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1.2 Row Reduction and Echelon Forms

Theory: Row Reduction and Echelon Forms

Echelon Forms (Lay §1.2)

A leading entry of a row is the leftmost nonzero entry of that (nonzero) row.

A matrix is in echelon form (REF) if:

  1. All nonzero rows are above any rows of all zeros.
  2. Each leading entry is in a column to the right of the leading entry of the row above it (“staircase”).
  3. All entries in a column below a leading entry are zeros.

It is in reduced echelon form (RREF) if, in addition:

  1. The leading entry in each nonzero row is 1.
  2. Each leading 1 is the only nonzero entry in its column (zeros above it too).

Schematically (■\blacksquare = nonzero leading entry, ∗\ast = any value):

REF: [■∗∗∗0■∗∗0000]RREF: [10∗∗01∗∗0000]\text{REF: } \begin{bmatrix} \blacksquare & \ast & \ast & \ast \\ 0 & \blacksquare & \ast & \ast \\ 0 & 0 & 0 & 0 \end{bmatrix} \qquad \text{RREF: } \begin{bmatrix} 1 & 0 & \ast & \ast \\ 0 & 1 & \ast & \ast \\ 0 & 0 & 0 & 0 \end{bmatrix}

Theorem 1 — Uniqueness of the Reduced Echelon Form (Lay §1.2): Each matrix is row equivalent to one and only one reduced echelon matrix.

REF vs RREF:

  • REF is not unique: different row operations give different echelon forms, and the pivot values can change — but the pivot locations are always the same.
  • RREF is unique: both the pivots (all 1) and their locations are fixed. That’s why we can talk about “the” RREF of AA.
  • To go from REF to RREF: scale all pivots to 1 and create zeros above the pivots.

Pivot Positions and Pivot Columns

Because pivot locations are the same in every echelon form, you can find them from any REF — you don’t need the full RREF.

The Row Reduction Algorithm

Forward phase (produces an echelon form):

  1. Begin with the leftmost nonzero column. This is a pivot column; the pivot position is at the top.
  2. Select a nonzero entry in the pivot column as the pivot. If necessary, interchange rows to move it to the pivot position.
  3. Use row replacement to create zeros in all positions below the pivot.
  4. Cover (ignore) the row containing the pivot and repeat steps 1–3 on the submatrix that remains, until there are no more nonzero rows.

Backward phase (produces the RREF):

  1. Starting with the rightmost pivot and working up and to the left, create zeros above each pivot. If a pivot is not 1, scale its row to make it 1.

Mini-example (Lay §1.2, Example 3)

[03−664−53−78−5893−912−9615]\begin{bmatrix} 0 & 3 & -6 & 6 & 4 & -5 \\ 3 & -7 & 8 & -5 & 8 & 9 \\ 3 & -9 & 12 & -9 & 6 & 15 \end{bmatrix}

Forward phase. Column 1 is the first pivot column, but its top entry is 0: swap R1↔R3R_1 \leftrightarrow R_3. Then R2←R2−R1R_2 \leftarrow R_2 - R_1, and R3←R3−32R2R_3 \leftarrow R_3 - \tfrac32 R_2:

∼[3−912−96153−78−58903−664−5]∼[3−912−961502−442−603−664−5]∼[3−912−961502−442−6000014]\sim \begin{bmatrix} 3 & -9 & 12 & -9 & 6 & 15 \\ 3 & -7 & 8 & -5 & 8 & 9 \\ 0 & 3 & -6 & 6 & 4 & -5 \end{bmatrix} \sim \begin{bmatrix} 3 & -9 & 12 & -9 & 6 & 15 \\ 0 & 2 & -4 & 4 & 2 & -6 \\ 0 & 3 & -6 & 6 & 4 & -5 \end{bmatrix} \sim \begin{bmatrix} 3 & -9 & 12 & -9 & 6 & 15 \\ 0 & 2 & -4 & 4 & 2 & -6 \\ 0 & 0 & 0 & 0 & 1 & 4 \end{bmatrix}

This is an echelon form; pivot columns are 1, 2 and 5.

