Uniqueness Depends on the Parameter
A system of linear equations can change its solvability as a parameter changes. For one parameter value, the coefficient matrix may be invertible and the system has exactly one solution; at another value, it may be singular. Cramer’s Rule makes the unique-solution case especially explicit: it expresses each coordinate as a quotient of determinants. The determinant in the denominator also identifies the parameter values where the rule cannot be used.
Consider a square system \(\mathbf{A}(t)\mathbf{x}=\mathbf{b}(t)\), where \(t\in\mathbb{R}\), \(\mathbf{A}(t)\in\mathbb{R}^{n\times n}\), and \(\mathbf{b}(t)\in\mathbb{R}^n\). For each \(i\in\{1,\dots,n\}\), let \(\mathbf{A}_i(t)\) be the matrix obtained from \(\mathbf{A}(t)\) by replacing its \(i\)th column with \(\mathbf{b}(t)\). A parameter value is exceptional for the Cramer calculation when \(\det\mathbf{A}(t)=0\).
Proof. Fix \(t\). The determinant test for invertibility says that the square matrix \(\mathbf{A}(t)\) is invertible if and only if \(\det\mathbf{A}(t)\ne0\). An invertible coefficient matrix gives exactly one solution to \(\mathbf{A}(t)\mathbf{x}=\mathbf{b}(t)\), and a singular square coefficient matrix does not give a unique solution. This proves the first claim. For the coordinate formula, suppose \(\det\mathbf{A}(t)\ne0\), and write the columns of \(\mathbf{A}(t)\) as \(\mathbf{a}_1,\dots,\mathbf{a}_n\). Since \(\mathbf{b}(t)=\sum_{j=1}^n x_j\mathbf{a}_j\), linearity of the determinant in column \(i\) gives $$ \det\mathbf{A}_i(t)=\sum_{j=1}^n x_j\det\begin{bmatrix}\mathbf{a}_1&\cdots&\mathbf{a}_{i-1}&\mathbf{a}_j&\mathbf{a}_{i+1}&\cdots&\mathbf{a}_n\end{bmatrix}. $$ For \(j\ne i\), the displayed matrix has two equal columns, so its determinant is zero. For \(j=i\), its determinant is \(\det\mathbf{A}(t)\). Therefore \(\det\mathbf{A}_i(t)=x_i\det\mathbf{A}(t)\). Dividing by the nonzero determinant gives the stated formula. Finally, if the determinant is zero, the determinant test says the columns are linearly dependent, so there is a nonzero vector \(\mathbf{z}\) with \(\mathbf{A}(t)\mathbf{z}=\mathbf{0}\). If \(\mathbf{x}_0\) is any solution, then \(\mathbf{x}_0+s\mathbf{z}\) is also a solution for every \(s\in\mathbb{R}\), and choosing \(s\ne0\) gives a different solution. Thus a singular system either has no solution or has multiple solutions. \(\square\)
A Two-Variable System and Its Exceptional Values
Worked Example: A Consistent and an Inconsistent Singular Case
For each \(t\in\mathbb{R}\), consider $$ \begin{bmatrix}1&t\\t&1\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}1\\1\end{bmatrix}. $$ Find the parameter values giving a unique solution, calculate that solution, and examine the other values.
The coefficient determinant is $$ \det\begin{bmatrix}1&t\\t&1\end{bmatrix}=1-t^2=(1-t)(1+t). $$ It is nonzero exactly when \(t\ne1\) and \(t\ne-1\), so those are precisely the values with a unique solution. Replacing the first column and then the second column gives $$ \det\begin{bmatrix}1&t\\1&1\end{bmatrix}=1-t,\qquad \det\begin{bmatrix}1&1\\t&1\end{bmatrix}=1-t. $$ Thus, for \(t\ne\pm1\), $$ x_1=\frac{1-t}{1-t^2}=\frac{1}{1+t},\qquad x_2=\frac{1-t}{1-t^2}=\frac{1}{1+t}. $$ The simplification is valid on this domain, where \(1-t\ne0\) as well as \(1+t\ne0\).
