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Gauss-Jordan and the uniqueness of the reduced form, pivot and free columns, and the general solution as a particular solution plus the homogeneous ones.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to carry an elimination on to reduced row echelon form by scaling each pivot to one and clearing its whole column, decide whether a given matrix is in that form, say why the form is unique to the matrix and why that uniqueness matters, identify the pivot and free columns and the basic and free variables, detect an inconsistent system from a row that reads zero equals something non-zero, and write the general solution in parametric vector form as a particular solution plus one direction vector for each free variable.
Echelon form is enough to solve a system, one unknown at a time, from the bottom up. It is not enough to describe a solution set, because two people sweeping the same matrix can stop at two different echelon forms. Carrying the elimination further removes that freedom, and what comes out is a matrix belonging to the original one and to no other.
A column of the coefficient matrix is a pivot column if the reduced form has a pivot in it, and a free column otherwise. The variables belonging to pivot columns are basic; the rest are free. The general solution written as a fixed vector plus free-variable multiples of fixed vectors is the parametric vector form. The fixed vector is a particular solution; the vectors multiplied by the free variables span the solution set of the same system with every right-hand side set to zero.
Gauss-Jordan is Gaussian elimination that does not stop at echelon form. Two more things are done to each pivot: it is scaled to $1$, and the entries above it are cleared as well as those below. When there is nothing left to do, the matrix is in reduced row echelon form:
The reduced form is unique. Whatever order the operations were done in, whatever multiples were chosen, every route from a given matrix ends at the same reduced form. Echelon form has no such property, and that difference is why every definition later in this course — rank, the basis of a null space, which columns are independent — is stated against the reduced form. A quantity read off a non-unique object would depend on who did the arithmetic.
Reading the solution set is then mechanical. A row reading $0 = c$ with $c$ non-zero means there are no solutions and the reading stops. Otherwise each pivot row expresses one basic variable in terms of the free ones; each free variable is set to a parameter; and the result, collected into vectors, is the parametric form.
Another way: picture
Think of the reduced form as a completed crossword rather than a set of clues. Echelon form is a set of clues that have to be worked in a particular order: the last one first, then the one above with the answer filled in, and so on. The reduced form has the answers already written in the squares. The work is the same work; it has simply been done in advance, once, instead of being redone every time somebody wants a different free variable set to a different value.
Another way: steps
Take the reduced augmented matrix
$$\left(\begin{array}{rrrr|r} 1 & 0 & 2 & 0 & 3 \\ 0 & 1 & -1 & 0 & 4 \\ 0 & 0 & 0 & 1 & 5 \end{array}\right).$$
Pivots sit in columns 1, 2 and 4, so $x_1$, $x_2$ and $x_4$ are basic and $x_3$ is free. Write $x_3 = t$. The rows give $x_1 = 3 - 2t$, $x_2 = 4 + t$, $x_4 = 5$, and $x_3 = t$ is its own statement. Collecting the four coordinates,
$$x = \begin{pmatrix} 3 \\ 4 \\ 0 \\ 5 \end{pmatrix} + t\begin{pmatrix} -2 \\ 1 \\ 1 \\ 0 \end{pmatrix}.$$
Two details are worth pausing on. The free variable's own entry is $0$ in the particular solution and $1$ in the direction vector, always — that is what setting $t$ to $0$ and then to $1$ does. And the signs in the direction vector are the negatives of the entries in the free column, because the term crossed the equals sign.
With two free variables there are two direction vectors and the solution set is a plane; with $k$ of them, $k$ vectors. The particular solution is unaffected by how many there are.
Set every right-hand side to zero and the same coefficient matrix gives the homogeneous system $Ax = 0$. Reducing it is the same arithmetic — the zero column stays a zero column throughout — and the reading is simpler, because the particular solution is $0$.
What comes out is exactly the direction vectors. So the general solution of $Ax = b$ is
$$\{\text{one particular solution}\} + \{\text{every solution of } Ax = 0\},$$
which is worth stating as a habit rather than a formula. Two consequences follow immediately:
The right-hand side decides only whether there is any solution. How many there are, once there is one, is settled by the coefficients alone — and that separation is the whole content of the next lesson.
A free variable is not an unsolved variable. The solution set has been described completely; it simply is not a single point. "$x_3$ is anything, and then $x_1$ and $x_2$ are these" is a full answer, and rounding it off to one convenient choice throws away most of it.
A zero row is not an inconsistency. $0 = 0$ is a true statement and means one equation was redundant. Only $0 = c$ with $c$ non-zero kills the system, and the difference is entirely in the augmented column.
