Matrices and Systems of Equations

Learning goals

Write an augmented matrix with the variables stripped away
Apply the three row operations, each reversible
Reduce to row-echelon form, then back-substitute
Read a contradiction row as no solution
Recognize a zero row as a dependent system
Require a pivot in every variable column for uniqueness

From equations to a grid of numbers

A matrix is a rectangular array of numbers, written inside square brackets. Nothing more mysterious than that. Here is one with two rows and three columns:

\left[\begin{array}{ccc} 2 & 3 & 8 \\ 5 & 4 & 7 \end{array}\right]

The reason a matrix is useful right now is that a system of linear equations is already a grid of numbers in disguise. Take the system

\begin{aligned} 2x + 3y &= 8, \\ 5x + 4y &= 7. \end{aligned}

Line up the variables in the same order in every equation, keep the constants on the right, and read off the coefficients row by row. The two coefficients on the left of each equation become the first two columns, and the constant becomes a third column. To remember that the third column plays a different role, we draw a vertical bar in front of it. The result is the augmented matrix of the system:

\left[\begin{array}{cc|c} 2 & 3 & 8 \\ 5 & 4 & 7 \end{array}\right]

The bar stands for the equals signs. Everything left of it is a coefficient of a variable; the single column to its right holds the constants. Drop the bar and the first two columns on their own form the coefficient matrix. The augmented matrix carries strictly more information, because it remembers the right-hand sides too, which is why it is the one we reduce.

The system on the left and its augmented matrix on the right hold exactly the same information. Every step of the solution works on the numbers alone, because the letters x and y never did any of the arithmetic.

Two small conventions keep the translation honest. A variable that is written with no number in front of it has coefficient $1$ , so $x - y = 1$ contributes a row of $1$ and $-1$ . A variable that is missing from an equation has coefficient $0$ . That $0$ is not optional: it holds the column open so the other coefficients stay in their correct places, exactly the placeholder job a zero does in place value. Write $x + z = 4$ inside a three-variable system and its row is $1, 0, 1$ , with the $0$ marking the absent $y$ .

Dimensions and entries

A matrix with $m$ rows and $n$ columns is called an $m \times n$ matrix (say ” $m$ by $n$ ”), always rows first. The augmented matrix above is $2 \times 3$ . To point at one number inside a matrix $A$ , we name its row and its column: the entry in row $i$ and column $j$ is written $a_{ij}$ . That notation uses the same letter as the matrix, but lowercase, with the row number first and the column number second. So in

A = \left[\begin{array}{cc|c} 2 & 3 & 8 \\ 5 & 4 & 7 \end{array}\right],

the entry $a_{11} = 2$ sits in the top left, $a_{13} = 8$ is the first constant, and $a_{21} = 5$ is the coefficient of $x$ in the second equation. Row first, column second, every time. That is the entire vocabulary you need; matrices carry much more machinery in later chapters, but here the matrix exists only to organize elimination, so this is where we stop.

Check your understanding

A system has the equations $2x + 3y = 8$ and $5x + 4y = 7$ . In its augmented matrix, what is the entry $a_{21}$ (row 2, column 1)?

4

7

5

2

The three row operations

Elimination lets you do three things to a system without changing which values of the variables solve it. Written on the augmented matrix, where each row is one equation, they become the three row operations:

Swap two rows. ( $R_i \leftrightarrow R_j$ .) You are just listing the equations in a different order.
Multiply a row by a nonzero constant. ( $R_i \to kR_i$ with $k \ne 0$ .) You are scaling one whole equation, both sides at once.
Add a multiple of one row to another. ( $R_i \to R_i + kR_j$ .) This is the elimination step itself: you add a chosen multiple of one equation to a second so that a variable cancels in the second.

These are the only moves you are allowed, and every one of them is reversible, which is the property that matters. Swapping back undoes a swap; dividing by $k$ undoes a multiply; subtracting the same multiple undoes an addition. Because you can always walk backward, no information is lost, and the reduced system has neither gained nor lost any solutions. Here is the argument in full.

Why the three row operations preserve the solution set#

A solution of a system is a choice of values for the variables that makes every equation true at once. We check each operation against that definition.

Reordering the rows changes nothing about the equations themselves, only the order in which they are listed. So a choice of values that satisfies all of them before the swap satisfies all of them after. The solution set is untouched.

Multiplying row $R_i$ by a nonzero constant $k$ replaces the equation $E$ with $kE$ . Any values that make $E$ true make $kE$ true, since multiplying both sides of a true equation by $k$ keeps it true. Conversely, any values that make $kE$ true also make $E$ true, because $k \ne 0$ lets you divide both sides by $k$ and recover $E$ . So $E$ and $kE$ have precisely the same solutions, and the rest of the system is unchanged. This is exactly where the word nonzero earns its place. Multiplying by $0$ would turn $E$ into $0 = 0$ , which every choice of values satisfies, and the original equation could never be recovered. That step would throw a genuine constraint away.

Adding $k$ times row $R_j$ to row $R_i$ replaces the equation $E_i$ with $E_i + kE_j$ , while $E_j$ itself stays in the system. Suppose some values satisfy the original system. Then they make both $E_i$ and $E_j$ true, so they make $E_i + kE_j$ true as well, and they still satisfy the untouched $E_j$ . So every old solution is a solution of the new system. For the reverse, suppose some values satisfy the new system. They make $E_i + kE_j$ true and, because $E_j$ is still present, they make $E_j$ true too. Subtracting $k$ times $E_j$ from $E_i + kE_j$ returns $E_i$ , so those values satisfy $E_i$ as well. Every new solution is therefore an old solution. The two systems have identical solution sets.

Since each operation leaves the solution set exactly as it was, any sequence of them does too. Whatever you reduce the matrix to, you may read its solutions as the solutions of the system you started with.

Check your understanding

Which of these is not a legal row operation on an augmented matrix?

A Swap the first and third rows B Multiply a row by

-2

C Add

3

times the first row to the second row D Multiply a row by

0

Row-echelon form and back-substitution

The goal of the row operations is to reach row-echelon form, the matrix version of the triangular shape you drove systems toward in the previous lessons. The leading entry of a row is its first nonzero number, and we call it a pivot. A matrix is in row-echelon form when each pivot sits strictly to the right of the pivot in the row above, and any all-zero rows lie at the bottom. For a square system with a single solution this makes a clean staircase: the entries below the diagonal are all zero.

Row-echelon (triangular) form. Each circled leading entry sits one column to the right of the one above it, and every entry in the triangle below them is zero. The bottom row pins the last variable down, and from there you climb back up.

Once the matrix is triangular, you translate the bottom row back into an equation and solve it, then work upward. Each row you climb into has only one new variable in it, because everything below its pivot is zero. So you substitute the values you already have and solve for that one new variable. Working from the bottom up like this is called back-substitution, and it is the reason the triangular shape is worth chasing.

Row reduction in action

Worked example 1 Solve a $2\times2$ system by row reduction

Solve the system

\begin{aligned} x + 2y &= 5, \\ 3x - y &= 1. \end{aligned}

Write the augmented matrix, keeping the variables in the order $x, y$ :

\left[\begin{array}{cc|c} 1 & 2 & 5 \\ 3 & -1 & 1 \end{array}\right]

The first column already has a $1$ on top, a convenient pivot. Clear the $3$ beneath it by replacing $R_2$ with $R_2 - 3R_1$ . Subtract $3$ times each entry of row $1$ from row $2$ :

\left[\begin{array}{cc|c} 1 & 2 & 5 \\ 0 & -7 & -14 \end{array}\right]

Now scale row $2$ so its pivot becomes $1$ , replacing $R_2$ with $-\tfrac{1}{7}R_2$ :

\left[\begin{array}{cc|c} 1 & 2 & 5 \\ 0 & 1 & 2 \end{array}\right]

The matrix is triangular. Read the bottom row back as an equation: it says $y = 2$ . Back-substitute into the top row, which reads $x + 2y = 5$ :

x + 2(2) = 5 \;\Longrightarrow\; x = 1.

The solution is $(x, y) = (1, 2)$ . A quick check in the original second equation confirms it: $3(1) - 2 = 1$ .

Worked example 2 Solve a $3\times3$ system by row reduction

Solve the system

\begin{aligned} x + y + z &= 6, \\ 2x + 3y + z &= 11, \\ x - y + 2z &= 5. \end{aligned}

Build the augmented matrix:

\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 2 & 3 & 1 & 11 \\ 1 & -1 & 2 & 5 \end{array}\right]

Use the pivot $1$ in the top left to clear the rest of column $1$ . Replace $R_2$ with $R_2 - 2R_1$ and $R_3$ with $R_3 - R_1$ :

\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & -1 & -1 \\ 0 & -2 & 1 & -1 \end{array}\right]

Now the pivot $1$ in row $2$ , column $2$ clears the $-2$ below it. Replace $R_3$ with $R_3 + 2R_2$ :

\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & -1 & -3 \end{array}\right]

Scale row $3$ to make its pivot $1$ , replacing $R_3$ with $-R_3$ :

\left[\begin{array}{ccc|c} 1 & 1 & 1 & 6 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 1 & 3 \end{array}\right]

This is triangular, so back-substitute from the bottom. Row $3$ gives $z = 3$ . Row $2$ reads $y - z = -1$ :

y - 3 = -1 \;\Longrightarrow\; y = 2.

Row $1$ reads $x + y + z = 6$ , and both other values are known:

x + 2 + 3 = 6 \;\Longrightarrow\; x = 1.

The solution is $(x, y, z) = (1, 2, 3)$ , the values the echelon matrix in the figure above was built from.

When a row gives it away

Not every system has a single solution. In the previous lessons you met the trichotomy: a linear system has exactly one solution, no solution, or infinitely many. In matrix form these last two cases announce themselves during reduction, in a single tell-tale row.