Check: With \(x_1=x_2=1/(1+t)\), the left side is $$ \begin{bmatrix}1&t\\t&1\end{bmatrix} \begin{bmatrix}\frac{1}{1+t}\\\frac{1}{1+t}\end{bmatrix} = \begin{bmatrix}\frac{1+t}{1+t}\\\frac{t+1}{1+t}\end{bmatrix} = \begin{bmatrix}1\\1\end{bmatrix}. $$ At \(t=1\), both equations reduce to \(x_1+x_2=1\), so there are multiple solutions. At \(t=-1\), the equations are \(x_1-x_2=1\) and \(-x_1+x_2=1\). Adding them gives \(0=2\), so there is no solution.
The determinant identifies the values where uniqueness fails, but it does not by itself say whether a singular system is inconsistent or has multiple solutions. At each exceptional value, substitute into the original system and check consistency. The Cramer quotients have zero denominators there, so they cannot decide that question.
Determinant Quotients as Rational Functions
If the entries of \(\mathbf{A}(t)\) and \(\mathbf{b}(t)\) are polynomials in \(t\), then each determinant in Cramer’s Rule is also a polynomial in \(t\). Consequently, on the set where \(\det\mathbf{A}(t)\ne0\), each solution coordinate is a rational function: a quotient of polynomials whose denominator is nonzero on the stated domain. The formula describes the unique solution only on that domain, even if algebraic simplification appears to remove a factor from the denominator.
Proof. The determinant is a finite sum of signed products of matrix entries. If each entry is a polynomial, every such product is a polynomial, and their sum is a polynomial. This applies to \(\det\mathbf{A}(t)\) and to every \(\det\mathbf{A}_i(t)\), since replacing a column with \(\mathbf{b}(t)\) still leaves only polynomial entries. Cramer’s Rule therefore writes each coordinate as a quotient of polynomials. The rule requires its denominator \(\det\mathbf{A}(t)\) to be nonzero; by the determinant test, these and only these parameter values give a unique solution. \(\square\)
Worked Example: A Three-Variable System
For \(t\in\mathbb{R}\), solve the system $$ \begin{bmatrix} 1&t&0\\ 0&t-2&1\\ 0&0&t+1 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}2\\1\\3\end{bmatrix} $$ whenever the solution is unique, and classify the exceptional values.
The coefficient matrix is upper triangular, so its determinant is the product of its diagonal entries: $$ \det\mathbf{A}(t)=1\cdot(t-2)\cdot(t+1)=(t-2)(t+1). $$ There is a unique solution exactly when \(t\ne2,-1\). The three column-replacement determinants are $$ \det\mathbf{A}_1(t)= \det\begin{bmatrix}2&t&0\\1&t-2&1\\3&0&t+1\end{bmatrix} =2(t-2)(t+1)-t\big((t+1)-3\big)=t^2-4, $$ $$ \det\mathbf{A}_2(t)= \det\begin{bmatrix}1&2&0\\0&1&1\\0&3&t+1\end{bmatrix} =(t+1)-3=t-2, $$ and $$ \det\mathbf{A}_3(t)= \det\begin{bmatrix}1&t&2\\0&t-2&1\\0&0&3\end{bmatrix} =3(t-2). $$ For \(t\ne2,-1\), Cramer’s Rule gives $$ x_1=\frac{t^2-4}{(t-2)(t+1)}=\frac{t+2}{t+1},\qquad x_2=\frac{1}{t+1},\qquad x_3=\frac{3}{t+1}. $$ The cancellation in \(x_1\) uses \(t\ne2\), which is part of the required domain.
Check: Substitution into the three rows gives $$ x_1+tx_2=\frac{t+2+t}{t+1}=2,\qquad (t-2)x_2+x_3=\frac{t-2+3}{t+1}=1,\qquad (t+1)x_3=3. $$ At \(t=2\), the equations are \(x_1+2x_2=2\), \(x_3=1\), and \(3x_3=3\). The last two both imply \(x_3=1\), while the first leaves \(x_2\) free and sets \(x_1=2-2x_2\); there are infinitely many solutions. At \(t=-1\), the third equation is \(0=3\), so the system is inconsistent.
A Cancelled Factor Does Not Restore Uniqueness
Worked Example: A Quotient with a Removable Factor
Consider $$ \begin{bmatrix}t&1\\1&1\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}1\\t\end{bmatrix}. $$ Find the solution for the values where it is unique, and explain why simplification cannot include the singular value.
The coefficient determinant is \(t-1\), so the system has a unique solution when \(t\ne1\). The determinants after replacing the first and second columns are $$ \det\begin{bmatrix}1&1\\t&1\end{bmatrix}=1-t,\qquad \det\begin{bmatrix}t&1\\1&t\end{bmatrix}=t^2-1. $$ Thus, for \(t\ne1\), $$ x_1=\frac{1-t}{t-1}=-1,\qquad x_2=\frac{t^2-1}{t-1}=t+1. $$ Check: \(t(-1)+(t+1)=1\), and \((-1)+(t+1)=t\), as required.
At \(t=1\), both equations become \(x_1+x_2=1\), so there are infinitely many solutions. A tempting but wrong step is to cancel \(t-1\) and then claim the formula \(x_1=-1,\ x_2=t+1\) gives the unique solution at \(t=1\). Cancellation gives an equivalent quotient only where \(t-1\ne0\); at \(t=1\), the original Cramer expressions have denominator zero and the coefficient matrix is singular. The simplified pair \((-1,2)\) is one solution there, but it is not the only one.
A Reliable Procedure
For a parameterized square system, keep the determinant test and the solution calculation in separate steps. This prevents a valid formula on one parameter domain from being mistakenly used at a singular value.
Find \(\det\mathbf{A}(t)\), factor it when possible, and solve \(\det\mathbf{A}(t)=0\). These roots are the exceptional values.
The solution is unique precisely for the real parameter values where \(\det\mathbf{A}(t)\ne0\).
Replace one column at a time by \(\mathbf{b}(t)\), and divide each replacement determinant by \(\det\mathbf{A}(t)\). Keep the excluded values visible even if factors cancel.
Substitute the coordinate formulas into the original equations. At each zero of the coefficient determinant, test the original system for consistency; it may have no solution or multiple solutions.
Common Pitfall: Dividing by a Parameter Expression
Every division in a Cramer calculation requires the divisor to be nonzero. For example, the step \((1-t)/(1-t^2)=1/(1+t)\) requires both \(1-t\ne0\) and \(1+t\ne0\), because the original denominator is \((1-t)(1+t)\). The cancelled expression \(1/(1+t)\) is not a valid Cramer solution at \(t=1\); in the first example, that parameter value gives infinitely many solutions. Always record the domain before simplifying, and return to the original equations to analyze excluded values.
The determinant also answers a focused question: it tells exactly when there is a unique solution, not what happens in every singular case. To decide between inconsistency and multiple solutions, the original equations must be checked at the exceptional parameter values.
Check Your Understanding
For each question, state the parameter restrictions before simplifying any quotient, and check any proposed solution in the original equations.
- For \(\begin{bmatrix}t&0\\0&1\end{bmatrix}\mathbf{x}=\begin{bmatrix}2\\3\end{bmatrix}\), determine the unique-solution values and solve on that domain.
- For \(\begin{bmatrix}1&t\\0&t-3\end{bmatrix}\mathbf{x}=\begin{bmatrix}1\\0\end{bmatrix}\), find the exceptional values and classify the system at each one.
- Explain why cancelling a factor from a Cramer quotient does not show that the original system has a unique solution where that factor is zero.
- State the condition under which Cramer’s Rule gives \(x_i=\det(\mathbf{A}_i)/\det(\mathbf{A})\).
- If the entries of \(\mathbf{A}(t)\) and \(\mathbf{b}(t)\) are polynomials, explain why each Cramer coordinate is a rational function wherever the system has a unique solution.
