01 / 20

Invertible
Matrices

Discover when a square matrix has an inverse, how to find it, and why invertibility determines whether a system \( A\mathbf{x} = \mathbf{b} \) has a unique solution.

Linear Algebra

What Is a Matrix Inverse?

A square matrix \( A \) is invertible (also called nonsingular) if there exists a matrix \( A^{-1} \) such that:

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

Here \( I \) is the identity matrix: the matrix that acts like the number \( 1 \) in multiplication. Multiplying any matrix by \( I \) leaves it unchanged.

The Inverse Undoes the Transformation

\( \mathbf{x} \) \( A \) \( A\mathbf{x} \) \( A^{-1} \) \( \mathbf{x} \) \( A^{-1}(A\mathbf{x}) = \mathbf{x} \)

The Identity Matrix \( I \)

The identity matrix \( I_n \) is the \( n \times n \) matrix with 1s on the diagonal and 0s everywhere else. It is the multiplicative identity for matrices:

$$ I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad I_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$
\( AI = A \)
Multiplying on the right by \( I \) returns the original matrix.
\( IA = A \)
Multiplying on the left by \( I \) also returns the original matrix.

Why Does Invertibility Matter?

Consider the linear system \( A\mathbf{x} = \mathbf{b} \). If \( A \) is invertible, we can solve directly:

$$ \mathbf{x} = A^{-1}\mathbf{b} $$
1
Existence: A solution always exists for every \( \mathbf{b} \).
2
Uniqueness: That solution is the only one: no free variables.
3
Stability: Small changes in \( \mathbf{b} \) produce small changes in \( \mathbf{x} \).

The Invertible Matrix Theorem

For an \( n \times n \) matrix \( A \), the following statements are all equivalent: each one implies all the others.

1
\( A \) is invertible.
2
\( \det(A) \neq 0 \).
3
The columns of \( A \) are linearly independent.
4
\( A\mathbf{x} = \mathbf{b} \) has a unique solution for every \( \mathbf{b} \in \mathbb{R}^n \).
5
The RREF of \( A \) is the identity matrix \( I_n \).

The Determinant Test

For a \( 2 \times 2 \) matrix, the determinant gives a quick yes/no answer:

$$ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \quad \det(A) = ad - bc $$
\( \det(A) \neq 0 \)
\( A \) is invertible. The system \( A\mathbf{x} = \mathbf{b} \) has a unique solution.
\( \det(A) = 0 \)
\( A \) is singular. The system has no solution or infinitely many.

Invertible vs Singular

Invertible Area > 0 \( \det(A) \neq 0 \) Singular Area = 0 \( \det(A) = 0 \)

The \( 2 \times 2 \) Inverse Formula

When \( \det(A) \neq 0 \), the inverse of a \( 2 \times 2 \) matrix has a closed-form formula:

$$ A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} $$
1
Swap the diagonal entries \( a \) and \( d \).
2
Negate the off-diagonal entries \( b \) and \( c \).
3
Divide every entry by \( \det(A) = ad - bc \).

How to Check Invertibility

1
Compute \( \det(A) \): If \( \det(A) = 0 \), stop: \( A \) is not invertible.
2
Row-reduce \( [A \mid I] \): Augment \( A \) with the identity and apply Gauss-Jordan elimination.
3
Read off the inverse: If the left side becomes \( I \), the right side is \( A^{-1} \).
4
Verify: Check that \( A A^{-1} = I \) (or at least spot-check a column).

Example 1: Finding \( A^{-1} \)

Find the inverse of \( A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \).

1
Determinant: \( \det(A) = (3)(2) - (1)(5) = 6 - 5 = \) \( 1 \). Since \( \det(A) \neq 0 \), \( A \) is invertible.
2
Apply formula: Swap \( 3, 2 \); negate \( 1, 5 \).
3
Result: \( A^{-1} = \frac{1}{1}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} \).

Example 2: A Singular Matrix

Is \( B = \begin{pmatrix} 2 & 4 \\ 1 & 2 \end{pmatrix} \) invertible?

1
Determinant: \( \det(B) = (2)(2) - (4)(1) = 4 - 4 = \) \( 0 \).
2
Conclusion: \( B \) is not invertible (singular).

Example 3: Solving \( A\mathbf{x} = \mathbf{b} \)

Solve the system using the inverse from Example 1:

$$ \begin{cases} 3x + y = 7 \\ 5x + 2y = 11 \end{cases} \implies A\mathbf{x} = \mathbf{b} $$
1
Identify: \( A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \), \( \mathbf{b} = \begin{pmatrix} 7 \\ 11 \end{pmatrix} \).
2
Multiply: \( \mathbf{x} = A^{-1}\mathbf{b} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}\begin{pmatrix} 7 \\ 11 \end{pmatrix} \).
3
Compute: \( \mathbf{x} = \begin{pmatrix} 14 - 11 \\ -35 + 33 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} \). So \( x = 3, \; y = -2 \).

Key Properties of Inverses

\( (A^{-1})^{-1} = A \)
Inverting twice returns the original matrix.
\( (AB)^{-1} = B^{-1}A^{-1} \)
The order reverses when inverting a product.
\( (A^T)^{-1} = (A^{-1})^T \)
Transpose and inverse commute.

Common Mistakes to Avoid

✕ Wrong product order
Writing \( (AB)^{-1} = A^{-1}B^{-1} \) instead of \( B^{-1}A^{-1} \). Matrix multiplication is not commutative.
✕ Forgetting the determinant check
Applying the inverse formula when \( \det(A) = 0 \). Always check the determinant first.
✕ Confusing \( A^{-1} \) with \( \frac{1}{A} \)
\( A^{-1} \) is not \( \frac{1}{A} \). Division by a matrix is not defined: you multiply by the inverse.

Summary Comparison

Property Invertible (\( \det \neq 0 \)) Singular (\( \det = 0 \))
Solutions to \( A\mathbf{x} = \mathbf{b} \) Unique for every \( \mathbf{b} \) None or infinitely many
RREF of \( A \) \( I_n \) Has a row of zeros
Columns Linearly independent Linearly dependent
Null space Only \( \{ \mathbf{0} \} \) Contains nonzero vectors
Rank \( n \) (full rank) \( < n \)

Question 1 of 5

What is the determinant of \( \begin{pmatrix} 4 & 6 \\ 2 & 3 \end{pmatrix} \)?

Question 2 of 5

If \( A \) is invertible, how many solutions does \( A\mathbf{x} = \mathbf{b} \) have for a given \( \mathbf{b} \)?

Question 3 of 5

Find \( A^{-1} \) for \( A = \begin{pmatrix} 1 & 2 \\ 3 & 7 \end{pmatrix} \).

Question 4 of 5

If \( A \) and \( B \) are both invertible, what is \( (AB)^{-1} \)?

Question 5 of 5

Which of the following is NOT equivalent to "\( A \) is invertible"?