Backward phase. R2←R2−2R3R_2 \leftarrow R_2 - 2R_3, R1←R1−6R3R_1 \leftarrow R_1 - 6R_3; then R2←12R2R_2 \leftarrow \tfrac12 R_2; then R1←R1+9R2R_1 \leftarrow R_1 + 9R_2; finally R1←13R1R_1 \leftarrow \tfrac13 R_1:

∼[3−912−90−902−440−14000014]∼[30−690−7201−220−7000014]∼[10−230−2401−220−7000014]\sim \begin{bmatrix} 3 & -9 & 12 & -9 & 0 & -9 \\ 0 & 2 & -4 & 4 & 0 & -14 \\ 0 & 0 & 0 & 0 & 1 & 4 \end{bmatrix} \sim \begin{bmatrix} 3 & 0 & -6 & 9 & 0 & -72 \\ 0 & 1 & -2 & 2 & 0 & -7 \\ 0 & 0 & 0 & 0 & 1 & 4 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & -2 & 3 & 0 & -24 \\ 0 & 1 & -2 & 2 & 0 & -7 \\ 0 & 0 & 0 & 0 & 1 & 4 \end{bmatrix}

Basic vs Free Variables and the Parametric Description

When the RREF of an augmented matrix is reached, read off the solution:

Each nonzero row of the RREF gives one equation that expresses a basic variable in terms of the free variables. Writing the solution set this way is a parametric description: the free variables act as parameters.

If the matrix in the mini-example above is the augmented matrix of a system in x1,…,x5x_1, \dots, x_5, the RREF gives

x1−2x3+3x4=−24x2−2x3+2x4=−7x5=4⟹{x1=−24+2x3−3x4x2=−7+2x3−2x4x3 is freex4 is freex5=4\begin{aligned} x_1 - 2x_3 + 3x_4 &= -24 \\ x_2 - 2x_3 + 2x_4 &= -7 \\ x_5 &= 4 \end{aligned} \qquad\Longrightarrow\qquad \begin{cases} x_1 = -24 + 2x_3 - 3x_4 \\ x_2 = -7 + 2x_3 - 2x_4 \\ x_3 \text{ is free} \\ x_4 \text{ is free} \\ x_5 = 4 \end{cases}

Every choice of x3,x4x_3, x_4 gives a solution, so there are infinitely many.

Common mistakes with free variables:

  • A variable that does not appear in any equation of the RREF (its column is all zeros) is still a variable — it is free, not zero. (See Exercise 2 below, textbook Exercise 8: x3x_3 is free, not x3=0x_3 = 0.)
  • A basic variable can be a plain number with no free variables in its formula (like x5=4x_5 = 4 above) — it is still basic.
  • Always solve for the basic variables in terms of the free ones, never the other way around.

Existence and Uniqueness

Theorem 2 — Existence and Uniqueness Theorem (Lay §1.2): A linear system is consistent if and only if the rightmost column of the augmented matrix is not a pivot column — that is, if and only if an echelon form of the augmented matrix has no row of the form [ 0 ⋯ 0  b ]with b≠0.[\,0 \ \cdots \ 0 \ \ b\,] \quad \text{with } b \ne 0. If the system is consistent, then the solution set contains either

  • (i) a unique solution, when there are no free variables, or
  • (ii) infinitely many solutions, when there is at least one free variable.

How to answer “consistent? unique?” questions — only an echelon form is needed:

  1. Row reduce the augmented matrix to echelon form (RREF not required).
  2. Is the last (augmented) column a pivot column? Yes → inconsistent. Stop.
  3. Otherwise consistent. Is every coefficient column a pivot column? Yes → unique solution. No → free variable(s) → infinitely many solutions.

Using Row Reduction to Solve a Linear System (Lay §1.2 summary)

  1. Write the augmented matrix.
  2. Row reduce to echelon form; decide consistency. If inconsistent, stop.
  3. Continue to the RREF.
  4. Write the system of equations corresponding to the RREF.
  5. Rewrite each nonzero equation to express its basic variable in terms of the free variables.

Solved Examples (Textbook Questions)

Exercise 1

(Adapted from Lay §1.2, Exercises 3 & 4)

Row reduce each matrix to reduced echelon form. Identify the pivot positions in the final matrix and in the original matrix, and list the pivot columns.

