Solve 3 linear equations in 3 unknowns by row reduction: -x - z = 2, 2x + 2y + z = 6, 3x + 2y + z = 6, step by step

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The problem

Solve this system of linear equations.

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

The result

\left( x, y, z \right) = \left( 0, \; 4, \; -2 \right)

The system has exactly one solution.

checked 49 lines of this solution were evaluated with exact arithmetic before it was shown to you. A solution with a false line is never served.

Picture 1 of 2Step 1: Write the system as one block of numbers

The system, as one block of numbers

how much xhow much yhow much zequals

equation 1equation 2equation 3

-1

0

-1

2

2

2

1

6

3

2

1

6

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.

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} -x - z &= 2 \\ 2 x + 2 y + z &= 6 \\ 3 x + 2 y + z &= 6 \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} -1 & 0 & -1 & 2 \\ 2 & 2 & 1 & 6 \\ 3 & 2 & 1 & 6 \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

-1

0

-1

2

2

2

1

6

3

2

1

6

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 $-1$ , which turns the pivot $-1$ 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
$1$	$-1 \cdot \left( -1 \right)$	$1$
$2$	$-1 \cdot 0$	$0$
$3$	$-1 \cdot \left( -1 \right)$	$1$
$4$	$-1 \cdot 2$	$-2$

Row 1, entry by entry.

\left[ \begin{array}{rrr|r} \boxed{1} & 0 & 1 & -2 \\ 2 & 2 & 1 & 6 \\ 3 & 2 & 1 & 6 \end{array} \right]

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

Replace row 2 by $R_{2} - 2 R_{1}$ . 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
$1$	$2 - 2 \cdot 1$	$0$
$2$	$2 - 2 \cdot 0$	$2$
$3$	$1 - 2 \cdot 1$	$-1$
$4$	$6 - 2 \cdot \left( -2 \right)$	$10$

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

\left[ \begin{array}{rrr|r} \boxed{1} & 0 & 1 & -2 \\ 0 & 2 & -1 & 10 \\ 3 & 2 & 1 & 6 \end{array} \right]

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

Replace row 3 by $R_{3} - 3 R_{1}$ . 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
$1$	$3 - 3 \cdot 1$	$0$
$2$	$2 - 3 \cdot 0$	$2$
$3$	$1 - 3 \cdot 1$	$-2$
$4$	$6 - 3 \cdot \left( -2 \right)$	$12$

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

\left[ \begin{array}{rrr|r} \boxed{1} & 0 & 1 & -2 \\ 0 & 2 & -1 & 10 \\ 0 & 2 & -2 & 12 \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

-1

0

-1

2

2

2

1

6

3

2

1

6

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.

Solve a system of linear equations by row reduction

Write the system as one block of numbers

Pivot 1: column 1, row 1

Pivot 2: column 2, row 2

Pivot 3: column 3, row 3

Read the answer off the reduced matrix

Check the answer in every original equation