Invertible matrices

Some square matrices have a true "opposite" under multiplication — a partner matrix that undoes them completely, sending them back to the identity. That partner is the inverse, and deciding when it exists (and proving it is one specific matrix, not several) is the capstone idea of the Matrices chapter. The definition looks deceptively simple: $AB = BA = I$ . But hidden inside it are three separate claims worth separating — that only square matrices can even qualify, that the inverse, when it exists, is unique, and that plenty of nonzero square matrices still fail to have one at all. This section builds all three carefully with worked verification, so that when the Determinants chapter hands you the formula $A^{-1} = \frac{1}{|A|}\operatorname{adj}(A)$ , you already know exactly what that formula is claiming to produce.

The core insight

A square matrix $A$ is invertible if some square matrix $B$ of the same order satisfies $AB = BA = I$ , and whenever such a $B$ exists it is the unique inverse $A^{-1}$ .

The core idea

Definition. A square matrix $A$ of order $n$ is invertible (or non-singular) if there exists a square matrix $B$ of order $n$ such that

AB = BA = I_n

where: $I_n$ is the identity matrix of order $n$ , and $B$ — when it exists — is called the inverse of $A$ , written $A^{-1}$ . Notice both products are demanded, and both must equal the identity; checking only one side is not a complete verification (see Common traps).

Why only square matrices qualify. If $A$ has order $m \times n$ with $m \neq n$ , then for $AB$ to even be defined $B$ must have $n$ rows, and for $BA$ to be defined $B$ must have $m$ columns — but for both $AB$ and $BA$ to equal identity matrices they would additionally need to be square themselves, of orders $m$ and $n$ respectively. Since $m \neq n$ , no single matrix $B$ can satisfy every one of these size constraints at once. So invertibility, in this sense, is a concept reserved for square matrices from the outset.

Singular vs. non-singular. Not every square matrix has an inverse. A square matrix with an inverse is called non-singular; one without an inverse is singular. The syllabus does not yet give you a quick test for which is which — that arrives in the Determinants chapter as the condition $|A| \neq 0$ — but you can already prove singularity directly for a specific matrix, as the worked example below shows for a matrix with proportional rows.

Why the identity matrix is the natural target. The identity matrix $I_n$ is the multiplicative "do nothing" matrix — for any matrix $A$ of compatible order, $AI = IA = A$ . Demanding $AB = BA = I$ is therefore demanding that multiplying by $B$ exactly undoes whatever $A$ did, in both directions, so that the pair together behaves like doing nothing at all. This is the same intuition as reciprocals of ordinary numbers, where $a \cdot a^{-1} = 1$ undoes multiplication by $a$ — except here order can matter for matrices in general, which is exactly why the definition insists on checking $AB$ and $BA$ separately rather than assuming one gives the other for free.

Figure

Two boxes labelled matrix A and matrix B connected by arrows both ways each producing the identity matrix showing the two-sided condition AB equals BA equals I

Derivation AThe inverse, if it exists, is unique

Prove that a square matrix $A$ cannot have two different inverses — that is, if $B$ and $C$ are both square matrices satisfying the defining condition $AB = BA = I$ and $AC = CA = I$ , then $B$ and $C$ must in fact be the same matrix, so the symbol $A^{-1}$ names one specific matrix and never an ambiguous choice among several candidates.

Step 1. Assume $B$ and $C$ are both inverses of $A$ . By definition this means $AB = BA = I$ and also $AC = CA = I$ , where $I$ is the identity matrix of the same order as $A$ .

Step 2. Start from $B$ and use the fact that multiplying any matrix by the identity leaves it unchanged: $B = BI$ .

Step 3. Replace $I$ using the second assumption, $I = AC$ , so $B = BI = B(AC)$ .

Step 4. Matrix multiplication is associative, so the grouping of the brackets can be shifted freely: $B(AC) = (BA)C$ .

Step 5. From the first assumption, $BA = I$ , so $(BA)C = IC$ .

Step 6. Multiplying by the identity again leaves the matrix unchanged: $IC = C$ .

Step 7. Chaining Steps 2 through 6 gives $B = BI = B(AC) = (BA)C = IC = C$ , so $B = C$ . Two matrices that were each assumed to be an inverse of $A$ turn out to be identical, which proves the inverse — when it exists — is unique.

Worked example

Verify that $A = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}$ are inverses of each other.

Step 1. Compute $AB$ entry by entry: $(1,1) = 2(1)+1(-1) = 1$ ; $(1,2) = 2(-1)+1(2) = 0$ ; $(2,1) = 1(1)+1(-1) = 0$ ; $(2,2) = 1(-1)+1(2) = 1$ . So $AB = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I_2$ .

Step 2. Compute $BA$ entry by entry: $(1,1) = 1(2)+(-1)(1) = 1$ ; $(1,2) = 1(1)+(-1)(1) = 0$ ; $(2,1) = -1(2)+2(1) = 0$ ; $(2,2) = -1(1)+2(1) = 1$ . So $BA = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I_2$ .

Step 3. Both $AB = I_2$ and $BA = I_2$ hold, so the full two-sided condition is satisfied. Therefore $B$ is the inverse of $A$ , and by the uniqueness result above, $B = A^{-1}$ is the only matrix with this property — $A^{-1} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}$ .

Competitive stretch

Show that $M = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$ has no inverse. Suppose $B = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ satisfies $MB = I_2$ . Multiplying out gives four equations: $a+2c=1$ , $b+2d=0$ , $2a+4c=0$ , $2b+4d=1$ . Doubling the first equation gives $2a+4c=2$ , which directly contradicts the third equation, $2a+4c=0$ . No values of $a,c$ can satisfy both, so no such $B$ exists — $M$ is singular. The underlying reason is visible in $M$ itself: row 2 is exactly $2\times$ row 1, a proportionality that recurs as the real test for singularity once determinants are introduced.

Common traps

Trying to talk about "the inverse" of a non-square matrix — invertibility in this sense is only defined for square matrices.
Checking only $AB = I$ and stopping there — a complete verification needs both $AB = I$ and $BA = I$ .
Assuming every nonzero square matrix has an inverse — the competitive-stretch example is a nonzero matrix that is still singular.
Trying to compute an inverse here using row operations or the adjoint formula — that machinery belongs to the Determinants chapter; this section only covers the definition, uniqueness, and direct verification.

Ask the tutor: Want to walk through the uniqueness proof again step by step, or try verifying a fresh pair of candidate inverse matrices together?

Why can a rectangular matrix never have an inverse

Because the defining condition $AB = BA = I$ forces $AB$ and $BA$ to both be square identity matrices of specific, and different, orders whenever $A$ is $m \times n$ with $m \neq n$ — $AB$ would need order $m$ and $BA$ would need order $n$ . No single matrix $B$ can satisfy both size requirements simultaneously, so the concept of an inverse is simply never set up for non-square matrices.

Do I need to check both AB equals I and BA equals I, or is one enough

For this chapter, always check both, since that is the definition you are asked to verify and it is what an exam expects. It is a deeper theorem (proved later, once determinants are available) that for square matrices, $AB=I$ alone actually forces $BA=I$ too — but relying on that shortcut without justification is not accepted as a complete verification at this stage.

If a matrix is singular does that mean it equals the zero matrix

No. The competitive-stretch example, $M = \begin{bmatrix}1&2\\2&4\end{bmatrix}$ , is singular but clearly nonzero. Singularity just means no inverse exists — commonly because one row (or column) is a scalar multiple of another, as here, which collapses the system $MB=I$ into a contradiction. The precise numerical test, $|A|=0$ , comes in the Determinants chapter.

Invertible matrices

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads