Solve 3 linear equations in 3 unknowns by row reduction: 2x + y - z = 8, -3x - y + 2z = -11, -2x + y + 2z = -3, step by step

1

Write the system as one block of numbers

Why this step

While we solve, the letters $x, y, z$ and the $=$ signs never change; only the numbers do. Carrying the letters through every line is pure copying, and copying is where slips happen. So we strip the system down to its numbers, keeping each number's POSITION to remember what it multiplies. That block is the augmented matrix.

\begin{aligned} 2 x + y - z &= 8 \\ -3 x - y + 2 z &= -11 \\ -2 x + y + 2 z &= -3 \end{aligned}

The system. Each unknown stands for a number we are trying to find; a solution is a choice of all of them that makes every equation true at once.

\left[ \begin{array}{rrr|r} 2 & 1 & -1 & 8 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3 \end{array} \right]

The augmented matrix. Column

1

holds the numbers in front of

x

, column

2

those in front of

y

, and so on. The upright bar stands where the equals signs were. A missing unknown is a

0

.

The system, as one block of numbers

how much xhow much yhow much zequals

equation 1equation 2equation 3

2

1

-1

8

-3

-1

2

-11

-2

1

2

-3

The same system with the letters stripped out. Every column holds the numbers in front of one unknown, every row is one equation, and the upright rule stands where the equals signs were. Nothing has been lost: the POSITION of each number now carries what the letter used to say.

2

Pivot 1: column 1, row 1

Why this step

We work one column at a time, left to right. The goal for a column is a single $1$ (the pivot) with zeros everywhere else in it. A column in that state says something very plain: its variable appears in exactly one equation, with nothing in front of it. Each cleared column is one variable pinned down.

Multiply row 1 by $\frac{1}{2}$ , which turns the pivot $2$ into $1$ . Multiplying an equation through by a number that is not zero keeps it true for exactly the same values.

column	working	new entry
$1$	$\frac{1}{2} \cdot 2$	$1$
$2$	$\frac{1}{2} \cdot 1$	$\frac{1}{2}$
$3$	$\frac{1}{2} \cdot \left( -1 \right)$	$-\frac{1}{2}$
$4$	$\frac{1}{2} \cdot 8$	$4$

Row 1, entry by entry.

\left[ \begin{array}{rrr|r} \boxed{1} & \frac{1}{2} & -\frac{1}{2} & 4 \\ -3 & -1 & 2 & -11 \\ -2 & 1 & 2 & -3 \end{array} \right]

The matrix now. The boxed entry is the pivot being worked on. checked

Replace row 2 by $R_{2} + 3 R_{1}$ . The multiplier is chosen so that the entry under attack, $-3$ , is cancelled exactly by the pivot: $-3 + 3 \cdot 1 = 0$ .

column	working	new entry
$1$	$-3 + 3 \cdot 1$	$0$
$2$	$-1 + 3 \cdot \frac{1}{2}$	$\frac{1}{2}$
$3$	$2 + 3 \cdot \left( -\frac{1}{2} \right)$	$\frac{1}{2}$
$4$	$-11 + 3 \cdot 4$	$1$

Row 2, entry by entry: old entry plus the multiple of row 1.

\left[ \begin{array}{rrr|r} \boxed{1} & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & \frac{1}{2} & \frac{1}{2} & 1 \\ -2 & 1 & 2 & -3 \end{array} \right]

The matrix now. The boxed entry is the pivot being worked on. checked

Replace row 3 by $R_{3} + 2 R_{1}$ . The multiplier is chosen so that the entry under attack, $-2$ , is cancelled exactly by the pivot: $-2 + 2 \cdot 1 = 0$ .

column	working	new entry
$1$	$-2 + 2 \cdot 1$	$0$
$2$	$1 + 2 \cdot \frac{1}{2}$	$2$
$3$	$2 + 2 \cdot \left( -\frac{1}{2} \right)$	$1$
$4$	$-3 + 2 \cdot 4$	$5$

Row 3, entry by entry: old entry plus the multiple of row 1.

\left[ \begin{array}{rrr|r} \boxed{1} & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & \frac{1}{2} & \frac{1}{2} & 1 \\ 0 & 2 & 1 & 5 \end{array} \right]

The matrix now. The boxed entry is the pivot being worked on. checked

Column 1: the pivot, and what it has to clear

2

1

-1

8

-3

-1

2

-11

-2

1

2

-3

the pivot: the entry being used the entries it is driving to zero the row the pivot lives in

The matrix before this column is dealt with. The boxed entry is the pivot, the entry this step works with; the marked entries below it are the ones being driven to zero. Every other number is carried along unchanged by the moves.

