Row echelon form and reduced row echelon form: Exercises

Before we start, one shared piece of notation, built from zero.

Figure — Row echelon form and reduced row echelon form

The figure above is your reference card for the whole page. Read it left-to-right:

Left grid (REF). The three shaded diagonal squares are the pivots — the first non-zero entry of each row. They "march right and down" like a staircase, which is exactly what REF demands. The plain numbers above the diagonal ( $2$ and $3$ ) are allowed to stay in REF.
Right grid (RREF). Same pivots, but now the entries above each pivot have been forced to $0$ (the shaded off-diagonal cells). RREF requires each pivot to be the only non-zero entry in its column, so this grid is the identity.
The arrow between them is the extra Gauss–Jordan work: "clean above." REF → RREF is precisely the act of zeroing everything above each pivot.

(The colour coding — pivots in one accent, cleaned zeros in another — is just a visual aid; the text above states every cell's role so the figure is not needed to follow the logic.)

Read the rules as a checklist. REF needs: (i) zero rows at the bottom, (ii) each pivot strictly right of the one above, (iii) zeros below each pivot. RREF adds: (iv) every pivot $= 1$ , (v) each pivot alone in its column (zeros above too).

$A$ : Pivots sit at positions $(1, 1), (2, 2), (3, 3)$ — each further right, no zero rows, zeros below. ✅ REF. But look above pivots: the $2$ and the $3$ are non-zero entries above pivots. So (v) fails. $A$ is in REF only.

$B$ : The zero row is in the middle, not the bottom → rule (i) fails immediately. $B$ is neither.

$C$ : Identity matrix. Every pivot is $1$ , alone in its column, staircase perfect. ✅ RREF (and therefore also REF).

Level 2 — Application

(Actually run Gaussian elimination.)

Exercise 2.1

Reduce to REF and solve by back-substitution: $⎩ ⎨ ⎧ x + y + z = 6 2 x + 3 y + z = 11 2 x + y + 3 z = 14$

Recall Solution 2.1

Augmented matrix: $12213111361114$ Step 1 — kill below the top-left pivot ( $x$ out of eqns 2,3): $R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - 2 R_{1}$ : $100 11 - 1 1 - 1 1 6 - 1 2$ Step 2 — next pivot is the $1$ at $(2, 2)$ ; clear below it: $R_{3} \to R_{3} + R_{2}$ : $100110 1 - 1 0 6 - 1 1$ The bottom row reads $0 = 1$ . Contradiction → the system has no solution (inconsistent). A pivot landed in the constants column — the algebraic "shout" of inconsistency.

Exercise 2.2

Reduce to REF and solve: $⎩ ⎨ ⎧ 2 x + 4 y - 2 z = 2 x + 3 y + z = 5 3 x + 5 y - 5 z = 1$

Recall Solution 2.2

$213435 - 2 1 - 5 251$ Step 1 — make the top-left pivot friendly. Scale $R_{1} \to \frac{1}{2} R_{1}$ : $113235 - 1 1 - 5 151$ Step 2 — kill below: $R_{2} \to R_{2} - R_{1}$ , $R_{3} \to R_{3} - 3 R_{1}$ : $100 21 - 1 - 1 2 - 2 14 - 2$ Step 3 — clear below pivot $(2, 2)$ : $R_{3} \to R_{3} + R_{2}$ : $100210 - 1 20 142$ Bottom row: $0 = 2$ → inconsistent, no solution. (Notice: two different-looking systems in this level both turned out inconsistent — the tell is always a pivot in the last column.)

Level 3 — Analysis

(Reach RREF, then interpret pivots, free variables and rank.)

Exercise 3.1

Bring to RREF and describe the full solution set: $120240011382153$

Recall Solution 3.1

Step 1 — clear below pivot $(1, 1)$ : $R_{2} \to R_{2} - 2 R_{1}$ : $100200011322133$ Step 2 — $R_{2}$ and $R_{3}$ are identical; clear below pivot in column 3: $R_{3} \to R_{3} - R_{2}$ : $100200010320130 (REF)$ Pivots already equal $1$ and column above the row-2 pivot (column 3) is already clean above → this is also RREF. Pivot columns: $1$ and $3$ . Non-pivot columns: $2$ and $4$ → variables $y$ and $w$ are free. Read the equations:

Row 1: $x + 2 y + 3 w = 1 \Rightarrow x = 1 - 2 y - 3 w$
Row 2: $z + 2 w = 3 \Rightarrow z = 3 - 2 w$

Solution set (with $y, w$ free): $(x, y, z, w) = (1 - 2 y - 3 w, y, 3 - 2 w, w) .$ Rank = number of pivots = == $2$ == (see Rank of a Matrix). With $4$ variables and rank $2$ , there are $4 - 2 = 2$ free variables → a whole 2-D plane of solutions in solution space.

Exercise 3.2

For which value(s) of $k$ does the system have (a) a unique solution, (b) infinitely many, (c) none? $⎩ ⎨ ⎧ x + y + z = 1 x + 2 y + 3 z = 2 x + y + k z = k$

Recall Solution 3.2

$111121 13 k 12 k$ $R_{2} \to R_{2} - R_{1}$ , $R_{3} \to R_{3} - R_{1}$ : $100110 12 k - 1 11 k - 1$ The decisive entry is the $(3, 3)$ pivot slot: $k - 1$ , with constant $k - 1$ .

If $k \neq = 1$ : the pivot $k - 1 \neq = 0$ , so all three columns have pivots → unique solution. Solve: $z = \frac{k - 1}{k - 1} = 1$ , then $y + 2 z = 1 \Rightarrow y = - 1$ , then $x + y + z = 1 \Rightarrow x = 1$ . So $(x, y, z) = (1, - 1, 1)$ for every $k \neq = 1$ .
If $k = 1$ : bottom row becomes $0 = 0$ — a zero row, no contradiction, and only $2$ pivots for $3$ variables → infinitely many solutions (one free variable $z$ ). From row 2: $y + 2 z = 1 \Rightarrow y = 1 - 2 z$ . From row 1: $x + y + z = 1 \Rightarrow x = 1 - y - z = 1 - (1 - 2 z) - z = z$ . So $(x, y, z) = (z, 1 - 2 z, z)$ .
No value of $k$ gives no solution, because the degenerate case ( $k = 1$ ) produces $0 = 0$ , never $0 = nonzero$ . So case (c) is empty.

Level 4 — Synthesis

(Combine RREF with a second idea: independence, or the inverse.)

Exercise 4.1

Use RREF to decide whether these vectors are linearly independent, and find the rank: $v_{1} = (1, 2, 1), v_{2} = (2, 4, 3), v_{3} = (3, 6, 4) .$

Recall Solution 4.1

Put them as columns of a matrix and reduce (pivot columns mark an independent set — see Linear Independence): $121243364$ $R_{2} \to R_{2} - 2 R_{1}$ , $R_{3} \to R_{3} - R_{1}$ : $100201301$ Swap $R_{2} \leftrightarrow R_{3}$ so zero row sinks: $100210310$ Pivots in columns $1$ and $2$ only → rank $= 2$ . Three vectors but only $2$ pivots ⇒ they are linearly dependent. Concretely the dependency shows up because column 3 is not a pivot column: $v_{3} = v_{1} + v_{2}$ (check: $(1, 2, 1) + (2, 4, 3) = (3, 6, 4)$ ✅).

Exercise 4.2

Find the inverse of $A = [2513]$ using the augment-with-$I$ method.

