LINEAR-ALGEBRA Β· Unit 7 Β· Video 3 Β· Interactive Practice

Why RREF Lets You Skip Back Substitution

IKey Formulas

Formula Name Description
R=[IF00]R = \begin{bmatrix} I & F \\ 0 & 0 \end{bmatrix} RREF block structure Identity II in pivot columns, FF in free columns (conceptually rearranged)
N=[βˆ’FI]N = \begin{bmatrix} -F \\ I \end{bmatrix} Null space matrix Columns are special solutions; rows interleave by variable type
xpivot=βˆ’F xfreex_{\text{pivot}} = -F \, x_{\text{free}} Pivot–free relation From the top block of Rx=0Rx = 0
dim(null space) =nβˆ’r= n - r Counting solutions nn columns minus rr pivots = number of special solutions

IIVisualization 1 β€” From Echelon Form to RREF

Echelon form already has zeros below the pivots β€” what two extra moves finish the job to RREF?

IIIVisualization 2 β€” The Block Structure of RREF

Read FF off an RREF, negate it, and the null-space matrix N=[βˆ’FI]N = \begin{bmatrix} -F \\ I \end{bmatrix} falls out β€” but do its rows stack, or interleave?

πŸ’‘ Each entry of βˆ’F-F lands in its pivot variable's row of NN. When a free column sits between two pivots β€” as in the video example β€” the βˆ’F-F and II rows interleave instead of stacking into clean blocks.

IVVisualization 3 β€” Every Combination Stays in the Null Space

If s1s_1 and s2s_2 each solve Ux=0Ux = 0, must every c1s1+c2s2c_1 s_1 + c_2 s_2 solve it too?

VQuiz Questions

Question 1 Β· Identifying the F Block

Given the RREF matrix: R=[10301βˆ’2000]R = \begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix}

Pivot columns are 1 and 2. The free column is 3. What is the matrix FF?

βœ… Correct! F is read directly from the pivot rows in the free columns β€” no negation yet.

❌ Not quite. Remember: F uses only the r pivot rows (not zero rows), and is read directly from R (don't negate).

Show solution

Solution:

FF consists of the entries in the pivot rows (rows 1 and 2) at the free columns (column 3):

  • Row 1, Column 3: 33
  • Row 2, Column 3: βˆ’2-2

F=[3βˆ’2]F = \begin{bmatrix} 3 \\ -2 \end{bmatrix}

Common mistakes:

  • Don't include the zero row β€” FF uses only the rr pivot rows, so it's 2Γ—12 \times 1, not 3Γ—13 \times 1
  • Don't negate β€” FF is read directly from RR; negation happens when building N=[βˆ’FI]N = \begin{bmatrix} -F \\ I \end{bmatrix}
  • Don't confuse with II β€” the identity block lives in the pivot columns, not the free columns

Question 2 Β· Row Interleaving (True / False)

Consider a system with 5 columns where the RREF has pivot columns 1, 3, and 4, and free columns 2 and 5.

Statement: In the null space matrix NN, the rows from βˆ’F-F always occupy the first three rows (1, 2, 3) and the rows from II always occupy the last two rows (4, 5).

βœ… Correct! The rows interleave based on which variables are pivot vs. free β€” they don't stack neatly into blocks.

❌ Not quite. Think about which variable each row represents. Rows are ordered by variable index, so pivot and free rows interleave.

Show solution

Solution: False

The rows of NN correspond to the original variable ordering x1,x2,x3,x4,x5x_1, x_2, x_3, x_4, x_5:

Row Variable Type Source
1 x1x_1 Pivot βˆ’F-F row 1
2 x2x_2 Free II row 1
3 x3x_3 Pivot βˆ’F-F row 2
4 x4x_4 Pivot βˆ’F-F row 3
5 x5x_5 Free II row 2

The rows interleave: βˆ’F,I,βˆ’F,βˆ’F,I-F, I, -F, -F, I β€” they don't stack neatly!

The formula N=[βˆ’FI]N = \begin{bmatrix} -F \\ I \end{bmatrix} shows the conceptual block structure (as if columns were rearranged to put pivots first). In the actual vector, you must slot entries into each variable's correct row.

Question 3 Β· Computing a Special Solution

Given the RREF: R=[10201βˆ’3000]R = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \end{bmatrix}

Pivot columns: 1 and 2. Free column: 3. What is the special solution for the null space?

βœ… Correct! Negate F for the pivot entries, and the free variable gets 1 from the identity.

❌ Close! You read F directly but forgot to negate. Pivot entries come from βˆ’F, not F.

❌ Not quite. Remember: pivot entries come from βˆ’F, and the free variable gets 1 (from the identity block).

Show solution

Solution:

Step 1 β€” Extract FF from pivot rows Γ—\times free column: F=[2βˆ’3]F = \begin{bmatrix} 2 \\ -3 \end{bmatrix}

Step 2 β€” Negate: βˆ’F=[βˆ’23]-F = \begin{bmatrix} -2 \\ 3 \end{bmatrix}

Step 3 β€” Build NN with correct row placement:

  • x1x_1 (pivot) ←\leftarrow row 1 of βˆ’F-F: βˆ’2\mathbf{-2}
  • x2x_2 (pivot) ←\leftarrow row 2 of βˆ’F-F: 3\mathbf{3}
  • x3x_3 (free) ←\leftarrow identity: 1\mathbf{1}

s=(βˆ’231)s = \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}

Verify: Rβ‹…s=[1(βˆ’2)+0(3)+2(1)0(βˆ’2)+1(3)+(βˆ’3)(1)0]=[000]R \cdot s = \begin{bmatrix} 1(-2) + 0(3) + 2(1) \\ 0(-2) + 1(3) + (-3)(1) \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} βœ“

Question 4 Β· Counting Special Solutions

A 5Γ—75 \times 7 matrix has an RREF with pivots in columns 1, 3, 4, and 5.

How many special solutions does the null space have?

βœ… Correct! Number of special solutions = n βˆ’ r = 7 βˆ’ 4 = 3.

❌ Not quite. Count: n = 7 columns, r = 4 pivots. Special solutions = n βˆ’ r.

Show solution

Solution:

  • Number of columns: n=7n = 7
  • Number of pivots (rank): r=4r = 4 (columns 1, 3, 4, 5)
  • Number of free variables: nβˆ’r=7βˆ’4=3n - r = 7 - 4 = \mathbf{3}

Each free variable produces one special solution, so there are 3 special solutions.

The free columns are 2, 6, and 7. The matrix FF would have size 4Γ—34 \times 3 (4 pivot rows Γ—\times 3 free columns), and the null space matrix NN would have size 7Γ—37 \times 3.

Key point: The number of rows (m=5m = 5) does not affect this count. It's always nβˆ’rn - r that determines the dimension of the null space.

Solved: 0 / 4