Why the Homogeneous System Matters
A square matrix can be tested for invertibility by examining a system with zero on the right-hand side. The connection is useful because the homogeneous system has an immediate solution, \(\mathbf{x}=\mathbf{0}\), and the question becomes whether any other solution is possible. For square matrices, having no other solution is not merely a statement about one system: it guarantees that every right-hand side produces exactly one solution.
Throughout, let \(n\in\mathbb{N}\), let \(\mathbf{A}\in\mathbb{R}^{n\times n}\), and let \(\mathbf{b},\mathbf{x}\in\mathbb{R}^n\) be column vectors. A system \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is homogeneous when \(\mathbf{b}=\mathbf{0}\). It is consistent when it has at least one solution. A matrix \(\mathbf{A}\) is invertible if there is a matrix \(\mathbf{A}^{-1}\) such that \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}_n\).
Proof. First suppose that \(\mathbf{A}\) is invertible. If \(\mathbf{A}\mathbf{x}=\mathbf{0}\), multiply both sides on the left by \(\mathbf{A}^{-1}\). Then \(\mathbf{x}=\mathbf{A}^{-1}\mathbf{0}=\mathbf{0}\), so (2) holds. Also, for any \(\mathbf{b}\in\mathbb{R}^n\), the vector \(\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}\) satisfies \(\mathbf{A}\mathbf{x}=\mathbf{A}\mathbf{A}^{-1}\mathbf{b}=\mathbf{b}\). If \(\mathbf{y}\) is another solution, then \(\mathbf{A}\mathbf{y}=\mathbf{b}\); multiplying by \(\mathbf{A}^{-1}\) gives \(\mathbf{y}=\mathbf{A}^{-1}\mathbf{b}=\mathbf{x}\). Thus (3) holds.
Now suppose (2) holds. Row reduction preserves the solutions of the homogeneous system. If the reduced row echelon form of \(\mathbf{A}\) had a column without a pivot, the variable for that column would be free; assigning it a nonzero value and the other free variables zero would give a nonzero solution. Therefore every column has a pivot. Since \(\mathbf{A}\) is \(n\times n\), its reduced row echelon form is \(\mathbf{I}_n\). Each row operation is left multiplication by an invertible elementary matrix, so a product \(\mathbf{E}\) of these matrices satisfies \(\mathbf{E}\mathbf{A}=\mathbf{I}_n\). Hence \(\mathbf{A}=\mathbf{E}^{-1}\), which is invertible. This proves (2) implies (1). Finally, if (3) holds, take \(\mathbf{b}=\mathbf{0}\). The zero vector is a solution, and uniqueness says it is the only solution, so (2) holds. These implications establish the equivalence. \(\square\)
The square-matrix condition matters in the proof: \(n\) columns with a pivot in every column yield the identity after reduction because there are also \(n\) rows. The theorem does not assert the same equivalence for a rectangular matrix. For a particular \(\mathbf{b}\), the equation might have no solution, one solution, or many; the theorem’s third statement requires exactly one solution for every right-hand side.
Testing the Homogeneous System in Practice
Worked Example: A Trivial Homogeneous Solution
Consider \(\mathbf{A}=\begin{bmatrix}2&1\\1&1\end{bmatrix}\). Determine whether the homogeneous system has only the zero solution, then solve \(\mathbf{A}\mathbf{x}=\begin{bmatrix}5\\3\end{bmatrix}\).
First row-reduce the coefficient matrix. Swap the rows, eliminate the entry below the first pivot, scale the second row, and clear above its pivot:
There is a pivot in each column, so the homogeneous system has no free variables. Its reduced equations are \(x_1=0\) and \(x_2=0\), and its only solution is \(\mathbf{x}=\mathbf{0}\). By the theorem, \(\mathbf{A}\) is invertible and every right-hand side has exactly one solution.
