1.1 Systems of linear equations

Big idea

A linear equation in the unknowns $x_1, x_2, \dots, x_n$ looks like $a_1x_1 + a_2x_2 + \cdots + a_nx_n = b$. The coefficients $a_i$ and the constant $b$ are fixed numbers. No unknown is squared, inverted, multiplied by another unknown, or fed into a function. A system of linear equations is a finite list of such equations in the same unknowns, and a solution is a list of numbers that satisfies every equation at once.

The section answers two questions about any system: does a solution exist, and if one exists, is it the only one. Almost everything later in the chapter restates those two questions in new language. Span, independence, matrix equations and transformations are all ways of asking existence and uniqueness about the same object.

The tool is elimination. You replace the system by a simpler system with exactly the same solution set, repeat, and stop when the answer is visible. The bookkeeping moves out of the equations and into a grid of numbers, because the unknowns never change and rewriting them each line is wasted effort.

Decoder

The dense sentences in this section are all about what “the same” means for two systems.

Two systems count as equivalent when they have the same solution set, and each of the three row operations turns a system into an equivalent one.

Read it backwards. The operations are the point: swap two equations, scale an equation by a nonzero number, add a multiple of one equation to another. None of those three changes which lists of numbers satisfy the system, because each one can be undone by another operation of the same kind. So you may apply them freely and the answer you reach at the end is the answer you started with.

Two words get used in place of “the same”. Equivalent systems have the same solution set. Row equivalent matrices are connected by a chain of row operations. They say the same thing, one about equations and one about the grid.

The other word to pin down is consistent. A system is consistent when it has at least one solution, whether that is one solution or infinitely many. It is inconsistent when it has none. Consistency is the existence question; counting solutions is the uniqueness question.

Definitions and results

Linear equation. An equation of the form $a_1x_1 + \cdots + a_nx_n = b$, with the coefficients and $b$ fixed numbers.

System, solution, solution set. A system is a finite list of linear equations in the unknowns $x_1, \dots, x_n$. A solution is an ordered list $(s_1, \dots, s_n)$ that makes every equation true when $x_i = s_i$. The solution set is the collection of all solutions. Two systems are equivalent when their solution sets are equal.

How many solutions are possible. A linear system has no solution, exactly one solution, or infinitely many solutions. Nothing else can happen. Two lines in the plane miss each other, cross once, or coincide; the same trichotomy holds in every dimension.

Coefficient and augmented matrix. Strip the unknowns and the equal signs. The coefficient matrix holds the coefficients, one row per equation and one column per unknown. The augmented matrix appends one more column holding the constants on the right-hand side. An $m$-equation, $n$-unknown system gives an $m \times n$ coefficient matrix and an $m \times (n+1)$ augmented matrix.

Elementary row operations. Swap two rows. Multiply a row by a nonzero constant. Add a multiple of one row to another row. Each is reversible by an operation of the same type.

Row equivalence preserves solutions. If the augmented matrices of two systems are row equivalent, the two systems have the same solution set. This is the licence for the entire method: eliminate as far as you like, then read the answer off the simple system.

Existence and uniqueness. These are the two questions to ask about any system. Is it consistent? If so, is the solution unique? Elimination answers both. A row that reads $0 = c$ with $c$ nonzero means there is no solution. Otherwise there is at least one, and the solution is unique exactly when every unknown is pinned down by a row rather than left free.

Worked examples

A system with one solution

Solve

$$ x_1 + 2x_2 + x_3 = 9 $$

$$ 2x_1 + 3x_2 - x_3 = 4 $$

$$ 3x_1 - x_2 + 2x_3 = 9 $$

The augmented matrix, with the last column holding the right-hand sides, is

$$ \begin{bmatrix} 1 & 2 & 1 & 9 \\ 2 & 3 & -1 & 4 \\ 3 & -1 & 2 & 9 \end{bmatrix} $$

Clear the first column. Subtract $2$ times row 1 from row 2, then $3$ times row 1 from row 3:

$$ \begin{bmatrix} 1 & 2 & 1 & 9 \\ 0 & -1 & -3 & -14 \\ 0 & -7 & -1 & -18 \end{bmatrix} $$

Clear the second column below the leading $-1$. Subtract $7$ times row 2 from row 3:

$$ \begin{bmatrix} 1 & 2 & 1 & 9 \\ 0 & -1 & -3 & -14 \\ 0 & 0 & 20 & 80 \end{bmatrix} $$

The last row says $20x_3 = 80$, so $x_3 = 4$. Row 2 says $-x_2 - 3(4) = -14$, so $x_2 = 2$. Row 1 says $x_1 + 2(2) + 4 = 9$, so $x_1 = 1$.

Check $(1, 2, 4)$ in the original equations: $1 + 4 + 4 = 9$, $2 + 6 - 4 = 4$, $3 - 2 + 8 = 9$. All three hold, and no unknown was left free, so this is the only solution.

A system with no solution

Solve

$$ x_1 - 2x_2 + x_3 = 1 $$

$$ 2x_1 - 3x_2 + x_3 = 3 $$

$$ 3x_1 - 5x_2 + 2x_3 = 5 $$

Subtract $2$ times row 1 from row 2 and $3$ times row 1 from row 3:

$$ \begin{bmatrix} 1 & -2 & 1 & 1 \\ 0 & 1 & -1 & 1 \\ 0 & 1 & -1 & 2 \end{bmatrix} $$

Subtract row 2 from row 3:

$$ \begin{bmatrix} 1 & -2 & 1 & 1 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$

The last row reads $0x_1 + 0x_2 + 0x_3 = 1$, which no numbers satisfy. The system is inconsistent and the solution set is empty.

Check the reasoning without the matrix: the third equation minus the first equals $2x_1 - 3x_2 + x_3 = 4$, while the second equation says that same combination equals $3$. Two different values for one expression, so nothing can work.

A system with infinitely many solutions

Keep the same left-hand sides and change the constants to $1$, $3$, $4$. The same two operations give

$$ \begin{bmatrix} 1 & -2 & 1 & 1 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$

The bottom row is now $0 = 0$, which is true and carries no information. Two rows constrain three unknowns, so one unknown is free. Let $x_3$ take any value. Row 2 gives $x_2 = 1 + x_3$. Row 1 gives $x_1 = 1 + 2x_2 - x_3 = 3 + x_3$.

Check two members of the family. With $x_3 = 0$ you get $(3, 1, 0)$: $3 - 2 + 0 = 1$, $6 - 3 + 0 = 3$, $9 - 5 + 0 = 4$. With $x_3 = 1$ you get $(4, 2, 1)$: $4 - 4 + 1 = 1$, $8 - 6 + 1 = 3$, $12 - 10 + 2 = 4$. Both work, so the solution set is infinite.

Practice

Elimination first. These drills hand you a system and ask for the matrix after the forward sweep, so you get fast at clearing a column without arithmetic slips.

Then finish the job. These systems have exactly one solution, and you report the values of the unknowns after back substitution.

Last, the classification question on its own. You decide whether a system has no solution, one solution, or infinitely many, without being asked to write the solution down.

Videos

Watch the opening MIT lecture for the two pictures of a system: rows as intersecting planes, columns as vectors that must combine to reach the target. The column picture is the one that pays off in the next few sections.

Quiz

Six mixed items over elimination, solving, and classification.