Applying The Determinant Test

Using the Determinant Test in a Parameter Problem

For a square matrix, calculating its determinant can turn an invertibility question into a numerical one: a nonzero determinant means the matrix is invertible, while a zero determinant means it is singular. When a matrix contains a parameter, the calculation describes which parameter values give each outcome. At a singular value, the homogeneous system also gains a nonzero solution, so the determinant calculation tells us something about the equations as well as the matrix.

Throughout, matrices and vectors have real entries unless stated otherwise. A square matrix \(\mathbf{A}\in\mathbb{R}^{n\times n}\) is invertible if there is a matrix \(\mathbf{A}^{-1}\) with \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}_n\); otherwise it is singular. The determinant test states that \(\mathbf{A}\) is invertible if and only if \(\det\mathbf{A}\ne0\). In a parameter problem, the practical steps are to calculate \(\det\mathbf{A}(t)\), identify where it is zero, and apply this test to the resulting cases.

Zero Determinants and Homogeneous Solutions

The zero vector is always a solution of the homogeneous system \(\mathbf{A}\mathbf{x}=\mathbf{0}\). The important distinction is whether it is the only solution. The determinant test connects that distinction exactly to singularity.

Proof. If \(\det\mathbf{A}\ne0\), the determinant test makes \(\mathbf{A}\) invertible. Multiplying \(\mathbf{A}\mathbf{x}=\mathbf{0}\) by \(\mathbf{A}^{-1}\) gives \(\mathbf{x}=\mathbf{0}\). Thus a nonzero solution is impossible whenever the determinant is nonzero. For the converse, suppose \(\det\mathbf{A}=0\). The determinant test says that \(\mathbf{A}\) is singular. Assume, for contradiction, that \(\mathbf{A}\mathbf{x}=\mathbf{0}\) has only the zero solution. Row operations preserve the solution set of a homogeneous system. In the reduced row echelon form of \(\mathbf{A}\), a column without a pivot would give a free variable and hence a nonzero solution: set that free variable to \(1\), set any other free variables to \(0\), and solve for the pivot variables. Therefore every column must have a pivot. Since \(\mathbf{A}\) is \(n\times n\), its reduced row echelon form must then be \(\mathbf{I}_n\). Each row operation is reversible, so a sequence of row operations taking \(\mathbf{A}\) to \(\mathbf{I}_n\) shows that \(\mathbf{A}\) is invertible. This contradicts singularity. Hence the homogeneous system has a nonzero solution when \(\det\mathbf{A}=0\), proving both directions. \(\square\)

This result concerns the homogeneous system, whose right-hand side is \(\mathbf{0}\). It does not say that every system \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is inconsistent when \(\mathbf{A}\) is singular. Such a system may have no solutions or many solutions, depending on \(\mathbf{b}\).

Worked Parameter Classifications

Parameter dependence can sometimes be especially simple when only one column varies linearly with the parameter. The next result explains why a determinant can then be read as an affine expression. It is a useful way to check a calculation, not a replacement for the determinant test.

Proof. The determinant is linear in each individual column when the other columns are held fixed. Write \(\mathbf{A}_{\mathbf{b}}\) for the matrix with \(\mathbf{b}\) in the varying column and the original fixed columns elsewhere, and define \(\mathbf{A}_{\mathbf{c}}\) in the same way using \(\mathbf{c}\). Multilinearity in that column gives \(\det\mathbf{A}(t)=\det\mathbf{A}_{\mathbf{b}}+t\det\mathbf{A}_{\mathbf{c}}\). Taking \(\alpha=\det\mathbf{A}_{\mathbf{b}}\) and \(\beta=\det\mathbf{A}_{\mathbf{c}}\) proves the claim. \(\square\)

Reading the Classification Carefully

A determinant calculation classifies a matrix at each parameter value; it does not automatically give a solution to every associated system. At a singular value, the theorem guarantees a nonzero vector in the null space, meaning a nonzero \(\mathbf{x}\) with \(\mathbf{A}(t)\mathbf{x}=\mathbf{0}\). Finding such a vector usually requires solving the homogeneous equations. Substituting the proposed vector back into the original matrix equation is an effective check.

The determinant’s sign is not the criterion for invertibility. A positive and a negative determinant are both nonzero, so both indicate invertibility. Only whether the determinant equals zero matters here.

Common Pitfall: Losing a Cofactor Sign

For the matrix in the first example, a common incorrect calculation is \(\det\mathbf{A}(t)=2t+2\), obtained by adding \(1\cdot2\) instead of subtracting it. The two-by-two rule requires a minus sign between the diagonal products:

The wrong calculation would predict a singular value \(t=-1\). But at \(t=-1\), the correct determinant is \(2(-1)-2=-4\ne0\), so the matrix is invertible. In contrast, at \(t=1\), the correct determinant is zero and the homogeneous system has a nonzero solution. When a parameter classification seems surprising, recompute the determinant before interpreting its roots.