For the stated right-hand side, row-reduce the augmented matrix using the same operations:
Scale the second row by \(-1\), then subtract the new second row from the first:
Thus \(\mathbf{x}=\begin{bmatrix}2\\1\end{bmatrix}\). Check: \(\mathbf{A}\mathbf{x}=\begin{bmatrix}2&1\\1&1\end{bmatrix}\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}4+1\\2+1\end{bmatrix}=\begin{bmatrix}5\\3\end{bmatrix}\), as required.
The pivot test in this computation is a way to detect whether a nonzero homogeneous solution exists. A free variable supplies a nonzero choice; no free variables means every variable is forced to zero. The theorem turns this observation into a test for invertibility specifically because the matrix is square.
What a Nonzero Homogeneous Solution Tells Us
A nonzero solution of \(\mathbf{A}\mathbf{x}=\mathbf{0}\) rules out invertibility. It also explains what can happen in a nonhomogeneous system. If one solution is known, adding any homogeneous solution gives another solution. More precisely, the null space \(\operatorname{Nul}\mathbf{A}\) is the set of vectors \(\mathbf{z}\) satisfying \(\mathbf{A}\mathbf{z}=\mathbf{0}\).
Proof. Suppose \(\mathbf{x}\) is any solution of \(\mathbf{A}\mathbf{x}=\mathbf{b}\). Since \(\mathbf{A}\mathbf{x}_p=\mathbf{b}\), subtraction gives \(\mathbf{A}(\mathbf{x}-\mathbf{x}_p)=\mathbf{b}-\mathbf{b}=\mathbf{0}\). Thus \(\mathbf{x}-\mathbf{x}_p\in\operatorname{Nul}\mathbf{A}\), so \(\mathbf{x}=\mathbf{x}_p+\mathbf{z}\) for some \(\mathbf{z}\in\operatorname{Nul}\mathbf{A}\). Conversely, if \(\mathbf{z}\in\operatorname{Nul}\mathbf{A}\), then \(\mathbf{A}(\mathbf{x}_p+\mathbf{z})=\mathbf{A}\mathbf{x}_p+\mathbf{A}\mathbf{z}=\mathbf{b}+\mathbf{0}=\mathbf{b}\). This proves the description of all solutions. If there is a nonzero \(\mathbf{z}\) in the null space, then \(\mathbf{x}_p+\mathbf{z}\ne\mathbf{x}_p\), so the system has at least two solutions. \(\square\)
Worked Example: A Singular Matrix with Infinitely Many Solutions
Let \(\mathbf{A}=\begin{bmatrix}1&2\\2&4\end{bmatrix}\). Find the homogeneous solutions and solve \(\mathbf{A}\mathbf{x}=\begin{bmatrix}3\\6\end{bmatrix}\).
For the homogeneous system, subtract twice the first row from the second:
The reduced equation is \(x_1+2x_2=0\). Setting the free variable \(x_2=t\), where \(t\in\mathbb{R}\), gives \(\mathbf{x}=\begin{bmatrix}-2t\\t\end{bmatrix}\). For example, \(t=1\) gives a nonzero homogeneous solution \(\begin{bmatrix}-2\\1\end{bmatrix}\). Direct multiplication confirms it: \(\mathbf{A}\begin{bmatrix}-2\\1\end{bmatrix}=\begin{bmatrix}-2+2\\-4+4\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\). Thus \(\mathbf{A}\) is not invertible.
For the nonhomogeneous system, the same row operation gives
Again let \(x_2=t\). Then \(x_1=3-2t\), so all solutions are \(\mathbf{x}=\begin{bmatrix}3-2t\\t\end{bmatrix}\). Check: multiplying gives \(\mathbf{A}\mathbf{x}=\begin{bmatrix}3-2t+2t\\6-4t+4t\end{bmatrix}=\begin{bmatrix}3\\6\end{bmatrix}\) for every real \(t\). The system has infinitely many solutions, consistent with the nonzero homogeneous solution.
Not every right-hand side for this matrix is consistent. For \(\mathbf{b}=\begin{bmatrix}3\\5\end{bmatrix}\), row reduction gives
The last row requires \(0=-1\), which is impossible, so this system has no solution. The same noninvertible matrix can therefore produce infinitely many solutions for one right-hand side and no solutions for another.
