Writes a system as an augmented matrix and solves it by row operations. Establishes why the three operations preserve the solution set, drives the matrix to row echelon form by a mechanical column-by-column procedure, and reads the inconsistent and dependent cases off recognisable rows.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 9 — Systems of Equations and Inequalities
§9.6 Solving Systems with Gaussian Elimination, pp. 1130-1144
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1130-1144 — the pages these objectives are drawn from
Warm-up
Solving a system involves a lot of copying.
Discussion prompt
When you eliminate a variable from two equations, what actually changes?
Hint: Which symbols get rewritten unchanged?
Answer:
Only the numbers change. The letters and the equals signs are copied out again on every line, doing no work.
And each letter's identity is already fixed by its position — the first term is always the first variable.
So the letters can be dropped entirely, leaving a grid of numbers. That is what makes the elimination mechanical, and it is the whole idea of this section.
Concept
Writing the system as a matrix of coefficients and constants turns elimination into a sequence of operations on rows, with no letters to copy and no choices to make.
augmented matrix — the matrix of a system's coefficients with the constants appended as a final column, separated by a bar
\[ \left[\begin{array}{cc|c}2&3&8\\1&-1&1\end{array}\right] \]
The bar is a notational reminder that the last column plays a different role — it holds constants rather than coefficients of a variable. The row operations treat it exactly like any other column.
Figure (svg): A system of equations beside its augmented matrix, with a vertical bar separating the coefficients from the constants
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1130-1133
Section
Section 1
Concept
Each row of the augmented matrix holds one equation's coefficients, each column holds one variable's coefficients, and the final column holds the constants.
Writing zeros for missing variables is essential and easy to forget. A term absent from an equation means its coefficient is zero, and skipping the entry shifts every later column into the wrong variable's position.
Figure (svg): A system of equations beside its augmented matrix, with a vertical bar separating the coefficients from the constants
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1130-1134
Picture it
The same information, with the letters removed.
Figure (svg): A system of equations beside its augmented matrix, with a vertical bar separating the coefficients from the constants
Being able to read the matrix back as equations is what makes the method trustworthy. At any point in the reduction, the current matrix is a system equivalent to the original.
Worked example
Coefficients in, letters out.
\[ \text{Write } 2x+3y=8, \; x-y=1 \text{ as an augmented matrix.} \]
Take the first equation's coefficients
Why: In variable order.
\[ 2\text{ and } 3 \]
Append its constant
Why: After the bar.
\[ 8 \]
Repeat for the second
Why: Coefficients then constant.
\[ 1, -1, 1 \]
Assemble
Why: One row per equation.
\[ a 2\text{ by } 3\text{ array} \]
Figure (svg): A system of equations beside its augmented matrix, with a vertical bar separating the coefficients from the constants
\[ \left[\begin{array}{cc|c}2&3&8\\1&-1&1\end{array}\right] \]
Verify: read it back
Why: The first row reads as 2 times the first variable plus 3 times the second equals 8 — the original equation. Being able to reconstruct the system confirms nothing was dropped or misplaced.
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1131-1132
Prediction
A system of three equations in three variables.
Predict first
How many columns does its augmented matrix have?
Correct: Four, three for variables and one for constants.
Why: Each variable gets a column and the constants get one more. Checking that every row has that many entries catches a skipped zero for a missing variable.
Worked example
A zero, not a gap.
\[ \text{Write } x+2z=5, \; y-z=1, \; x+y=4 \text{ as an augmented matrix.} \]
Fix the variable order
Why: First, second, third.
First equation
Why: No second variable.
\[ 1, 0, 2, 5 \]
Second equation
Why: No first variable.
\[ 0, 1, -1, 1 \]
Third equation
Why: No third variable.
\[ 1, 1, 0, 4 \]
Figure (svg): The solution to Worked example a missing variable shown as a ladder of expressions, one row per legal move
\[ \left[\begin{array}{ccc|c}1&0&2&5\\0&1&-1&1\\1&1&0&4\end{array}\right] \]
Verify: check the column count
Why: Three variables plus one constant column gives four columns, and every row has four entries. A row with three would signal a missing zero, which is the standard slip when a variable is absent from an equation.
