Gaussian elimination — forward elimination, back substitution: Exercises

Quick reminder of the vocabulary we lean on (all from the parent):

Augmented matrix $[A ∣ b]$ — coefficients and right-hand side packed side by side, separated by a bar.
Pivot — the nonzero entry we use in each column to clear the entries below it.
Multiplier $m_{ik} = a_{ik} / a_{k k}$ — the exact amount of the pivot row we subtract to zero out entry $a_{ik}$ .
Upper-triangular $U$ — all zeros below the diagonal; the "staircase" that lets us solve bottom-up.

L1 — Recognition

Exercise 1.1

Which of these matrices is in upper-triangular form (ready for back substitution)? $A = [2013], B = [1402], C = 100530217$

Recall Solution

WHAT we check: upper-triangular means every entry strictly below the diagonal is zero. Look only below the diagonal line.

$A$ : below-diagonal entry is the $0$ in position $(2, 1)$ . ✓ Upper-triangular.
$B$ : position $(2, 1)$ is $4 \neq = 0$ . ✗ Not upper-triangular.
$C$ : below the diagonal we have positions $(2, 1), (3, 1), (3, 2)$ , all $0$ . ✓ Upper-triangular.

Answer: $A$ and $C$ are upper-triangular; $B$ is not.

Exercise 1.2

In the augmented matrix below, name the pivot for column 1 and the entry we must eliminate underneath it. $[46 - 2 1 69]$

Recall Solution

Pivot = the diagonal entry in column 1 = $a_{11} = 4$ . Entry to eliminate = the one below it in column 1 = $a_{21} = 6$ . The multiplier will be $m_{21} = a_{21} / a_{11} = 6/4 = \frac{3}{2}$ , and we do $R_{2} \to R_{2} - \frac{3}{2} R_{1}$ . (You are not asked to run it here — just to identify the players.)

L2 — Application

Exercise 2.1

Solve by forward elimination + back substitution: $x + 2 y 3 x + 4 y = 5 = 11$

Recall Solution

Augmented matrix: $[1324511]$ Forward elimination. Pivot $a_{11} = 1$ . Multiplier $m_{21} = 3/1 = 3$ , so $R_{2} \to R_{2} - 3 R_{1}$ . WHY: subtracting $3$ copies of row 1 turns the $3$ in position $(2, 1)$ into $3 - 3 (1) = 0$ . Row 2 becomes: $(3 - 3, 4 - 6 ∣ 11 - 15) = (0, - 2 ∣ - 4)$ . $[10 2 - 2 5 - 4]$ Back substitution.

Row 2: $- 2 y = - 4 \Rightarrow y = 2$ .
Row 1: $x + 2 (2) = 5 \Rightarrow x = 1$ .

Answer: $x = 1, y = 2$ . Check original row 2: $3 (1) + 4 (2) = 11$ ✓.

Exercise 2.2

Solve the $3 \times 3$ system: $x + y + z 2 x + 3 y + z x - y + 2 z = 6 = 11 = 5$

Recall Solution

Augmented matrix: $121 13 - 1 1126115$ Column 1, pivot $a_{11} = 1$ .

$m_{21} = 2/1 = 2$ : $R_{2} \to R_{2} - 2 R_{1} = (0, 1, - 1 ∣ - 1)$ .
$m_{31} = 1/1 = 1$ : $R_{3} \to R_{3} - 1 R_{1} = (0, - 2, 1 ∣ - 1)$ . $100 11 - 2 1 - 1 1 6 - 1 - 1$ Column 2, pivot $a_{22} = 1$ .
$m_{32} = - 2/1 = - 2$ : $R_{3} \to R_{3} - (- 2) R_{2} = R_{3} + 2 R_{2} = (0, 0, - 1 ∣ - 3)$ . $100110 1 - 1 - 1 6 - 1 - 3$ Back substitution.
Row 3: $- z = - 3 \Rightarrow z = 3$ .
Row 2: $y - z = - 1 \Rightarrow y - 3 = - 1 \Rightarrow y = 2$ .
Row 1: $x + y + z = 6 \Rightarrow x + 2 + 3 = 6 \Rightarrow x = 1$ .

Answer: $x = 1, y = 2, z = 3$ . Check original row 3: $1 - 2 + 6 = 5$ ✓.