Common Pitfall: Uniqueness Is About Differences
A frequent error is to reason that if \(\mathbf{x}\) and \(\mathbf{y}\) both solve \(\mathbf{A}\mathbf{x}=\mathbf{b}\), then \(\mathbf{A}(\mathbf{x}-\mathbf{y})=\mathbf{b}\). Subtracting the two equations correctly gives \(\mathbf{A}(\mathbf{x}-\mathbf{y})=\mathbf{b}-\mathbf{b}=\mathbf{0}\), not \(\mathbf{b}\). Thus two solutions differ by a homogeneous solution. They must be equal when the homogeneous system has only the zero solution.
For a concrete correction, in the preceding example with \(\mathbf{b}=\begin{bmatrix}3\\6\end{bmatrix}\), take \(\mathbf{x}=\begin{bmatrix}3\\0\end{bmatrix}\) and \(\mathbf{y}=\begin{bmatrix}1\\1\end{bmatrix}\). Both multiply by \(\mathbf{A}\) to give \(\begin{bmatrix}3\\6\end{bmatrix}\). Their difference is \(\mathbf{x}-\mathbf{y}=\begin{bmatrix}2\\-1\end{bmatrix}\), and \(\mathbf{A}(\mathbf{x}-\mathbf{y})=\begin{bmatrix}2-2\\4-4\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\). The error is assigning the right-hand side \(\mathbf{b}\) to the difference equation instead of zero.
Another distinction is equally important: a trivial homogeneous solution guarantees uniqueness for a consistent system, while the theorem guarantees consistency for every right-hand side only in the square invertible case. For a noninvertible square matrix, a given right-hand side need not have any solution, as the inconsistent example shows. The phrase “exactly one solution for every \(\mathbf{b}\)” includes both requirements: existence and uniqueness.
Worked Example: A Singular Matrix in Three Variables
Consider \(\mathbf{A}=\begin{bmatrix}1&0&1\\0&1&1\\1&1&2\end{bmatrix}\). Determine whether it is invertible and solve \(\mathbf{A}\mathbf{x}=\begin{bmatrix}2\\3\\5\end{bmatrix}\).
Subtract the first row and then the second row from the third:
The third variable is free. In the homogeneous system, setting \(x_3=t\) gives \(x_1=-t\), \(x_2=-t\), and hence \(\mathbf{x}=\begin{bmatrix}-t\\-t\\t\end{bmatrix}\). Taking \(t=1\) gives a nonzero homogeneous solution, so the matrix is not invertible.
For the stated right-hand side, apply the same row operations to the augmented matrix:
Set \(x_3=t\). The first two equations give \(x_1=2-t\) and \(x_2=3-t\), so the full solution set is \(\mathbf{x}=\begin{bmatrix}2-t\\3-t\\t\end{bmatrix}\). Check: the three entries of \(\mathbf{A}\mathbf{x}\) are \((2-t)+t=2\), \((3-t)+t=3\), and \((2-t)+(3-t)+2t=5\), as required.
Check Your Understanding
For each question, distinguish carefully between the homogeneous system, a particular right-hand side, and the claim about every right-hand side.
- State the three equivalent conditions in the theorem for a square real matrix.
- If \(\mathbf{A}\mathbf{x}=\mathbf{0}\) has a nonzero solution, what does this imply about invertibility?
- For a consistent system \(\mathbf{A}\mathbf{x}=\mathbf{b}\), explain why adding a homogeneous solution to a particular solution gives another solution.
- Can a noninvertible square matrix yield a consistent system with infinitely many solutions? Explain using the solution-set theorem.
- For \(\mathbf{A}=\begin{bmatrix}1&2\\2&4\end{bmatrix}\), solve \(\mathbf{A}\mathbf{x}=\begin{bmatrix}1\\2\end{bmatrix}\), and state whether the solution is unique.
