Staircases that solve systems

Mathematics · Linear Algebra · ages 18-19

Name ______________________ Date ____________

R1 is [1 2 | 5] and R2 is [3 4 | 7]. After R2 <- R2 - 3R1, what is the last entry of R2?
Answer: ______________
Solve x + y = 8 and x - y = 2. What is x?
Answer: ______________
A staircase row reading 0 = 5 means the system has no solution.
Circle one: True False
Which list gives the three allowed row operations?
- Swap columns, delete a row, or scale every solution
- Swap rows, scale by a nonzero number, or add a row multiple
- Add columns, scale by zero, or swap the two sides
A coefficient matrix has 4 columns and pivots in columns 1 and 3. Which columns are free?
- Columns 1 and 3
- Columns 2 and 4
- Columns 1 and 4
- No columns are free
A consistent system has 5 unknowns and its staircase form has 3 pivots. How many free variables are there?
Answer: ______________
R1 is [2 1 | 9] and R2 is [8 3 | 5]. Which replacement makes the first entry of R2 zero?
- R2 <- R2 - R1
- R2 <- R2 + 4R1
- R2 <- R2 - 4R1
Which statement describes reduced row echelon form?
- Every pivot column has a leading 1 and zeros elsewhere
- Every row must contain two leading entries
- Every free column must contain only zero entries
- Every solution must be multiplied by the same number
Use row operations to solve x + y + z = 6, 2x - y + z = 3 and x + 2y - z = 2. What is z?
Answer: ______________
A row was multiplied by -5. What multiplier reverses that row operation? Enter it as a decimal.
Answer: ______________

Answer key

For grown-ups. Fold this page away before handing over the rest.

-8 · 3R1 is [3 6 | 15]. Subtracting it from R2 gives [0 -2 | -8], so the last entry is -8.
5 · Add the equations to get 2x = 10, so x = 5. Then y = 3, and both equations check.
True · No values can satisfy it, so the system is inconsistent.
Swap rows, scale by a nonzero number, or add a row multiple · Swaps, nonzero scalings, and adding multiples are all reversible, so they preserve the solution set.
Columns 2 and 4 · Free columns are the columns without pivots. With pivots in 1 and 3, columns 2 and 4 are free.
2 · In a consistent system, free variables are unknowns minus pivots: 5 minus 3 is 2.
R2 <- R2 - 4R1 · Subtract 4R1 because 8 - 4(2) = 0. Subtracting R1 gives 6 in that entry; adding 4R1 gives 16. Only subtracting 4R1 clears it.
Every pivot column has a leading 1 and zeros elsewhere · In reduced row echelon form, each pivot is 1 and is the only nonzero entry in its column.
3 · Subtract 2R1 from R2 and R1 from R3. After swapping the lower rows and adding 3 times the new R2 to R3, the last row gives -7z = -21, so z = 3.
-0.2 · The reciprocal of -5 is -1/5, which is -0.2. Multiplying the changed row by -0.2 returns every entry to its former value.