Signs in the direction vector. The free column entry $2$ becomes $-2$ in the direction vector, because $x_1 + 2x_3 = 3$ rearranges to $x_1 = 3 - 2x_3$. Copying the column across unchanged is the commonest slip in this lesson.
Reduced form of the augmented matrix, not of the coefficient matrix. Pivots are counted in the coefficient columns. A pivot in the last column is not a pivot for any variable — it is the signature of an inconsistent system.
Echelon form is not unique; reduced form is. Two correct sweeps can disagree about echelon form and must agree about the reduced form, which is why definitions are pinned to the latter.
$\left(\begin{array}{rrr|r} 2 & 4 & 2 & 10 \\ 0 & 3 & 6 & 9 \end{array}\right)$: scale row 1 by $\tfrac{1}{2}$ and row 2 by $\tfrac{1}{3}$ to make both pivots $1$.
Condition three, first.
$\left(\begin{array}{rrr|r} 1 & 2 & 1 & 5 \\ 0 & 1 & 2 & 3 \end{array}\right)$: subtract $2 \times$ row 2 from row 1 to clear above the second pivot.
Condition four, working leftwards.
$\left(\begin{array}{rrr|r} 1 & 0 & -3 & -1 \\ 0 & 1 & 2 & 3 \end{array}\right)$: with $x_3 = t$, the solution is $(-1 + 3t,\; 3 - 2t,\; t)$.
Read it straight off, signs flipped.
$\left(\begin{array}{rr|r} 1 & 2 & 3 \\ 2 & 4 & 7 \end{array}\right)$ reduces to $\left(\begin{array}{rr|r} 1 & 2 & 3 \\ 0 & 0 & 1 \end{array}\right)$: the bottom row says $0 = 1$.
Stop. There are no solutions.
$\left(\begin{array}{rr|r} 1 & 2 & 3 \\ 2 & 4 & 6 \end{array}\right)$ reduces to $\left(\begin{array}{rr|r} 1 & 2 & 3 \\ 0 & 0 & 0 \end{array}\right)$: the bottom row says $0 = 0$.
Redundant, not contradictory.
The coefficient halves are identical, so the constants alone decided between no solutions and a whole line of them: $(3 - 2t,\; t)$.
Which is why the augmented column is never dropped.
Pivots in columns 1 and 2, none in column 3, so $x_3$ is free. Put $x_3 = t$.
Name the free variable first.
Row 1 gives $x_1 = 7 - 4t$ and row 2 gives $x_2 = 1 + 2t$.
Signs flip as the terms cross over.
So $x = (7, 1, 0) + t(-4, 2, 1)$: a line of solutions, described exactly.
The augmented matrix $\begin{pmatrix} 1 & 6 & 9 \\ 0 & 1 & 6 \end{pmatrix}$ is in echelon form. Write its reduced row echelon form.
This task has no paper form; do it on a device.
A system in three unknowns reduces to $\left(\begin{array}{rrr|r} 1 & 0 & 5 & 2 \\ 0 & 1 & 5 & 4 \end{array}\right)$. Its solutions are a particular solution plus a multiple of one direction vector. Fill both in.
| Particular solution | Direction vector | |
|---|---|---|
| $x_1$ | ||
| $x_2$ | ||
| $x_3$ |
A system reduces to $\left(\begin{array}{rrr|r} 1 & 0 & 3 & 6 \\ 0 & 1 & 4 & 5 \end{array}\right)$. Take the free variable to be $x_3 = 2$. What is $x_1$?
Answer:
Which of these matrices is in reduced row echelon form?
A system in $x$ and $y$ reduces to the single row $\left(\begin{array}{rr|r} 1 & 4 & 2 \end{array}\right)$, so $y$ is free. Plot the solution with $y = 0$ and the solution with $y = 1$.
Plot your answer on the grid:
A stall sells white rolls at $1$ penny, brown rolls at $3$ pence and rye rolls at $9$ pence. A customer took $16$ rolls altogether and paid $22$ pence. More than one basket fits. Write, as a column of three numbers in the order white, brown, rye, the basket that contains no rye rolls.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The augmented matrix $\begin{pmatrix} 1 & 3 & 6 \\ 0 & 1 & 8 \end{pmatrix}$ is in echelon form. Write its reduced row echelon form.
This task has no paper form; do it on a device.
You can reduce a matrix fully, name the free variables, and write the whole solution set in parametric form. Say in your own words why the general solution of a consistent system is one solution plus every solution of the homogeneous system. Next: counting pivots, and what the count decides.
10. Your turn: solve the system whose reduced form is $\left(\begin{array}{rrr|r} 1 & 0 & 4 & 7 \\ 0 & 1 & -2 & 1 \end{array}\right)$, step 3
Particular solution plus direction.