L3 — Analysis

Exercise 3.1

Perform elimination and classify the system (unique / none / infinite): $x + 2 y - z 2 x + 4 y - 2 z 3 x + 6 y - z = 1 = 2 = 5$

Recall Solution

Augmented matrix: $123246 - 1 - 2 - 1 125$ Column 1, pivot $1$ .

$m_{21} = 2$ : $R_{2} \to R_{2} - 2 R_{1} = (0, 0, 0 ∣ 0)$ .
$m_{31} = 3$ : $R_{3} \to R_{3} - 3 R_{1} = (0, 0, 2 ∣ 2)$ . $100200 - 1 02 102$ Row 2 is all zeros — no pivot available in column 2. Swap $R_{2} \leftrightarrow R_{3}$ to bring a nonzero entry up (this reorders rows without changing the solution set): $100200 - 1 20 120$ Read the structure:
Row 3 is $0 = 0$ — always true, gives no information → a free variable exists.
Pivots sit in columns 1 and 3. Column 2 has no pivot, so $y$ is free.

Classification: infinitely many solutions. Let $y = t$ .

Row 2: $2 z = 2 \Rightarrow z = 1$ .
Row 1: $x + 2 t - 1 = 1 \Rightarrow x = 2 - 2 t$ .

Answer: $(x, y, z) = (2 - 2 t, t, 1)$ for any real $t$ → infinitely many solutions.

Exercise 3.2

Determine, without fully solving, whether this system is consistent: $x - y + 2 z 2 x - 2 y + 4 z = 4 = 7$

Recall Solution

Column 1, pivot $1$ . $m_{21} = 2$ : $R_{2} \to R_{2} - 2 R_{1}$ . Row 2: $(2 - 2, - 2 + 2, 4 - 4 ∣ 7 - 8) = (0, 0, 0 ∣ - 1)$ . $[10 - 1 0 20 4 - 1]$ Row 2 reads $0 = - 1$ — impossible. Answer: the system is inconsistent (no solution). WHY elimination revealed it: the two equations describe parallel-but-shifted planes; subtracting one from twice-the-other cancels all variables and leaves a false numeric statement.

L4 — Synthesis

Exercise 4.1

The system below has a zero in the first pivot position. Use partial pivoting (swap in the row with the largest first-column entry in absolute value), then solve. $021 11 - 1 1 - 1 2 415$

Recall Solution

Pivot problem. Position $(1, 1)$ is $0$ — we cannot divide by it. Look at column-1 entries: $∣0∣, ∣2∣, ∣1∣$ . The largest is $∣2∣$ in row 2. Swap $R_{1} \leftrightarrow R_{2}$ : $201 11 - 1 - 1 12 145$ Column 1, pivot $2$ .

$m_{21} = 0/2 = 0$ : row 2 already clear, nothing to do.
$m_{31} = 1/2$ : $R_{3} \to R_{3} - \frac{1}{2} R_{1} = (0, - \frac{3}{2}, \frac{5}{2} ∣ \frac{9}{2})$ . $200 11 - \frac{3}{2} - 1 1 \frac{5}{2} 14 \frac{9}{2}$ Column 2, pivot $1$ .
$m_{32} = - \frac{3}{2} /1 = - \frac{3}{2}$ : $R_{3} \to R_{3} + \frac{3}{2} R_{2} = (0, 0, 4 ∣ \frac{21}{2})$ . $200110 - 1 14 14 \frac{21}{2}$ Back substitution.
Row 3: $4 z = \frac{21}{2} \Rightarrow z = \frac{21}{8}$ .
Row 2: $y + z = 4 \Rightarrow y = 4 - \frac{21}{8} = \frac{11}{8}$ .
Row 1: $2 x + y - z = 1 \Rightarrow 2 x = 1 - \frac{11}{8} + \frac{21}{8} = 1 + \frac{10}{8} = \frac{18}{8} = \frac{9}{4} \Rightarrow x = \frac{9}{8}$ .

Answer: $x = \frac{9}{8}, y = \frac{11}{8}, z = \frac{21}{8}$ . Check original row 1 ( $0 x + y + z = 4$ ): $\frac{11}{8} + \frac{21}{8} = \frac{32}{8} = 4$ ✓.

Exercise 4.2

Find the value(s) of $k$ for which the system has no solution: $x + 2 y 2 x + k y = 3 = 7$

Recall Solution

Augmented matrix: $[12 2 k 37]$ Eliminate. $m_{21} = 2/1 = 2$ : $R_{2} \to R_{2} - 2 R_{1}$ . Row 2: $(0, k - 4 ∣ 7 - 6) = (0, k - 4 ∣ 1)$ . $[10 2 k - 4 31]$ Analyse the last row $(0) (x) + (k - 4) y = 1$ .

