A Workshop in Inverse Computation
Finding a candidate inverse is only part of the task. A sound solution also explains why the computation applies, tracks the order of matrix products, and checks the result by multiplication. This workshop brings these habits together for square matrices over \(\mathbb{R}\). We will use row operations to compute inverses, relate those operations to elementary matrices, and prove two identities that often appear in matrix calculations.
A square matrix \(\mathbf{A}\in\mathbb{R}^{n\times n}\) is invertible if there is a matrix \(\mathbf{B}\in\mathbb{R}^{n\times n}\) such that \(\mathbf{A}\mathbf{B}=\mathbf{B}\mathbf{A}=\mathbf{I}_n\). In that case \(\mathbf{B}\) is the inverse, written \(\mathbf{A}^{-1}\). The inverse is unique. The pivot criterion for invertibility says that a square matrix is invertible exactly when its reduced row echelon form is \(\mathbf{I}_n\), or equivalently, when it has a pivot in every column. These criteria let us distinguish a computation that can produce an inverse from one that cannot.
Computing an Inverse with an Augmented Matrix
To use row reduction, place \(\mathbf{A}\) beside the identity matrix of the same size, forming \(\begin{bmatrix}\mathbf{A}&\mathbf{I}_n\end{bmatrix}\). Apply each row operation to the entire augmented matrix. If the left block becomes \(\mathbf{I}_n\), the right block is \(\mathbf{A}^{-1}\). The next example shows every operation and checks the result.
Worked Example: Reduce an Augmented Matrix
Find the inverse of \(\mathbf{A}=\begin{bmatrix}1&2\\3&7\end{bmatrix}\), if it exists.
Start with the augmented matrix. Apply \(R_2\to R_2-3R_1\), then apply \(R_1\to R_1-2R_2\) using the new second row:
The left block is \(\mathbf{I}_2\), so the right block gives the candidate \(\mathbf{A}^{-1}=\begin{bmatrix}7&-2\\-3&1\end{bmatrix}\). Check by multiplying in both orders:
Both checks confirm the proposed inverse.
The augmented procedure has a precise justification. A row operation on a matrix is left multiplication by an invertible elementary matrix. Thus, if a sequence of row operations changes the left block \(\mathbf{A}\) into \(\mathbf{I}_n\), there is a product \(\mathbf{E}\) of elementary matrices with \(\mathbf{E}\mathbf{A}=\mathbf{I}_n\). The same operations change the right block \(\mathbf{I}_n\) into \(\mathbf{E}\), so the augmented reduction displays the matrix that undoes \(\mathbf{A}\).
Proof. Suppose row operations produce \(\begin{bmatrix}\mathbf{I}_n&\mathbf{B}\end{bmatrix}\). The operations are left multiplication by a product \(\mathbf{E}\) of elementary matrices. Each elementary matrix is invertible, since the row operation it represents can be undone by another row operation. Therefore \(\mathbf{E}\) is invertible. The left block gives \(\mathbf{E}\mathbf{A}=\mathbf{I}_n\), so \(\mathbf{A}=\mathbf{E}^{-1}\) and is invertible. The right block gives \(\mathbf{E}\mathbf{I}_n=\mathbf{B}\), hence \(\mathbf{B}=\mathbf{E}=\mathbf{A}^{-1}\). Conversely, suppose \(\mathbf{A}\) is invertible. By the pivot criterion for invertibility, its RREF is \(\mathbf{I}_n\), so a sequence of row operations changes the left block to \(\mathbf{I}_n\). Applying that same sequence to the identity block yields some matrix \(\mathbf{B}\); the first part of the proof shows \(\mathbf{B}=\mathbf{A}^{-1}\). Thus the augmented reduction produces \(\begin{bmatrix}\mathbf{I}_n&\mathbf{A}^{-1}\end{bmatrix}\). \(\square\)
Elementary Matrices Record Row Operations
An elementary matrix is a matrix obtained by performing one elementary row operation on an identity matrix. Left multiplication by that matrix performs the corresponding row operation on any matrix with the same number of rows. Since every row operation can be undone, every elementary matrix is invertible. For example, the operation \(R_2\to R_2-3R_1\) is represented by \(\begin{bmatrix}1&0\\-3&1\end{bmatrix}\), and its inverse represents \(R_2\to R_2+3R_1\).
When row operations reduce \(\mathbf{A}\) to \(\mathbf{I}_n\), their elementary matrices multiply on the left. If \(\mathbf{E}_k\cdots\mathbf{E}_2\mathbf{E}_1\mathbf{A}=\mathbf{I}_n\), then \(\mathbf{E}_k\cdots\mathbf{E}_1\) is the inverse of \(\mathbf{A}\). The order records the order in which the operations are applied: the first operation is represented by the rightmost factor.
Worked Example: Read the Inverse from Row Operations
For \(\mathbf{A}=\begin{bmatrix}1&2\\3&7\end{bmatrix}\), the first operation is \(R_2\to R_2-3R_1\), represented by \(\mathbf{E}_1=\begin{bmatrix}1&0\\-3&1\end{bmatrix}\). The second operation is \(R_1\to R_1-2R_2\), represented by \(\mathbf{E}_2=\begin{bmatrix}1&-2\\0&1\end{bmatrix}\). Their effect is
Multiplying the elementary matrices gives
Therefore \(\mathbf{A}^{-1}=\mathbf{E}_2\mathbf{E}_1\). As a direct check, \(\mathbf{A}(\mathbf{E}_2\mathbf{E}_1)=\begin{bmatrix}1&2\\3&7\end{bmatrix}\begin{bmatrix}7&-2\\-3&1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\), consistent with the augmented calculation.
