-time solve	Small systems
Requires	\( \det(A) \neq 0 \)	Row reduction	\( \det(A) \neq 0 \)
Scalability	Great with tech	Good	Poor for large

Step	Action	Key Formula
1	Write \( AX = B \)	Identify \( A, X, B \)
2	Compute determinant	\( ad - bc \)
3	Find inverse	\( \frac{1}{\det(A)}\begin{bmatrix}d & -b\\-c & a\end{bmatrix} \)
4	Multiply	\( X = A^{-1}B \)
5	Verify	Substitute back

Solving Systems UsingMatrix Inverses

What Is a Matrix?

The Matrix Equation \( AX = B \)

What Is the Inverse of a Matrix?

The Determinant: Does the Inverse Exist?

The \( 2 \times 2 \) Inverse Formula

Solving \( AX = B \) with the Inverse

Matrix Inverse vs. Other Methods

How to Apply the Method

Example 1: A \( 2 \times 2 \) System

Example 1: Finding the Solution

Example 2: Fractional Inverse

Example 3: A \( 3 \times 3 \) System (with Technology)

When the Inverse Does Not Exist

Common Mistakes to Avoid

Summary: The Complete Process

Question 1 of 5

Question 2 of 5

Question 3 of 5

Question 4 of 5

Question 5 of 5

Solving Systems Using
Matrix Inverses