3

Pivot 2: column 2, row 2

Why this step

The same job for the next column. Working left to right matters: the rows we use now already have zeros in the earlier pivot columns, so clearing this column cannot spoil the columns already finished. That is why the method never has to go back.

Swap row 2 and row 3, to bring the entry $2$ up to the pivot position. Swapping two equations changes nothing about what they say together.

\left[ \begin{array}{rrr|r} 1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & \boxed{2} & 1 & 5 \\ 0 & \frac{1}{2} & \frac{1}{2} & 1 \end{array} \right]

The matrix now. The boxed entry is the pivot being worked on. checked

Multiply row 2 by $\frac{1}{2}$ , which turns the pivot $2$ into $1$ . Multiplying an equation through by a number that is not zero keeps it true for exactly the same values.

column	working	new entry
$1$	$\frac{1}{2} \cdot 0$	$0$
$2$	$\frac{1}{2} \cdot 2$	$1$
$3$	$\frac{1}{2} \cdot 1$	$\frac{1}{2}$
$4$	$\frac{1}{2} \cdot 5$	$\frac{5}{2}$

Row 2, entry by entry.

\left[ \begin{array}{rrr|r} 1 & \frac{1}{2} & -\frac{1}{2} & 4 \\ 0 & \boxed{1} & \frac{1}{2} & \frac{5}{2} \\ 0 & \frac{1}{2} & \frac{1}{2} & 1 \end{array} \right]

The matrix now. The boxed entry is the pivot being worked on. checked

Replace row 1 by $R_{1} - \frac{1}{2} R_{2}$ . The multiplier is chosen so that the entry under attack, $\frac{1}{2}$ , is cancelled exactly by the pivot: $\frac{1}{2} - \frac{1}{2} \cdot 1 = 0$ .

column	working	new entry
$1$	$1 - \frac{1}{2} \cdot 0$	$1$
$2$	$\frac{1}{2} - \frac{1}{2} \cdot 1$	$0$
$3$	$-\frac{1}{2} - \frac{1}{2} \cdot \frac{1}{2}$	$-\frac{3}{4}$
$4$	$4 - \frac{1}{2} \cdot \frac{5}{2}$	$\frac{11}{4}$

Row 1, entry by entry: old entry minus the multiple of row 2.

\left[ \begin{array}{rrr|r} 1 & 0 & -\frac{3}{4} & \frac{11}{4} \\ 0 & \boxed{1} & \frac{1}{2} & \frac{5}{2} \\ 0 & \frac{1}{2} & \frac{1}{2} & 1 \end{array} \right]

The matrix now. The boxed entry is the pivot being worked on. checked

Replace row 3 by $R_{3} - \frac{1}{2} R_{2}$ . The multiplier is chosen so that the entry under attack, $\frac{1}{2}$ , is cancelled exactly by the pivot: $\frac{1}{2} - \frac{1}{2} \cdot 1 = 0$ .

column	working	new entry
$1$	$0 - \frac{1}{2} \cdot 0$	$0$
$2$	$\frac{1}{2} - \frac{1}{2} \cdot 1$	$0$
$3$	$\frac{1}{2} - \frac{1}{2} \cdot \frac{1}{2}$	$\frac{1}{4}$
$4$	$1 - \frac{1}{2} \cdot \frac{5}{2}$	$-\frac{1}{4}$

Row 3, entry by entry: old entry minus the multiple of row 2.

\left[ \begin{array}{rrr|r} 1 & 0 & -\frac{3}{4} & \frac{11}{4} \\ 0 & \boxed{1} & \frac{1}{2} & \frac{5}{2} \\ 0 & 0 & \frac{1}{4} & -\frac{1}{4} \end{array} \right]

The matrix now. The boxed entry is the pivot being worked on. checked

Column 2: the pivot, and what it has to clear

1

\frac{1}{2}

-\frac{1}{2}

4

0

\frac{1}{2}

\frac{1}{2}

1

0

2

1

5

the pivot: the entry being used the entries it is driving to zero the row the pivot lives in

The matrix before this column is dealt with. The boxed entry is the pivot, the entry this step works with; the marked entries below it are the ones being driven to zero. Every other number is carried along unchanged by the moves.

4

Pivot 3: column 3, row 3

Why this step

The same job for the next column. Working left to right matters: the rows we use now already have zeros in the earlier pivot columns, so clearing this column cannot spoil the columns already finished. That is why the method never has to go back.

Multiply row 3 by $4$ , which turns the pivot $\frac{1}{4}$ into $1$ . Multiplying an equation through by a number that is not zero keeps it true for exactly the same values.

column	working	new entry
$1$	$4 \cdot 0$	$0$
$2$	$4 \cdot 0$	$0$
$3$	$4 \cdot \frac{1}{4}$	$1$
$4$	$4 \cdot \left( -\frac{1}{4} \right)$	$-1$

Row 3, entry by entry.