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1132-1134
Trap
\[ x+2z=5 \;\Longrightarrow\; \text{row reads }1,\;2,\;5 \]
Write only the coefficients that appear
Why: The absent second variable is skipped rather than recorded as zero.
The 2 now sits in the second variable's column, describing a different equation entirely.
Write a zero for every absent variable. Position is what identifies a variable now, so a gap changes the meaning.
Here the row should read one, zero, two, five — the zero holding the second variable's place.
Check every row has the same number of entries. A short row is the visible symptom of a skipped zero.
Faded example
For the equation with no second variable.
Fill in the blanks
x+2z=5 \;\Longrightarrow\; 1,\;0,\;2,\;5
Why: The absent variable's coefficient is zero and must occupy its column. Omitting it would shift the 2 into the wrong position and change what the row means.
Sorting
Rows, columns and the last column play different roles.
Sort into buckets
Sort each description.
Step zero
You are given a system to solve with matrices.
Discussion prompt
What do you do before any row operations?
Hint: Two things about the writing.
Answer:
Fix a variable order and write the augmented matrix with a column for each, including zeros for any that are absent.
Then check every row has the same number of entries. A short row means a skipped zero, which silently changes what the system says.
Both take seconds. The row operations that follow are mechanical, so all the care belongs at this stage, where a slip cannot be detected later.
Section
Section 2
Concept
Rows may be swapped, scaled by a nonzero number, or replaced by themselves plus a multiple of another row. Each is reversible, which is why the solution set survives.
Reversibility is the whole justification. If an operation can be undone, the new system implies the old one and vice versa, so they have exactly the same solutions. Multiplying by zero destroys a row and cannot be reversed, which is why it is excluded.
Figure (svg): Three cards giving the three legitimate row operations and why each preserves the solution set
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1134-1137
Picture it
Each card names one and its justification.
Figure (svg): Three cards giving the three legitimate row operations and why each preserves the solution set
The caption gives the single reason all three are legitimate. Any operation that cannot be undone would risk changing the solution set, which is why the list has exactly three members.
Worked example
The elimination step in matrix form.
\[ \text{Use row 1 to clear the leading entry of row 2 in } \left[\begin{array}{cc|c}1&2&5\\3&1&4\end{array}\right]. \]
Identify the target
Why: The 3 in row 2.
Choose the multiple
Why: Three times row 1 matches it.
\[ \text{subtract } 3 \times\text{ row } 1 \]
Apply across the whole row
Why: Including the constant.
\[ 3 - 3, 1 - 6, 4 - 15 \]
Write the new row
Why: The leading entry is now zero.
\[ 0, -5, -11 \]
Figure (svg): Three cards giving the three legitimate row operations and why each preserves the solution set
\[ \left[\begin{array}{cc|c}1&2&5\\0&-5&-11\end{array}\right] \]
Verify: read the new row back
Why: It says negative 5 times the second variable equals negative 11, an equation in one variable. That is exactly what elimination produces, and it was reached by one operation applied uniformly across the row.
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1135-1136
Sorting
It must be reversible.
Sort into buckets
Sort each operation.
Worked example
Undo the operation and the original returns.
\[ \text{Show that the previous step can be undone.} \]
Identify the inverse operation
Why: Add back what was subtracted.
\[ \text{add } 3 \times\text{ row } 1 \]
Apply it to the new row 2
Why: Entry by entry.
\[ 0 + 3, -5 + 6, -11 + 15 \]
Compare
Why: The original row returns.
\[ 3, 1, 4 \]
Conclude
Why: The two systems are equivalent.
Figure (svg): The solution to Worked example why reversibility matters shown as a ladder of expressions, one row per legal move
\[ \text{reversible, so equivalent} \]
Verify: state what reversibility proves
Why: Because each system can be obtained from the other, any solution of one is a solution of the other. That two-way implication is what equivalence means, and it is why the reduced matrix can be trusted.
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1136-1137
Error analysis
A student clears a leading entry.
Annotate
On: \( \text{subtract }3\times\text{row 1 from the coefficients of row 2, leaving its constant} \)
The bar separates coefficients from constants for reading, and nothing else. Every row operation runs the full width of the row, constants included.
