The system can be written as \( \begin{bmatrix} 3 & -2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 1\\ 5 \end{bmatrix}. \) (Beware that the variables in the second equation are given out of order.) The augmented matrix of the system is \[ \left[\begin{array}{rrr|r} 3 & -2 & -1 & 1\\ 1 & -1 & 2 & 5 \end{array}\right] \] Performing \(R_1 \leftrightarrow R_2\) gives \[ \left[\begin{array}{rrr|r} 1 & -1 & 2 & 5 \\ 3 & -2 & -1 & 1 \end{array}\right] \] Performing \(R_2 \leftarrow R_2 - 3 R_1\) gives \[ \left[\begin{array}{rrr|r} 1 & -1 & 2 & 5 \\ 0 & 1 & -7 & -14 \end{array}\right] \] Performing \(R_1 \leftarrow R_1 + R_2\) gives \[ \left[\begin{array}{rrr|r} 1 & 0 & -5 & -9 \\ 0 & 1 & -7 & -14 \end{array}\right] \] So \(z\) is a free variable. Setting \(z = s\), the second row gives \(y - 7s = -14\) and the first row gives \(x - 5s = -9\). Hence, the solutions are \(\begin{bmatrix} x\\y\\z \end{bmatrix} = \begin{bmatrix} -9 + 5s \\ -14 + 7s \\ s\end{bmatrix}\) for all \(s \in \mathbb{R}\).

The augmented matrix for the system is \(\left[\begin{array}{rrrr|r} 1 & 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 0 & 1\\ 0 & 1 & 1 & 1 & 0\\ 1 & 1 & 0 & 1 & 0 \end{array}\right].\) We row reduce it as follows (keeping in mind that \(1+1=0\) in \(GF(2)\)): \[\begin{array}{rl} & \left[\begin{array}{rrrr|r} 1 & 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 0 & 1\\ 0 & 1 & 1 & 1 & 0\\ 1 & 1 & 0 & 1 & 0 \end{array}\right] \\ \xrightarrow{R_2\leftarrow R_2 + R_1} & \left[\begin{array}{rrrr|r} 1 & 1 & 1 & 1 & 1\\ 0 & 0 & 0 & 1 & 0\\ 0 & 1 & 1 & 1 & 0\\ 1 & 1 & 0 & 1 & 0 \end{array}\right] \\ \xrightarrow{R_4 \leftarrow R_4 + R_1} & \left[\begin{array}{rrrr|r} 1 & 1 & 1 & 1 & 1\\ 0 & 0 & 0 & 1 & 0\\ 0 & 1 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 & 1 \end{array}\right] \\ \xrightarrow{R_1 \leftarrow R_1 + R_3} & \left[\begin{array}{rrrr|r} 1 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1 & 0\\ 0 & 1 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 & 1 \end{array}\right] \\ \xrightarrow{R_2 \leftrightarrow R_3} & \left[\begin{array}{rrrr|r} 1 & 0 & 0 & 0 & 1\\ 0 & 1 & 1 & 1 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 1 \end{array}\right] \\ \xrightarrow{R_3 \leftrightarrow R_4} & \left[\begin{array}{rrrr|r} 1 & 0 & 0 & 0 & 1\\ 0 & 1 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \end{array}\right] \\ \xrightarrow{R_2 \leftrightarrow R_2+R_3} & \left[\begin{array}{rrrr|r} 1 & 0 & 0 & 0 & 1\\ 0 & 1 & 0 & 1 & 1\\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \end{array}\right] \\ \xrightarrow{R_2 \leftrightarrow R_2+R_4} & \left[\begin{array}{rrrr|r} 1 & 0 & 0 & 0 & 1\\ 0 & 1 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \end{array}\right] \\ \end{array} \] Hence, \(\begin{bmatrix} x_1\\x_2\\x_3\\x_4 \end{bmatrix} = \begin{bmatrix} 1\\1\\1\\0 \end{bmatrix}\) is the unique solution.

Observe that \(A\) is already in RREF. The free variables are \(x_2\) and \(x_4\). Hence, we may set \(x_2 = s\) and \(x_4 = t\) where \(s,t \in \mathbb{R}\).

Note that the system is \begin{eqnarray*} x_1 + 2x_2 + x_4 & = & 0 \\ x_3 & = & 0. \end{eqnarray*} Hence, \(x_1 = -2s - t\) and \(x_3 = 0\). Thus, the solutions are \(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4\end{bmatrix} = \begin{bmatrix} -2s -t \\ s \\ 0 \\t \end{bmatrix} = s\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0\end{bmatrix} + t \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1\end{bmatrix}\) for all possible choices of real numbers \(s\) and \(t\).

Observe that \(A\) and \(B\) differ in two rows; namely, row 2 and row 3. The requirement that \(A\) be transformed into \(B\) with only two elementary row operations forces us to perform one operation that alters the second row and another that alters the third row.

Note that the third row of \(A\) has a nonzero in the first column and the only other row of \(A\) that has the same property is the first row. Thus, we can eliminate the leading \(1\) in the third row only using the operation \(R_3 \leftarrow R_3 - R_1\). The result is \(\begin{bmatrix} 1 & 2 & 0 \\ 0 & 3 & -6 \\0 & 1 & 5\end{bmatrix}\). Let us call this matrix \(A'\).

Notice that \(A'\) and \(B\) differ only in the second row. In fact, row 2 of \(A'\) is simply 3 times row 2 of \(B\). Thus, \(A' \xrightarrow{R_2 \leftarrow \frac{1}{3} R_2} B\).