\left[ \begin{array}{rrr|r} 1 & 0 & -\frac{3}{4} & \frac{11}{4} \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ 0 & 0 & \boxed{1} & -1 \end{array} \right]

The matrix now. The boxed entry is the pivot being worked on. checked

Replace row 1 by $R_{1} + \frac{3}{4} R_{3}$ . The multiplier is chosen so that the entry under attack, $-\frac{3}{4}$ , is cancelled exactly by the pivot: $-\frac{3}{4} + \frac{3}{4} \cdot 1 = 0$ .

column	working	new entry
$1$	$1 + \frac{3}{4} \cdot 0$	$1$
$2$	$0 + \frac{3}{4} \cdot 0$	$0$
$3$	$-\frac{3}{4} + \frac{3}{4} \cdot 1$	$0$
$4$	$\frac{11}{4} + \frac{3}{4} \cdot \left( -1 \right)$	$2$

Row 1, entry by entry: old entry plus the multiple of row 3.

\left[ \begin{array}{rrr|r} 1 & 0 & 0 & 2 \\ 0 & 1 & \frac{1}{2} & \frac{5}{2} \\ 0 & 0 & \boxed{1} & -1 \end{array} \right]

The matrix now. The boxed entry is the pivot being worked on. checked

Replace row 2 by $R_{2} - \frac{1}{2} R_{3}$ . The multiplier is chosen so that the entry under attack, $\frac{1}{2}$ , is cancelled exactly by the pivot: $\frac{1}{2} - \frac{1}{2} \cdot 1 = 0$ .

column	working	new entry
$1$	$0 - \frac{1}{2} \cdot 0$	$0$
$2$	$1 - \frac{1}{2} \cdot 0$	$1$
$3$	$\frac{1}{2} - \frac{1}{2} \cdot 1$	$0$
$4$	$\frac{5}{2} - \frac{1}{2} \cdot \left( -1 \right)$	$3$

Row 2, entry by entry: old entry minus the multiple of row 3.

\left[ \begin{array}{rrr|r} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & \boxed{1} & -1 \end{array} \right]

The matrix now. The boxed entry is the pivot being worked on. checked

Column 3: the pivot, and what it has to clear

1

0

-\frac{3}{4}

\frac{11}{4}

0

1

\frac{1}{2}

\frac{5}{2}

0

0

\frac{1}{4}

-\frac{1}{4}

the pivot: the entry being used the entries it is driving to zero the row the pivot lives in

The matrix before this column is dealt with. The boxed entry is the pivot, the entry this step works with; the marked entries below it are the ones being driven to zero. Every other number is carried along unchanged by the moves.

5

Read the answer off the reduced matrix

Why this step

The reduced matrix is the same system written in its plainest form, so now we turn the numbers back into equations and simply read them. Three things can happen, and the matrix shows at a glance which one: an impossible row, a pivot for every unknown, or an unknown left without a pivot.

\left[ \begin{array}{rrr|r} 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -1 \end{array} \right]

The reduced matrix: every pivot is

1

, each pivot is the only non-zero entry in its column, and the pivots step down to the right. checked

Every unknown has a pivot, so every row names one unknown outright. Reading the rows:

\begin{aligned} x &= 2 \\ y &= 3 \\ z &= -1 \end{aligned}

The solution.

The reduced matrix, which is the answer written plainly

how much xhow much yhow much zequals

equation 1equation 2equation 3

1

0

0

2

0

1

0

3

0

0

1

-1

Every pivot column now holds a single 1 with zeros above and below it, so each row says one unknown equals one number. The answer is read straight off; no algebra is left to do.

6

Check the answer in every original equation

Why this step

Row reduction involves dozens of small calculations, and one wrong sign anywhere gives a confident, wrong answer. The check costs little and is decisive: put the answer into the ORIGINAL equations, not the reduced ones. If every one balances, no slip can have survived.

2 \cdot 2 + 3 - \left( -1 \right) = 8

Equation 1, with each unknown replaced by its value. checked

-3 \cdot 2 - 3 + 2 \left( -1 \right) = -11

Equation 2, with each unknown replaced by its value. checked

-2 \cdot 2 + 3 + 2 \left( -1 \right) = -3

Equation 3, with each unknown replaced by its value. checked

Back to the equations we started from

how much xhow much yhow much zequals

equation 1equation 2equation 3

2

1

-1

8

-3

-1

2

-11

-2

1

2

-3

The ORIGINAL block of numbers again. The check puts the answer into these rows, not into the reduced ones, because a slip made during the reduction would be repeated by checking against its own result.

Solve a system of linear equations by row reduction