Prediction
Three operations, one justification.
Predict first
What do all three have in common?
Correct: Each can be undone.
Why: Reversibility means the new system implies the old and vice versa, so the two have exactly the same solutions. That is why multiplying by zero is excluded — it destroys information irrecoverably.
Faded example
Subtracting three times the first row from the second.
Fill in the blanks
(3,\;1,\;4)-3(1,\;2,\;5)=(0,\;-5,\;-11)
Why: The operation applies entry by entry across the whole row, constant included. The leading entry becomes zero, which is the point of choosing that particular multiple.
Explain it to yourself
Three operations are allowed and others are not.
Discussion prompt
Explain what reversibility proves.
Hint: What can be deduced in each direction?
Answer:
If an operation can be undone, the new system can be obtained from the old and the old from the new. Each implies the other.
So any solution of one is a solution of the other — the solution sets are identical, not merely overlapping.
An irreversible step would only give implication in one direction, risking lost or invented solutions. That is the whole reason the list has exactly three members, and a good explanation notes that multiplying by zero is excluded for precisely this reason.
Section
Section 3
Concept
The goal is a matrix where each row's first nonzero entry is a one, further right than the row above, with zeros beneath it. That is §9.2's triangular form in matrix notation.
Working column by column is what makes the procedure mechanical: establish a leading one in the first column, clear everything below it, move to the second column, and repeat. There are no choices to make and no judgement to exercise.
Figure (svg): A matrix in row echelon form, with a staircase of leading ones and zeros below each
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1137-1140
Picture it
Leading ones descending to the right, zeros below each.
Figure (svg): A matrix in row echelon form, with a staircase of leading ones and zeros below each
This is exactly the triangular form of §9.2, reached without any decision about which variable to eliminate. The column order makes those decisions automatically.
Worked example
Column by column.
\[ \text{Reduce } \left[\begin{array}{cc|c}2&4&6\\1&3&5\end{array}\right]. \]
Make a leading one
Why: Halve the first row.
\[ 1, 2, 3 \]
Clear below it
Why: Subtract row 1 from row 2.
\[ 0, 1, 2 \]
Check the second column
Why: The leading entry is already one.
Read the form
Why: Staircase achieved.
Figure (svg): A matrix in row echelon form, with a staircase of leading ones and zeros below each
\[ \left[\begin{array}{cc|c}1&2&3\\0&1&2\end{array}\right] \]
Verify: read the bottom row back
Why: It says the second variable equals 2, an equation in one unknown — which is what echelon form guarantees for the last row. Back-substitution can start immediately from there.
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1138-1139
Sorting
Leading ones descending, zeros below.
Sort into buckets
Sort each description.
Worked example
From the bottom row upward.
\[ \text{Solve from } \left[\begin{array}{cc|c}1&2&3\\0&1&2\end{array}\right]. \]
Read the bottom row
Why: One variable.
\[ y = 2 \]
Read the top row
Why: Both variables.
\[ x + 2 y = 3 \]
Substitute
Why: The known value.
\[ x + 4 = 3 \]
Solve
Why: One step.
\[ x = -1 \]
Figure (svg): The solution to Worked example back-substitute shown as a ladder of expressions, one row per legal move
\[ (x,y)=(-1,2) \]
Verify: check in the original system
Why: The original first equation was 2x plus 4y equals 6, and negative 2 plus 8 is 6 — correct. Checking against the original rather than the reduced matrix confirms that the row operations preserved the system.
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1139-1140
Trap
\[ \text{clear the entry above the second pivot before clearing below the first} \]
Work on whichever entry looks convenient
Why: The column-by-column order is abandoned.
Entries already cleared get refilled, and the reduction goes in circles.
Finish each column before moving right. Establish the leading one, clear everything below it, then move on.
Working below the pivot only means later operations cannot disturb the zeros already created.
Clearing above the pivots is a separate later stage if reduced echelon form is wanted, and it is done after the staircase is complete.
Prediction
The first column has a leading one and zeros below it.
Predict first
What comes next?
Correct: Move to the second column and make a leading one there.