(a) [123445676789](b) [135735795791]\text{(a)}\ \begin{bmatrix} 1 & 2 & 3 & 4 \\ 4 & 5 & 6 & 7 \\ 6 & 7 & 8 & 9 \end{bmatrix} \qquad \text{(b)}\ \begin{bmatrix} 1 & 3 & 5 & 7 \\ 3 & 5 & 7 & 9 \\ 5 & 7 & 9 & 1 \end{bmatrix}
Show solution ↓
Solution

(a) R2←R2−4R1R_2 \leftarrow R_2 - 4R_1, R3←R3−6R1R_3 \leftarrow R_3 - 6R_1; then R2←−13R2R_2 \leftarrow -\tfrac13 R_2:

[123445676789]∼[12340−3−6−90−5−10−15]∼[123401230−5−10−15]\begin{bmatrix} 1 & 2 & 3 & 4 \\ 4 & 5 & 6 & 7 \\ 6 & 7 & 8 & 9 \end{bmatrix} \sim \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & -3 & -6 & -9 \\ 0 & -5 & -10 & -15 \end{bmatrix} \sim \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & -5 & -10 & -15 \end{bmatrix}

R3←R3+5R2R_3 \leftarrow R_3 + 5R_2 (echelon form), then R1←R1−2R2R_1 \leftarrow R_1 - 2R_2 (RREF):

∼[123401230000]∼[10−1−201230000]\sim \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix} \sim \begin{bmatrix} \boxed{1} & 0 & -1 & -2 \\ 0 & \boxed{1} & 2 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}

Pivot positions in the original matrix are at the same places: [123445676789]\begin{bmatrix} \boxed{1} & 2 & 3 & 4 \\ 4 & \boxed{5} & 6 & 7 \\ 6 & 7 & 8 & 9 \end{bmatrix}. Pivot columns: 1 and 2.

(b) R2←R2−3R1R_2 \leftarrow R_2 - 3R_1, R3←R3−5R1R_3 \leftarrow R_3 - 5R_1; then R2←−14R2R_2 \leftarrow -\tfrac14 R_2; then R3←R3+8R2R_3 \leftarrow R_3 + 8R_2:

[135735795791]∼[13570−4−8−120−8−16−34]∼[135701230−8−16−34]∼[13570123000−10]\begin{bmatrix} 1 & 3 & 5 & 7 \\ 3 & 5 & 7 & 9 \\ 5 & 7 & 9 & 1 \end{bmatrix} \sim \begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & -4 & -8 & -12 \\ 0 & -8 & -16 & -34 \end{bmatrix} \sim \begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & 1 & 2 & 3 \\ 0 & -8 & -16 & -34 \end{bmatrix} \sim \begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & -10 \end{bmatrix}

Backward phase: R3←−110R3R_3 \leftarrow -\tfrac{1}{10}R_3; then R1←R1−7R3R_1 \leftarrow R_1 - 7R_3, R2←R2−3R3R_2 \leftarrow R_2 - 3R_3; then R1←R1−3R2R_1 \leftarrow R_1 - 3R_2:

∼[135701230001]∼[135001200001]∼[10−1001200001]\sim \begin{bmatrix} 1 & 3 & 5 & 7 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 1 \end{bmatrix} \sim \begin{bmatrix} 1 & 3 & 5 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \sim \begin{bmatrix} \boxed{1} & 0 & -1 & 0 \\ 0 & \boxed{1} & 2 & 0 \\ 0 & 0 & 0 & \boxed{1} \end{bmatrix}

In the original matrix: [135735795791]\begin{bmatrix} \boxed{1} & 3 & 5 & 7 \\ 3 & \boxed{5} & 7 & 9 \\ 5 & 7 & 9 & \boxed{1} \end{bmatrix}. Pivot columns: 1, 2 and 4.

Note that the pivot positions in the original matrix need not contain a 1 or even a “nice” number — only their location matters.

Exercise 2

(Adapted from Lay §1.2, Exercises 7–10)

Find the general solutions of the systems whose augmented matrices are given.