The augmented matrix for the system is \[\left[\begin{array}{rcc|c} 1 & -1 & 1 & 1 \\ -1 & 2 & 1 & -1 \\ 0 & 1 & a & 0 \end{array}\right].\] Performing \(R_2 \leftarrow R_2 + R_1\) gives \[\left[\begin{array}{rcr|c} 1 & -1 & 1 & 1 \\ 0 & 1 & 2 & 0 \\ 0 & 1 & a & 0 \end{array}\right].\] Performing \(R_3 \leftarrow R_3 - R_2\) gives \[\left[\begin{array}{rcc|c} 1 & -1 & 1 & 1 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & a-2 & 0 \end{array}\right].\] Performing \(R_1 \leftarrow R_1 + R_2\) gives \[\left[\begin{array}{rrc|c} 1 & 0 & 3 & 1 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & a-2 & 0 \end{array}\right].\] If \(a \neq 2\), then \(a-2 \neq 0\) and so the third row gives \((a-2)z = 0\) implying that \(z = 0\). Substituting this into the equation corresponding to the second row, we get \(y = 0\). Hence, \(x = 1\). So there is a unique solution in this case.

If \(a = 2\), then \(a-2 = 0\) and the augmented matrix is in RREF. We have \(z\) as a free variable and we get infinitely many solutions with \(y = -2z\) and \(x = 1 - 3z\).

In summary, the system has infinitely many solutions if and only if \(a = 2\).

The augmented matrix for the system is \[\left[\begin{array}{rcc|c} 1 & 1 & -4 & 1 \\ -2 & 1 & -1 & -1 \\ 1 & -1 & 2 & 0 \end{array}\right].\] Performing \(R_2 \leftarrow R_2 + 2R_1\) gives \[\left[\begin{array}{rcr|c} 1 & 1 & -4 & 1 \\ 0 & 3 & -9 & 1 \\ 1 & -1 & 2 & 0 \end{array}\right].\] Performing \(R_3 \leftarrow R_3 - R_1\) gives \[\left[\begin{array}{rcc|c} 1 & 1 & -4 & 1 \\ 0 & 3 & -9 & 1 \\ 0 & -2 & 6 & -1 \\ \end{array}\right].\] Performing \(R_2 \leftarrow \frac{1}{3} R_2\) gives \[\left[\begin{array}{rcc|c} 1 & 1 & -4 & 1 \\ 0 & 1 & -3 & \frac{1}{3} \\ 0 & -2 & 6 & -1 \\ \end{array}\right].\] Performing \(R_3 \leftarrow R_3 + 2R_2 \) gives \[\left[\begin{array}{rrc|c} 1 & 1 & -4 & 1 \\ 0 & 1 & -3 & \frac{1}{3} \\ 0 & 0 & 0 & - \frac{2}{3} \end{array}\right].\] The last row represents the equation \(0 = -\frac{2}{3}\), which is impossible. Hence, the system has no solution.

The answer is “yes”. But before we look at the solution, we first take a look at how one could approach such a question.

Unless you have seen this question before, you probably would not know which way the answer goes. You might have a guess but it has to be verified.

Since there are more variables than there are equations, we know right away that not every column can be a pivot column in the reduced row-echelon form of the augmented matrix. Therefore, there must be at least one free variable and we are tempted to answer in the negative.

However, we must be careful here and ask, “Does the existence of a free variable imply that there must be a solution?” The criterion for being inconsistent is for the row-reduced augmented matrix to have a row of the form \([0 \cdots 0 \mid 1]\). It does not involve the relationship between the number of variables and the number of equations. Hence, if our system is simply \begin{eqnarray*} x + y + z & = &0 \\ 0x +0 y +0 z & = &1, \end{eqnarray*} then we have a system with more variables than equations that is inconsistent. If we don't like the look of the second equation, we can simply add the first equation to the second to obtain the system \begin{eqnarray*} x + y + z & = & 0 \\ x + y + z & = & 1. \end{eqnarray*} Clearly, \(x+y+z\) cannot equal to both 1 and 0 at the same time and so the system is inconsistent. The system also happens to have more variables than equations.

This result means that one can talk about the reduced row-echelon form of \(A\).

Suppose that \(A\) can be row reduced to matrices \(B\) and \(C\) in RREF via different sequences of elementary row operations. We want to show that \(B = C\).

First, note that the homogeneous systems \(Bx = 0\) and \(Cx = 0\) must have the same solutions as \(Ax = 0\).

Suppose that the earliest column at which \(B\) and \(C\) differ is column \(i\). Because \(B\) and \(C\) are in RREF, \(B_i\) (column \(i\) of \(B\)) and \(C_i\) cannot both be pivot columns. Without loss of generality, we may assume that \(C_i\) is not a pivot column. Hence, \(x_i\) is a free variable in \(Cx = 0\). Now, solving for \(x\) in \(Cx = 0\) after setting \(x_i = 1\) and all the remaining free variables to \(0\), we obtain a solution \(x'\) such that \(x'_i = 1\) and \(x'_j = 0\) for all \(j \gt i\). Since \(Cx' = 0\) and \(B_j = C_j\) for \(j = 1,\ldots,i-1\), we have \[Bx' = Bx' - Cx' = \sum_{j=1}^n B_jx'_j-\sum_{j=1}^n C_jx'_j = \sum_{j=1}^n (B_j-C_j)x'_j = \sum_{j=1}^i (B_j-C_j)x'_j = (B_i - C_i)x'_i = B_i - C_i \neq 0.\] So \(x'\) is not a solution of \(Bx = 0\), contradicting that \(Bx = 0\) and \(Cx = 0\) have the same solutions.

(The above proof is an example of a minimality argument. Choosing the smallest column index \(i\) at which the two matrices differ allows us to ignore the rest of the matrices. The technique is often useful in proving counterexamples cannot exist by considering a smallest counterexample and arriving at a contradiction.)