The Determinant Test For Invertibility

Invertibility and the Determinant

A square matrix is invertible when its action can be undone by another 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 \(\mathbf{A}\mathbf{B}=\mathbf{B}\mathbf{A}=\mathbf{I}_n\). The determinant offers a single scalar test for this property. The key point is not merely that the test works, but why its two directions follow from different ideas: the inverse identity gives one direction, while row reduction gives the other.

We will use two facts about determinants established by their defining properties. First, for square matrices of the same size, \(\det(\mathbf{A}\mathbf{B})=\det\mathbf{A}\det\mathbf{B}\), and \(\det\mathbf{I}_n=1\). Second, an elementary row operation has a nonzero effect factor on the determinant: swapping two rows multiplies it by \(-1\), scaling one row by \(c\ne0\) multiplies it by \(c\), and adding a multiple of one row to another leaves it unchanged. Thus each allowed row operation preserves whether the determinant is zero.

What Row Reduction Says About Invertibility

Recall that the reduced row echelon form (RREF) of a matrix is obtained by elementary row operations. In a homogeneous system, each row operation gives an equivalent system, so row reduction preserves the set of solutions to \(\mathbf{A}\mathbf{x}=\mathbf{0}\). A pivot variable is determined by the other variables; a variable in a column without a pivot is free.

Proof. Suppose first that \(\mathbf{R}=\mathbf{I}_n\). The row operations taking \(\mathbf{A}\) to \(\mathbf{R}\) are reversible: a row swap is undone by the same swap, scaling a row by \(c\ne0\) is undone by scaling it by \(1/c\), and adding \(c\) times one row to another is undone by adding \(-c\) times that row. Each operation is left multiplication by a matrix representing that operation. The product of these matrices, say \(\mathbf{E}\), is invertible because reversing the operations gives its inverse. Since \(\mathbf{E}\mathbf{A}=\mathbf{I}_n\), we have \(\mathbf{A}=\mathbf{E}^{-1}\), so \(\mathbf{A}\) is invertible.

Now suppose \(\mathbf{R}\ne\mathbf{I}_n\). A square RREF with a pivot in every column must be \(\mathbf{I}_n\): its \(n\) pivots must occupy \(n\) different rows, and the pivot entries and all other entries in those columns have the defining RREF values. Therefore \(\mathbf{R}\ne\mathbf{I}_n\) has at least one column without a pivot. Assign the value \(1\) to one free variable and \(0\) to all the other free variables, then determine the pivot variables from \(\mathbf{R}\mathbf{x}=\mathbf{0}\). This gives a nonzero solution \(\mathbf{x}\), and row operations preserve homogeneous solution sets, so \(\mathbf{A}\mathbf{x}=\mathbf{0}\) as well. If \(\mathbf{A}\) had an inverse, multiplying this equation by \(\mathbf{A}^{-1}\) would give \(\mathbf{x}=\mathbf{0}\), a contradiction. Thus \(\mathbf{A}\) is not invertible. \(\square\)

Proof of the Determinant Test

Proof. Let \(\mathbf{A}\in\mathbb{R}^{n\times n}\). First suppose \(\mathbf{A}\) is invertible, with inverse \(\mathbf{A}^{-1}\). The multiplicative determinant law applied to \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}_n\) gives

A product equal to \(1\) cannot have a zero factor, so \(\det\mathbf{A}\ne0\). This also gives the inverse identity \(\det(\mathbf{A}^{-1})=1/\det\mathbf{A}\).

Conversely, suppose \(\det\mathbf{A}\ne0\), and row-reduce \(\mathbf{A}\) to \(\mathbf{R}=\operatorname{rref}(\mathbf{A})\). At every step, the determinant is either unchanged, multiplied by \(-1\), or multiplied by a nonzero scalar. Hence \(\det\mathbf{R}\ne0\). A matrix with a zero row has determinant zero, by expansion along that row. If \(\mathbf{R}\) had fewer than \(n\) pivots, it would have a zero row: its pivots occupy distinct rows, leaving at least one row without a pivot, and in RREF such a row is zero. Thus \(\mathbf{R}\) has a pivot in every column, which for a square RREF means \(\mathbf{R}=\mathbf{I}_n\). By the row-reduction criterion, \(\mathbf{A}\) is invertible. Both implications hold, proving the theorem. \(\square\)

The two directions have different proof mechanisms. The inverse identity immediately rules out a zero determinant for an invertible matrix. For the converse, nonzero determinant survives row reduction; the resulting RREF must be the identity, so the row operations can be reversed to produce an inverse.

Worked Applications of the Test

A Common Row-Operation Error

Mistake: Treating row scaling as though it leaves the determinant unchanged. Starting from \(\mathbf{I}_2\), the incorrect calculation might claim that scaling row 1 by \(2\) preserves the determinant, and therefore that \(\det\begin{bmatrix}2&0\\0&1\end{bmatrix}=1\).

Correction: Scaling a row by \(c\) multiplies the determinant by \(c\). Here \(c=2\), so the new determinant is \(2\det\mathbf{I}_2=2\). Direct calculation confirms this: \(\det\begin{bmatrix}2&0\\0&1\end{bmatrix}=2\cdot1-0=2\). The determinant test then identifies this matrix as invertible, consistent with its inverse \(\begin{bmatrix}1/2&0\\0&1\end{bmatrix}\).

When using row reduction to test invertibility, track whether each row-operation factor is zero or nonzero; the exact determinant value is not always needed. Every permitted row operation has a nonzero factor, so it preserves the distinction between zero and nonzero determinants. That is precisely what allows the RREF to settle the test.