(a) [13473976](b) [140727010]\text{(a)}\ \left[\begin{array}{ccc|c} 1 & 3 & 4 & 7 \\ 3 & 9 & 7 & 6 \end{array}\right] \qquad \text{(b)}\ \left[\begin{array}{ccc|c} 1 & 4 & 0 & 7 \\ 2 & 7 & 0 & 10 \end{array}\right](c) [01−651−27−6](d) [1−2−133−6−22]\text{(c)}\ \left[\begin{array}{ccc|c} 0 & 1 & -6 & 5 \\ 1 & -2 & 7 & -6 \end{array}\right] \qquad \text{(d)}\ \left[\begin{array}{ccc|c} 1 & -2 & -1 & 3 \\ 3 & -6 & -2 & 2 \end{array}\right]
Show solution ↓
Solution

(a) Exercise 7. R2←R2−3R1R_2 \leftarrow R_2 - 3R_1, then R2←−15R2R_2 \leftarrow -\tfrac15 R_2, then R1←R1−4R2R_1 \leftarrow R_1 - 4R_2:

[13473976]∼[134700−5−15]∼[13470013]∼[130−50013]\left[\begin{array}{ccc|c} 1 & 3 & 4 & 7 \\ 3 & 9 & 7 & 6 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 3 & 4 & 7 \\ 0 & 0 & -5 & -15 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 3 & 4 & 7 \\ 0 & 0 & 1 & 3 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 3 & 0 & -5 \\ 0 & 0 & 1 & 3 \end{array}\right]

Pivots in columns 1 and 3 → basic x1,x3x_1, x_3; free x2x_2. Equations: x1+3x2=−5x_1 + 3x_2 = -5, x3=3x_3 = 3.

{x1=−5−3x2x2 is freex3=3\begin{cases} x_1 = -5 - 3x_2 \\ x_2 \text{ is free} \\ x_3 = 3 \end{cases}

(b) Exercise 8. R2←R2−2R1R_2 \leftarrow R_2 - 2R_1, then R2←−R2R_2 \leftarrow -R_2, then R1←R1−4R2R_1 \leftarrow R_1 - 4R_2:

[140727010]∼[14070−10−4]∼[14070104]∼[100−90104]\left[\begin{array}{ccc|c} 1 & 4 & 0 & 7 \\ 2 & 7 & 0 & 10 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 4 & 0 & 7 \\ 0 & -1 & 0 & -4 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 4 & 0 & 7 \\ 0 & 1 & 0 & 4 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 0 & 0 & -9 \\ 0 & 1 & 0 & 4 \end{array}\right]

Basic x1,x2x_1, x_2; free x3x_3:

{x1=−9x2=4x3 is free\begin{cases} x_1 = -9 \\ x_2 = 4 \\ x_3 \text{ is free} \end{cases}

Classic error: writing x3=0x_3 = 0 because x3x_3 “doesn’t appear”. Column 3 has no pivot, so x3x_3 is free — the basic variables just happen not to depend on it.

(c) Exercise 9. Swap R1↔R2R_1 \leftrightarrow R_2, then R1←R1+2R2R_1 \leftarrow R_1 + 2R_2:

[01−651−27−6]∼[1−27−601−65]∼[10−5401−65]\left[\begin{array}{ccc|c} 0 & 1 & -6 & 5 \\ 1 & -2 & 7 & -6 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & -2 & 7 & -6 \\ 0 & 1 & -6 & 5 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 0 & -5 & 4 \\ 0 & 1 & -6 & 5 \end{array}\right]

Basic x1,x2x_1, x_2; free x3x_3. Equations x1−5x3=4x_1 - 5x_3 = 4, x2−6x3=5x_2 - 6x_3 = 5:

{x1=4+5x3x2=5+6x3x3 is free\begin{cases} x_1 = 4 + 5x_3 \\ x_2 = 5 + 6x_3 \\ x_3 \text{ is free} \end{cases}

(d) Exercise 10. R2←R2−3R1R_2 \leftarrow R_2 - 3R_1, then R1←R1+R2R_1 \leftarrow R_1 + R_2:

[1−2−133−6−22]∼[1−2−13001−7]∼[1−20−4001−7]\left[\begin{array}{ccc|c} 1 & -2 & -1 & 3 \\ 3 & -6 & -2 & 2 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & -2 & -1 & 3 \\ 0 & 0 & 1 & -7 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & -2 & 0 & -4 \\ 0 & 0 & 1 & -7 \end{array}\right]

Basic x1,x3x_1, x_3; free x2x_2. Equations x1−2x2=−4x_1 - 2x_2 = -4, x3=−7x_3 = -7:

{x1=−4+2x2x2 is freex3=−7\begin{cases} x_1 = -4 + 2x_2 \\ x_2 \text{ is free} \\ x_3 = -7 \end{cases}
Exercise 3

(Adapted from Lay §1.2, Exercises 11 & 12)

Find the general solutions of the systems whose augmented matrices are given.

