2.2: When There is No Unique Solution
Given \(n\) equations and \(n\) unknowns, one usually expects a unique solution. But two other possibilities exist: there could be no solution, or an infinite number of solutions. We will illustrate what happens during Gaussian elimination in these two cases. Consider
\[\begin{aligned}-3x_1+2x+2-x_3&=-1, \\ 6x_1-6x_2+7x_3&=-7, \\ 3x_1-4x_2+6x_3&=b.\end{aligned} \nonumber \]
Note that the first two equations are the same as in (2.1.1) , but the left-hand-side of the third equation has been replaced by the sum of the left-hand-sides of the first two equations, and the right-hand-side has been replaced by the parameter \(b\). If \(b = −8\), then the third equation is just the sum of the first two equations and adds no new information to the system. In this case, the equations should admit an infinite number of solutions. However, if \(b\neq −8\), then the third equation is inconsistent with the first two equations and there should be no solution.
We solve by Gaussian elimination to see how it plays out. Writing the augmented matrix and doing row elimination, we have
\[\left(\begin{array}{rrrr}-3&2&-1&-1\\6&-6&7&-7\\3&-4&6&b\end{array}\right)\to\left(\begin{array}{rrrc}-3&2&-1&-1\\0&-2&5&-9\\0&-2&5&b-1\end{array}\right)\to\left(\begin{array}{rrrc}-3&2&-1&-1 \\ 0&-2&5&-9\\0&0&0&b+8\end{array}\right).\nonumber \]
Evidently, Gaussian elimination has reduced the last row of the matrix \(\text{A}\) to zeros, and the last equation becomes
\[0=b+8.\nonumber \]
If \(b\neq −8\), there will be no solution, and if \(b = −8\), the under-determined systems of equations becomes
\[\begin{aligned}-3x_1+2x_2-x_3&=-1 \\ -2x_2+5x_3&=-9.\end{aligned} \nonumber \]
The unknowns \(x_1\) and \(x_2\) can be solved in terms of \(x_3\) as
\[x_1=\frac{10}{3}+\frac{4}{3}x_3,\quad x_2=\frac{9}{2}+\frac{5}{2}x_3,\nonumber \]
indicating an infinite family of solutions dependent on the free choice of \(x_3\).
To be clear, for a linear system represented by \(\text{Ax} = \text{b}\), if there is a unique solution then \(\text{A}\) is invertible and the solution is given formally by
\[\text{x}=\text{A}^{-1}\text{b}.\nonumber \]
If there is not a unique solution, then \(\text{A}\) is not invertible. We then say that the matrix \(\text{A}\) is singular. Whether or not an \(n\)-by-\(n\) matrix \(\text{A}\) is singular can be determined by row reduction on \(\text{A}\). After row reduction, if the last row of \(\text{A}\) is all zeros, then \(\text{A}\) is a singular matrix; if not, then \(\text{A}\) is an invertible matrix. We have already shown in the two-by-two case, that \(\text{A}\) is invertible if and only if \(\det\text{ A}\neq 0\), and we will later show that this is also true for \(n\)-by-\(n\) matrices.