Why: The procedure works column by column: finish one completely, then move right. Clearing above the pivots is a later stage, and doing it early would refill entries the next column's work is about to change.
Faded example
The first row begins with a 2.
Fill in the blanks
\tfrac23}\times(2,\;4,\;6)=(1,\;2,\;___)
Why: Scaling the whole row by the reciprocal of its leading entry produces a leading one. The operation is reversible since the scale factor is nonzero, so the solution set is unchanged.
Explain it
Echelon form is reached without decisions.
Discussion prompt
Explain to a classmate what removes the choices.
Hint: What fixes the order of work?
Answer:
The columns fix the order. Work on the first column until it has a leading one with zeros below, then move to the second, and so on.
There is no choosing which variable to eliminate — §9.2's one real decision — because the column order decides it.
And each column's work cannot disturb the previous columns' zeros, since operations below a pivot leave the columns to its left alone. That is what makes the procedure safe to run without thinking, and a good explanation notes it is exactly why a computer can do it.
Section
Section 4
Concept
A row of zeros with a nonzero constant means no solution. A row of zeros throughout means the equations were dependent and there are infinitely many.
This is the practical gain over §9.1's algebra. There the special cases appeared as a strange final line that had to be interpreted; here they appear as a recognisable row partway through, and the reduction can stop as soon as one shows up.
Figure (svg): A contrast between the row that signals no solution and the row that signals infinitely many
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1140-1143
Picture it
Identical except for one entry.
Figure (svg): A contrast between the row that signals no solution and the row that signals infinitely many
One entry separates no solution from infinitely many, which is why reading the last column carefully matters. The rest of the row is the same in both cases.
Worked example
A false row appears.
\[ \text{Reduce } \left[\begin{array}{cc|c}1&2&3\\2&4&9\end{array}\right]. \]
Clear below the pivot
Why: Subtract twice row 1.
\[ 0, 0, 3 \]
Read the new row
Why: Zeros with a nonzero constant.
\[ 0 = 3 \]
Interpret
Why: A false statement.
Stop
Why: No further work is useful.
Figure (svg): A contrast between the row that signals no solution and the row that signals infinitely many
\[ \text{inconsistent, no solution} \]
Verify: check the original equations
Why: The second equation is twice the first on the left but its constant is 9 rather than 6, so the two describe parallel lines. The row form made that visible in one operation, where comparing the equations directly requires noticing the proportionality.
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1141-1142
Sorting
The last column decides.
Sort into buckets
Sort each row.
Worked example
A zero row appears instead.
\[ \text{Reduce } \left[\begin{array}{cc|c}1&2&3\\2&4&6\end{array}\right]. \]
Clear below the pivot
Why: Subtract twice row 1.
\[ 0, 0, 0 \]
Read the new row
Why: All zeros.
\[ 0 = 0 \]
Interpret
Why: A true statement.
Describe the solutions
Why: One equation, two unknowns.
Figure (svg): The solution to Worked example a dependent system shown as a ladder of expressions, one row per legal move
\[ x=3-2t,\; y=t \]
Verify: count the surviving equations
Why: One nonzero row remains for two unknowns, leaving one free variable — a line. Had two nonzero rows survived, there would be a unique solution instead, so counting the nonzero rows is the quickest read on the outcome.
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1142-1143
Error analysis
A student reaches a row of zeros with a 3 in the last column.
Annotate
On: \( 0\;\;0\;\;|\;\;3 \;\Longrightarrow\; \text{infinitely many solutions} \)
One entry separates the two cases entirely. Reading the last column before interpreting the row is what distinguishes an impossible system from an under-determined one.
Prediction
Reducing a three-variable system leaves two nonzero rows.
Predict first
What is the solution set?
Correct: A line, since one variable is free.
Why: Three unknowns with two independent equations leaves one degree of freedom. Counting the nonzero rows against the number of variables gives the dimension of the solution set directly.
Faded example
A row of zeros with a nonzero constant.
Fill in the blanks
0x+0y=3 \;\Longrightarrow\; 0=3, \text___
Why: Whatever values the variables take, the left side is zero — so the equation asserts that zero equals three. No assignment can satisfy it, which is what makes the system inconsistent.
Explain it to yourself
The same two cases appeared there too.