If $k - 4 \neq = 0$ (i.e. $k \neq = 4$ ): we can solve $y = \frac{1}{k - 4}$ → unique solution.
If $k - 4 = 0$ (i.e. $k = 4$ ): the row reads $0 = 1$ → contradiction → no solution.

Answer: the system has no solution exactly when $k = 4$ .

L5 — Mastery

Exercise 5.1

During elimination you record the multipliers you used. For the matrix $A = [2643],$ run forward elimination to get $U$ , collect the multiplier into the matrix $L$ , and verify $A = LU$ (the LU Decomposition idea).

Recall Solution

Forward elimination. Column 1, pivot $2$ . $m_{21} = 6/2 = 3$ : $R_{2} \to R_{2} - 3 R_{1}$ . Row 2: $(6 - 6, 3 - 12) = (0, - 9)$ . $U = [20 4 - 9] .$ Build $L$ . $L$ is lower-triangular with $1$ s on the diagonal and the multiplier $m_{21} = 3$ placed in position $(2, 1)$ (the exact spot we cleared): $L = [1301] .$ Verify $LU = A$ . $LU = [1301] [20 4 - 9] = [1 \cdot 2 + 0 \cdot 0 3 \cdot 2 + 1 \cdot 0 1 \cdot 4 + 0 \cdot (- 9) 3 \cdot 4 + 1 \cdot (- 9)] = [2643] = A . ✓$ WHY this works: each elimination step " $R_{2} \to R_{2} - 3 R_{1}$ " is undone by " $R_{2} \to R_{2} + 3 R_{1}$ ", which is exactly what multiplying by $L$ does. Storing the multiplier $3$ is recording the recipe to rebuild $A$ from $U$ .

Exercise 5.2

Use forward elimination to compute the Determinant of $A = 1232583814 .$ (Fact you may use: for an upper-triangular matrix the determinant is the product of the diagonal, and the replacement operation $R_{i} \to R_{i} - m R_{j}$ does not change the determinant.)

Recall Solution

Column 1, pivot $1$ .

$m_{21} = 2$ : $R_{2} \to R_{2} - 2 R_{1} = (0, 1, 2)$ .
$m_{31} = 3$ : $R_{3} \to R_{3} - 3 R_{1} = (0, 2, 5)$ . $100212325$ Column 2, pivot $1$ .
$m_{32} = 2$ : $R_{3} \to R_{3} - 2 R_{2} = (0, 0, 1)$ . $U = 100210321$ Determinant. No row swaps were used (each step was a replacement, which leaves $det$ unchanged), so $det A = det U = 1 \cdot 1 \cdot 1 = 1.$ Answer: $det A = 1$ . Note on sign: if we HAD swapped rows, each swap flips the sign of the determinant — so the rule is $det A = (- 1)^{# swaps} \times (product of pivots)$ .

Exercise 5.3

A student claims: "Forward elimination on any $3 \times 3$ system always needs exactly 3 multipliers." Give a system where it needs fewer, and explain why.

Recall Solution

Counterexample. Take a matrix that already has zeros where we would eliminate: $A = 100240356 .$ It is already upper-triangular. The entries below each pivot are already $0$ , so every multiplier would be $m_{ik} = 0/ a_{k k} = 0$ — meaning no actual elimination is performed. Zero multipliers are needed. General principle: the number of nonzero eliminations depends on how many below-pivot entries are already zero. The count "3" is the maximum for a full $3 \times 3$ (positions $(2, 1), (3, 1), (3, 2)$ ), not a guarantee. Any pre-existing zero below a pivot saves a step. Answer: the diagonal/triangular system above needs 0 multipliers; the claim of "always 3" is false.

Connections

LU Decomposition — Exercise 5.1 is the seed: stored multipliers become $L$ .
Row Echelon Form / Reduced Row Echelon Form — the staircase these exercises build toward.
Pivoting and Numerical Stability — Exercise 4.1's swap rule, made rigorous.
Rank of a Matrix — count the pivots in Exercise 3.1 (two pivots → rank 2).
Determinant — Exercise 5.2's product-of-pivots trick.
Solving Linear Systems — Consistency — the unique/none/infinite verdicts of L3.