Recall Solution 4.2

Augment with the $2 \times 2$ identity and drive the left block to RREF; the right block becomes $A^{- 1}$ : $[25131001]$ $R_{1} \to \frac{1}{2} R_{1}$ : $[15 \frac{1}{2} 3 \frac{1}{2} 0 01]$ $R_{2} \to R_{2} - 5 R_{1}$ : $[10 \frac{1}{2} \frac{1}{2} \frac{1}{2} - \frac{5}{2} 01]$ $R_{2} \to 2 R_{2}$ : $[10 \frac{1}{2} 1 \frac{1}{2} - 5 02]$ $R_{1} \to R_{1} - \frac{1}{2} R_{2}$ : $[1001 3 - 5 - 1 2]$ So $A^{- 1} = [3 - 5 - 1 2]$ . Check: $A A^{- 1} = [2513] [3 - 5 - 1 2] = [1001]$ ✅.

Level 5 — Mastery

(Build systems / prove structural facts.)

Exercise 5.1

Construct a $3 \times 4$ augmented matrix (3 equations, 3 unknowns $x, y, z$ ) that is consistent with exactly a one-parameter family of solutions, and whose RREF has pivots in columns $1$ and $3$ . Then write its solution.

Recall Solution 5.1

One free variable ⇒ rank $2$ , pivots in columns $1, 3$ (so column $2$ , i.e. $y$ , is free). A clean RREF meeting this is: $100200010530$

Pivots: columns $1, 3$ ✅; column $2$ pivot-free → $y$ free ✅.
Last (constant) column has no pivot → consistent ✅; rank $2 < 3$ variables → one-parameter family ✅. Solution: from row 1, $x = 5 - 2 y$ ; from row 2, $z = 3$ ; $y$ free: $(x, y, z) = (5 - 2 y, y, 3) .$ A valid original system (un-reduce it): $x + 2 y = 5, z = 3, 0 = 0$ — or dress it up as $x + 2 y = 5, 2 x + 4 y + z = 13, z = 3$ .

Exercise 5.2

Prove (by contradiction, at the level of pivots) that a square $n \times n$ matrix $A$ is invertible iff its RREF is the identity $I_{n}$ .

Recall Solution 5.2

Setup. The RREF of a square matrix is either $I_{n}$ (a pivot in every column) or has at least one pivot-free column (hence at least one zero row, since $n$ pivots can't fit if a column is skipped). These are the only two possibilities for a square matrix.

( $\Leftarrow$ ) RREF $= I_{n} \Rightarrow$ invertible. Each row operation equals left-multiplication by an elementary matrix $E_{k}$ , and each $E_{k}$ is invertible. If $E_{m} \dots E_{1} A = I_{n}$ , then $A = (E_{m} \dots E_{1})^{- 1} = E_{1}^{- 1} \dots E_{m}^{- 1}$ , a product of invertibles → $A$ invertible, with $A^{- 1} = E_{m} \dots E_{1}$ .

( $\Rightarrow$ ) invertible $\Rightarrow$ RREF $= I_{n}$ . Suppose not: the RREF $R$ has a zero row. Then $det R = 0$ , but $R = E_{m} \dots E_{1} A$ with all factors invertible, so $det R = (\prod det E_{k}) det A$ . Since each $det E_{k} \neq = 0$ and (by assumption) $det A \neq = 0$ , the product $det R \neq = 0$ — contradiction with the zero row forcing $det R = 0$ . Hence the RREF has no zero row → a pivot in every column → RREF $= I_{n}$ . $■$

Recall Self-check: rank ↔ solution-count cheat sheet

Each line below is written as Prompt ::: Answer — cover the part after the ::: and test yourself, then reveal it. Let $r$ = rank of coefficient block, $r^{'}$ = rank of augmented matrix, $n$ = number of unknowns. If $r^{'} > r$ (pivot in last column) ::: no solution (inconsistent) If $r^{'} = r = n$ ::: unique solution If $r^{'} = r < n$ ::: infinitely many, with $n - r$ free variables

Connections

Gaussian Elimination · Gauss-Jordan Elimination
Rank of a Matrix · Free and Basic Variables · Solving Systems of Linear Equations
Matrix Inverse via RREF · Linear Independence · Elementary Matrices

Exercises — Row echelon form and reduced row echelon form

Level 1 — Recognition

Exercise 1.1