Discussion prompt
Explain what the matrix form improves.
Hint: When do you find out?
Answer:
In §9.1 the special cases appeared as a strange final statement after the algebra was finished, which had to be recognised and interpreted.
Here they appear as a recognisable row partway through, and the reduction can stop as soon as one shows up — no further work is useful.
They are also visually distinct: a row of zeros looks different from a row with entries. A good explanation notes that this is a general benefit of the matrix form — structure that was implicit in the algebra becomes visible on the page.
Section
Section 5
Concept
Continuing past echelon form to clear the entries above each leading one gives reduced row echelon form, in which the solution can be read off directly with no back-substitution.
Whether to go to reduced form is a matter of taste for small systems, since the extra row operations cost about what back-substitution would. For large systems and for computer implementation the reduced form's uniformity is worth having.
Figure (svg): A matrix in row echelon form, with a staircase of leading ones and zeros below each
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1143-1144
Picture it
Reduced form additionally clears the starred entries.
Figure (svg): A matrix in row echelon form, with a staircase of leading ones and zeros below each
With those entries cleared, each row reads as one variable equalling one constant, and no substitution is needed at all.
Worked example
Clear above the pivots.
\[ \text{Reduce } \left[\begin{array}{cc|c}1&2&3\\0&1&2\end{array}\right] \text{ further.} \]
Identify the entry above a pivot
Why: The 2 in row 1.
Subtract twice row 2 from row 1
Why: Across the whole row.
\[ 1, 0, -1 \]
Read the result
Why: Each row has one variable.
Read the solution
Why: Directly from the last column.
\[ (-1, 2) \]
Figure (svg): A matrix in row echelon form, with a staircase of leading ones and zeros below each
\[ \left[\begin{array}{cc|c}1&0&-1\\0&1&2\end{array}\right] \]
Verify: compare with the back-substitution answer
Why: Back-substituting from echelon form gave the same pair, negative one and two. The extra row operation replaced the substitution step, doing the same work in a different order.
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1143-1144
Prediction
The matrix has zeros above and below every pivot.
Predict first
What can you do?
Correct: Read the solution directly from the last column.
Why: Each row now reads as one variable equalling one constant, so no substitution is needed. The extra row operations did the work back-substitution would otherwise have done.
Worked example
Four variables, same procedure.
\[ \text{What changes when a system has four variables instead of two?} \]
Count the columns
Why: One per variable plus constants.
Note the procedure
Why: Column by column as before.
Count the pivots
Why: One per variable, at most.
Conclude
Why: More arithmetic, no new ideas.
Figure (svg): The solution to Worked example the method does not change with size shown as a ladder of expressions, one row per legal move
\[ \text{the same procedure, more columns} \]
Verify: contrast with substitution
Why: Substitution in four variables would require choosing which variable to isolate at each of several stages, with the expressions growing at each step. The row procedure has no such choices and no growing expressions, which is why it is the one that scales.
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1144-1144
Trap
\[ \text{clear above the first pivot before the second column has a pivot} \]
Start the reduced-form work before the staircase is complete
Why: Entries above are cleared while lower columns are unfinished.
Later operations on lower rows change those entries again, wasting the work.
Complete the echelon staircase first, working strictly downward and rightward.
Only then work back up the pivots, clearing above each. At that point nothing below will change again.
The two stages are separate and ordered. Interleaving them means redoing work rather than making progress.
Sorting
Echelon first, reduced second.
Sort into buckets
Sort each operation.
Faded example
Subtracting twice the second row from the first.
Fill in the blanks
(1,\;2,\;3)-2(0,\;1,\;2)=(1,\;0,\;-1)
Why: The entry above the second pivot becomes zero and the constant updates accordingly. The first row now reads as the first variable equalling negative one, which is the solution directly.
Explain it
Substitution gets harder with more variables and this does not.
Discussion prompt
Explain the difference to a classmate.
Hint: What has to be decided at each step?
Answer:
Substitution requires choosing which variable to isolate at each stage, and the substituted expressions grow longer as more variables are eliminated.
The row procedure has no choices: work down the columns in order, one pivot at a time. Each entry stays a single number rather than growing into an expression.