When the Augmented Reduction Cannot Work
If \(\mathbf{A}\) is not invertible, its left block cannot be reduced to the identity. The reason is not that a zero row must stay zero: adding one row to another can change which rows are zero. The relevant fact is that row operations preserve rank. In particular, a matrix of rank less than \(n\) cannot be row-equivalent to \(\mathbf{I}_n\), whose rank is \(n\).
Worked Example: A Singular Matrix
Consider \(\mathbf{A}=\begin{bmatrix}1&2\\2&4\end{bmatrix}\). Its second row is twice its first, so its rank is \(1\), not \(2\). Reduce the augmented matrix using \(R_2\to R_2-2R_1\):
The left block still has rank \(1\), because row operations preserve rank. It cannot become \(\mathbf{I}_2\), which has rank \(2\), so \(\mathbf{A}\) is not invertible. Notice that the zero row in the left block is not required to remain a zero row under every possible later operation; the rank argument, not the zero row alone, establishes the obstruction.
Inverse of a Product: Order Matters
Suppose \(\mathbf{A}\) and \(\mathbf{B}\) are invertible \(n\times n\) matrices. To undo the effect of \(\mathbf{A}\mathbf{B}\), first undo \(\mathbf{A}\) and then undo \(\mathbf{B}\). This reverses the order of the factors. The identity follows directly from associativity and the defining equations for inverses.
Proof. Multiply \(\mathbf{A}\mathbf{B}\) by \(\mathbf{B}^{-1}\mathbf{A}^{-1}\) in each order. Associativity and the inverse equations give
Thus \(\mathbf{B}^{-1}\mathbf{A}^{-1}\) is an inverse of \(\mathbf{A}\mathbf{B}\), and uniqueness of the inverse gives the formula. \(\square\)
Worked Example: Check the Reversed Order
Let \(\mathbf{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}1&0\\2&1\end{bmatrix}\). Their inverses, checked by multiplication, are \(\mathbf{A}^{-1}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\) and \(\mathbf{B}^{-1}=\begin{bmatrix}1&0\\-2&1\end{bmatrix}\). First compute
Check the proposed product inverse:
The reversed order matters. The other order is \(\mathbf{A}^{-1}\mathbf{B}^{-1}=\begin{bmatrix}3&-1\\-2&1\end{bmatrix}\), and multiplying it on the left by \(\mathbf{A}\mathbf{B}\) gives \(\begin{bmatrix}7&-2\\4&-1\end{bmatrix}\), not \(\mathbf{I}_2\).
Inverse of a Transpose
For a real matrix, transposition reverses the order of a product: \((\mathbf{A}\mathbf{B})^{\top}=\mathbf{B}^{\top}\mathbf{A}^{\top}\). If \(\mathbf{A}\) is invertible, this identity and the equations \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}_n\) show that transposing the inverse gives the inverse of the transpose.
Proof. Transpose each of the two inverse equations. Using the transpose-of-a-product identity gives
Therefore \((\mathbf{A}^{-1})^{\top}\) is a two-sided inverse of \(\mathbf{A}^{\top}\). By uniqueness, it equals \((\mathbf{A}^{\top})^{-1}\). \(\square\)
Common Mistakes and a Reliable Check
Do not reverse the product formula incorrectly. A tempting but wrong claim is \((\mathbf{A}\mathbf{B})^{-1}=\mathbf{A}^{-1}\mathbf{B}^{-1}\). In the product example above, that order gives \(\begin{bmatrix}3&-1\\-2&1\end{bmatrix}\); multiplying it by \(\mathbf{A}\mathbf{B}=\begin{bmatrix}3&1\\2&1\end{bmatrix}\) gives \(\begin{bmatrix}7&-2\\4&-1\end{bmatrix}\), not the identity. The correction is \((\mathbf{A}\mathbf{B})^{-1}=\mathbf{B}^{-1}\mathbf{A}^{-1}\), verified by multiplication in both orders.
Apply row operations to both blocks. In an augmented matrix, an operation changes the corresponding rows on the left and right. Reducing only the left block does not track the matrix that will become the inverse. Also, a zero row is not by itself an invariant: rows can be combined to produce a nonzero row. To show that a left block cannot become the identity, use the pivot criterion or the fact that row operations preserve rank.
Check dimensions and hypotheses. The product and transpose formulas above assume square invertible matrices of the same size. A non-square matrix does not have a two-sided inverse of the form defined here, and a singular square matrix has no inverse to compute. When a candidate is found, multiplying it by the original matrix is a direct check; for a square matrix, checking both orders confirms the two-sided inverse equations.
Check Your Understanding
For each question, justify the conclusion using a row reduction, an inverse identity, or a multiplication check.
- Use the augmented-matrix method to find the inverse of \(\begin{bmatrix}1&1\\2&3\end{bmatrix}\), then check by multiplication.
- What rank obstruction prevents a \(3\times3\) matrix of rank \(2\) from reducing to \(\mathbf{I}_3\)?
- If \(\mathbf{A}\) and \(\mathbf{B}\) are invertible, write the inverse of \(\mathbf{A}\mathbf{B}\mathbf{A}\) in the correct order.
- State the inverse of \(\mathbf{A}^{\top}\) when \(\mathbf{A}\) is invertible, and identify the product-transpose identity used to justify it.
- Explain why every row operation used on \(\begin{bmatrix}\mathbf{A}&\mathbf{I}_n\end{bmatrix}\) must also be applied to the right block.
