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Solving \(A\mathbf{x} = \mathbf{b}\)
via Inverses

Learn how to use the matrix inverse to solve systems of linear equations: find \(\mathbf{x} = A^{-1}\mathbf{b}\) directly and understand when this method works.

Linear Algebra

What Is \(A\mathbf{x} = \mathbf{b}\)?

A system of linear equations can be written in one compact matrix equation. The matrix \(A\) holds the coefficients, \(\mathbf{x}\) holds the unknowns, and \(\mathbf{b}\) holds the constants on the right-hand side.

\(a_{11}\)
\(a_{12}\)
\(a_{21}\)
\(a_{22}\)
 ⋮
\(A\)
\(x_1\)
\(x_2\)
 ⋮
\(\mathbf{x}\)
 =
\(b_1\)
\(b_2\)
 ⋮
\(\mathbf{b}\)

What Is a Matrix Inverse?

For a square matrix \(A\), its inverse \(A^{-1}\) is the unique matrix satisfying:

\(A\)
The original square matrix (must be invertible)
\(A^{-1}\)
The inverse: "undoes" multiplication by \(A\)
\(I\)
Identity matrix: diagonal of 1s, zeros elsewhere

Why Use Inverses to Solve Systems?

Think of it like solving \(5x = 15\) in ordinary algebra. You divide both sides by 5, which is the same as multiplying by the reciprocal \(\frac{1}{5}\). The matrix inverse is the matrix version of a reciprocal.

Scalar Analogy
\(5x = 15\)
\(x = \frac{1}{5} \cdot 15 = 3\)
 Matrix Version
\(A\mathbf{x} = \mathbf{b}\)
\(\mathbf{x} = A^{-1}\mathbf{b}\)

The Inverse Solution Theorem

1
Start: \(A\mathbf{x} = \mathbf{b}\)
2
Left-multiply both sides by \(A^{-1}\): \(A^{-1}(A\mathbf{x}) = A^{-1}\mathbf{b}\)
3
Simplify the left side: \((A^{-1}A)\mathbf{x} = I\mathbf{x} = \mathbf{x}\)
4
Result: \(\mathbf{x} = A^{-1}\mathbf{b}\)

When Does \(A^{-1}\) Exist?

A square matrix \(A\) is invertible (also called nonsingular) if and only if its determinant is nonzero:

\(\det(A) \neq 0\)
Invertible: exactly one solution to \(A\mathbf{x}=\mathbf{b}\) for every \(\mathbf{b}\)
\(\det(A) = 0\)
Singular: either no solution or infinitely many solutions

Invertible vs Singular

Invertible (\(\det \neq 0\)) Singular (\(\det = 0\))
\(A^{-1}\) exists? Yes No
Solutions to \(A\mathbf{x}=\mathbf{b}\) Exactly one None or infinitely many
Columns of \(A\) Linearly independent Linearly dependent
Rank Full rank (\(n\)) Less than \(n\)
Geometric view Maps to all of \(\mathbb{R}^n\) Collapses a dimension

The \(2 \times 2\) Inverse Formula

For a \(2 \times 2\) matrix, there is a direct formula:

1
Swap the main diagonal entries (\(a\) and \(d\))
2
Negate the off-diagonal entries (\(b\) and \(c\))
3
Divide every entry by the determinant \(ad - bc\)

How to Apply the Method

1
Write the system as \(A\mathbf{x} = \mathbf{b}\): identify the coefficient matrix \(A\) and the constant vector \(\mathbf{b}\).
2
Check invertibility: compute \(\det(A)\). If \(\det(A) = 0\), this method does not apply.
3
Find \(A^{-1}\): use the \(2\times2\) formula, cofactor expansion, or row reduction \([A \mid I] \to [I \mid A^{-1}]\).
4
Multiply: compute \(\mathbf{x} = A^{-1}\mathbf{b}\) via matrix-vector multiplication.
5
Verify: substitute \(\mathbf{x}\) back into the original equation to confirm \(A\mathbf{x} = \mathbf{b}\).

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

Solve the system: \(2x + 3y = 8\), \(x + 2y = 5\).

Example 1: Solution

Example 2: Larger Coefficients

Solve: \(4x - y = 7\), \(3x + 2y = 17\).

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

Solve: \(x + y + z = 6\), \(2y + 5z = -4\), \(2x + 5y - z = 27\).

Geometric Interpretation

\(x\)
\(y\)
\(2x + 3y = 8\)
\(x + 2y = 5\)
\((1, 2)\)
 1 2 3 4 1 2 3 4

Common Mistakes

✕ Multiplying on the Wrong Side
Writing \(\mathbf{x} = \mathbf{b}A^{-1}\) instead of \(\mathbf{x} = A^{-1}\mathbf{b}\). Matrix multiplication is not commutative: order matters.
✕ Forgetting to Check \(\det(A)\)
Attempting to find \(A^{-1}\) when \(\det(A) = 0\). The inverse does not exist for singular matrices.
✕ Sign Errors in the Formula
In the \(2\times2\) formula, forgetting to negate the off-diagonal entries or mis-swapping elements.

Summary: Methods for Solving \(A\mathbf{x} = \mathbf{b}\)

Method Best For Limitation
\(A^{-1}\mathbf{b}\) Small systems, multiple \(\mathbf{b}\) vectors Only if \(A\) is invertible
Row Reduction Any system, any size Labor-intensive for large matrices
Cramer's Rule Small systems with determinants Very slow for \(n > 3\)
LU Decomposition Large systems, repeated solves Requires factorization step

Question 1 of 5

If \(A\) is an invertible matrix, the solution to \(A\mathbf{x} = \mathbf{b}\) is:

Question 2 of 5

Given \(A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\), what is \(\det(A)\)?

Question 3 of 5

If \(A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} 4 \\ 7 \end{pmatrix}\), what is \(\mathbf{x} = A^{-1}\mathbf{b}\)?

Question 4 of 5

A matrix \(A\) has \(\det(A) = 0\). Which statement is true?

Question 5 of 5

What is the main advantage of computing \(A^{-1}\) once, rather than using row reduction each time?