So the work grows predictably with size and the method never changes. That is why it is the algorithm every computer uses, and a good explanation notes that the absence of decisions is exactly what makes it programmable.
Comparison
Fill the blanks from memory. The same method, differently written.
Comparison matrix
| §9.2's algebra | row reduction | |
|---|---|---|
| what is written | equations with letters | a grid of numbers |
| choices to make | which variable to eliminate | none; the columns decide |
| special cases appear as | a strange final statement | a recognisable row |
| scaling to more variables | expressions grow | more columns, same procedure |
The second row is the practical gain. Removing the decisions is what makes the procedure mechanical and therefore reliable.
Pattern
Five steps, and the middle three are a loop.
Steps 2 to 4 are the loop, and they contain no decisions — which is what makes the whole procedure mechanical and safe to run without thinking.
OpenStax Algebra and Trigonometry 2e, §11.6 Solving Systems with Gaussian Elimination §11.6
Check
Row operations.
Check your understanding
Which operation is not allowed on a matrix row?
Answer: A
Why: Multiplying by zero destroys the row's information and cannot be undone, so it may change the solution set. The other three are reversible, which is exactly why they preserve the solutions.
Check
Special rows.
Check your understanding
A reduced matrix has a row of zeros with a 4 in the last column. What does that mean?
Answer: A
Why: The row reads as zero equals four, which no assignment of values can satisfy. Had the constant been zero as well, the row would have been vacuous and the system would have infinitely many solutions instead.
Check
The procedure.
Check your understanding
After a column has a leading one and zeros below it, what comes next?
Answer: A
Why: The procedure works column by column downward and rightward. Clearing above the pivots is a later stage, and doing it now would waste work since lower columns are not yet finished.
Real world
Every large-scale numerical computation begins with a linear solve.
Discussion prompt
Why is row reduction the method computers use rather than substitution?
Hint: What does a computer need from an algorithm?
Answer:
It has no decisions in it. Work down the columns, one pivot at a time — a procedure that can be written as a loop with no judgement at any step.
And the work is predictable: for n variables the operation count is known in advance, where substitution's expressions grow in ways that depend on the particular system.
So it is programmable and its cost can be estimated before it runs, which matters when the system has thousands of variables — as it does in weather modelling, structural analysis and circuit simulation. The absence of choices is the feature, not a limitation.
Commit first
State your confidence along with your answer.
Predict first
Why do the three row operations preserve the solution set?
Correct: Each one is reversible, so the two systems imply each other.
Why: Reversibility means the new system can be obtained from the old and vice versa, so their solution sets are identical rather than merely overlapping. Multiplying by zero is excluded precisely because it cannot be undone.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why dropping the letters loses nothing.
Hint: What identified each variable?
Answer:
Each variable was already identified by its position in the equation — the first term was always the first variable, whether or not a letter was written.
So the letters were redundant labels, copied out on every line and doing no work. Removing them leaves the coefficients, which are what the elimination actually operates on.
The one thing to be careful about is absent variables: a missing term means a zero in that column, not a gap. A good explanation stresses that position now carries all the information, which is why the zero cannot be skipped.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The third is the procedure that has to become automatic, and the second is what makes it trustworthy rather than a set of moves to copy.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write one system beside its augmented matrix, marking which rows are equations and which columns are variables. List the three row operations with a one-line reason each. Then reduce one matrix to echelon form showing each operation, and finish by writing the two special rows with what each means.
If your three reasons all come down to reversibility, and your two special rows differ only in the last column, the section's ideas are on the page rather than its steps.
Recap
Five things, and the second is what justifies the rest.
| if you remember one thing | it should be this |
|---|---|
| about the matrix | position replaces the letters, so absent variables need zeros |
| about the operations | all three are reversible, which is why they are safe |
| about the procedure | column by column, downward and rightward, no decisions |
| about special rows | the last column separates impossible from under-determined |
Section 9.7 introduces the matrix inverse, which solves a whole system in one multiplication — and explains why some systems cannot be solved that way at all.
OpenStax, Precalculus, §9.6 Solving Systems with Gaussian Elimination §9.6, pp. 1130-1144 — everything on these slides traces back here
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