Worked example 3 Spot the no-solution and infinitely-many cases

Start with a system that has no solution:

\begin{aligned} x + y &= 2, \\ 2x + 2y &= 5. \end{aligned}

Reduce its augmented matrix by replacing $R_2$ with $R_2 - 2R_1$ :

\left[\begin{array}{cc|c} 1 & 1 & 2 \\ 2 & 2 & 5 \end{array}\right] \;\longrightarrow\; \left[\begin{array}{cc|c} 1 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right]

Translate the bottom row back into an equation. Its coefficients are all zero, but the constant is not: it reads $0x + 0y = 1$ , that is, $0 = 1$ . No values of $x$ and $y$ can make that true, so the system has no solution. A row of the form $[\,0 \ \ 0 \ \mid\ c\,]$ with $c \ne 0$ is the matrix shouting contradiction.

Now change a single number so the system has infinitely many solutions:

\begin{aligned} x + y &= 2, \\ 2x + 2y &= 4. \end{aligned}

The same operation $R_2 \to R_2 - 2R_1$ now gives:

\left[\begin{array}{cc|c} 1 & 1 & 2 \\ 2 & 2 & 4 \end{array}\right] \;\longrightarrow\; \left[\begin{array}{cc|c} 1 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right]

This time the bottom row is entirely zero, standing for $0 = 0$ . That is true for every choice of values, so it adds no constraint at all: the second equation was really just the first one doubled. What remains is the single equation $x + y = 2$ . Let $y = t$ be free; then $x = 2 - t$ , and every $(2 - t,\, t)$ is a solution. The system is dependent, with a whole line of solutions.

The pattern is worth stating plainly, because it is how you diagnose a system straight from its reduced matrix. A row of zeros on the left with a nonzero constant on the right, $[\,0 \ \cdots\ 0 \mid c\,]$ with $c \ne 0$ , means no solution. A row that is entirely zero, $[\,0 \ \cdots\ 0 \mid 0\,]$ , carries no information and signals a dependent system, which has infinitely many solutions as long as no contradiction row appears elsewhere. In that dependent case, each variable column without a pivot is a free variable. You assign each free variable a parameter, one parameter per pivot-free column, and solve for the pivoted variables in terms of those parameters. If neither kind of row shows up and every variable ends with its own pivot, the solution is unique.

Check your understanding

While reducing a system's augmented matrix, you reach a row whose left entries are all zero but whose constant is $4$ , standing for the equation $0 = 4$ . What does this tell you about the system?

A It has exactly one solution B It has no solution C It has infinitely many solutions D It has exactly two solutions

Common mistakes

Forgetting the constant column. The augmented matrix must keep the right-hand sides, separated by the bar. Reduce only the coefficient matrix and you lose track of what each equation equals.
Dropping a zero coefficient. A variable missing from an equation still needs a $0$ in its column, holding the place so the other coefficients line up. Skipping it shifts entries into the wrong columns.
Multiplying a row by zero. Scaling a row is legal only by a nonzero constant. Multiplying by $0$ wipes an equation out to $0 = 0$ and discards a real constraint.
Only partly updating a row. In $R_2 \to R_2 - 3R_1$ , every entry of row $2$ changes, including the constant. Change the coefficients but forget the constant (or the reverse) and the row no longer stands for a true equation.
Misreading the two special rows. $[\,0 \ \ 0 \mid c\,]$ with $c \ne 0$ means no solution, while a fully zero row $[\,0 \ \ 0 \mid 0\,]$ adds no new constraint. If the system is otherwise consistent and has a free variable, that zero row marks the dependent case, with infinitely many solutions. They look almost alike but say opposite things, so check the constant.
Reading a triangular matrix top-down. Back-substitution starts at the bottom row, where a single variable is pinned down, and climbs up. Starting at the top leaves you with too many unknowns at once.

Key takeaways

A matrix is a rectangular array of numbers; an $m \times n$ matrix has $m$ rows and $n$ columns, and $a_{ij}$ is the entry in row $i$ , column $j$ .
The augmented matrix of a system holds each equation’s coefficients as a row, with the constants in a final column set off by a bar. It is the coefficients of the system with the variables stripped away.
The three row operations are swapping two rows, multiplying a row by a nonzero constant, and adding a multiple of one row to another. All three are exactly the moves of elimination, and each is reversible, so none of them changes the solution set.
Reduce to row-echelon (triangular) form, then back-substitute from the bottom row upward to read off the solution.
A reduced row $[\,0 \ \cdots\ 0 \mid c\,]$ with $c \ne 0$ means no solution; a fully zero row means a dependent system with infinitely many solutions, provided no contradiction row appears. A pivot in every variable’s column means a unique solution when no contradiction row appears.

Matrices and Systems of Equations

Learning goals

From equations to a grid of numbers

Dimensions and entries

The three row operations

Row-echelon form and back-substitution

Row reduction in action

When a row gives it away

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)