(a) [3−420−912−60−68−40](b) [1−7065001−2−3−17−427]\text{(a)}\ \left[\begin{array}{ccc|c} 3 & -4 & 2 & 0 \\ -9 & 12 & -6 & 0 \\ -6 & 8 & -4 & 0 \end{array}\right] \qquad \text{(b)}\ \left[\begin{array}{cccc|c} 1 & -7 & 0 & 6 & 5 \\ 0 & 0 & 1 & -2 & -3 \\ -1 & 7 & -4 & 2 & 7 \end{array}\right]
Show solution ↓
Solution

(a) Exercise 11. Rows 2 and 3 are multiples of row 1 (R2=−3R1R_2 = -3R_1, R3=−2R1R_3 = -2R_1). So R2←R2+3R1R_2 \leftarrow R_2 + 3R_1, R3←R3+2R1R_3 \leftarrow R_3 + 2R_1, then R1←13R1R_1 \leftarrow \tfrac13 R_1:

[3−420−912−60−68−40]∼[3−42000000000]∼[1−4323000000000]\left[\begin{array}{ccc|c} 3 & -4 & 2 & 0 \\ -9 & 12 & -6 & 0 \\ -6 & 8 & -4 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} 3 & -4 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & -\tfrac43 & \tfrac23 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]

Basic x1x_1; free x2,x3x_2, x_3. Equation: x1−43x2+23x3=0x_1 - \tfrac43 x_2 + \tfrac23 x_3 = 0:

{x1=43x2−23x3x2 is freex3 is free\begin{cases} x_1 = \tfrac43 x_2 - \tfrac23 x_3 \\ x_2 \text{ is free} \\ x_3 \text{ is free} \end{cases}

(b) Exercise 12. R3←R3+R1R_3 \leftarrow R_3 + R_1, then R3←R3+4R2R_3 \leftarrow R_3 + 4R_2:

[1−7065001−2−3−17−427]∼[1−7065001−2−300−4812]∼[1−7065001−2−300000]\left[\begin{array}{cccc|c} 1 & -7 & 0 & 6 & 5 \\ 0 & 0 & 1 & -2 & -3 \\ -1 & 7 & -4 & 2 & 7 \end{array}\right] \sim \left[\begin{array}{cccc|c} 1 & -7 & 0 & 6 & 5 \\ 0 & 0 & 1 & -2 & -3 \\ 0 & 0 & -4 & 8 & 12 \end{array}\right] \sim \left[\begin{array}{cccc|c} 1 & -7 & 0 & 6 & 5 \\ 0 & 0 & 1 & -2 & -3 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]

This is already the RREF. Pivots in columns 1 and 3 → basic x1,x3x_1, x_3; free x2,x4x_2, x_4. Equations: x1−7x2+6x4=5x_1 - 7x_2 + 6x_4 = 5, x3−2x4=−3x_3 - 2x_4 = -3, 0=00 = 0:

{x1=5+7x2−6x4x2 is freex3=−3+2x4x4 is free\begin{cases} x_1 = 5 + 7x_2 - 6x_4 \\ x_2 \text{ is free} \\ x_3 = -3 + 2x_4 \\ x_4 \text{ is free} \end{cases}
Exercise 4

(Adapted from Lay §1.2, Exercises 13 & 14)

Find the general solutions of the systems whose augmented matrices are given.

(a) [1−30−10−20100−41000194000000](b) [12−5−60−501−6−302000010000000]\text{(a)}\ \left[\begin{array}{ccccc|c} 1 & -3 & 0 & -1 & 0 & -2 \\ 0 & 1 & 0 & 0 & -4 & 1 \\ 0 & 0 & 0 & 1 & 9 & 4 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] \qquad \text{(b)}\ \left[\begin{array}{ccccc|c} 1 & 2 & -5 & -6 & 0 & -5 \\ 0 & 1 & -6 & -3 & 0 & 2 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right]
Show solution ↓
Solution

Both matrices are already in echelon form, so only the backward phase is needed.

(a) Exercise 13. Pivots in columns 1, 2, 4. Clear above the pivots: R1←R1+R3R_1 \leftarrow R_1 + R_3 (kills the −1-1 in column 4), then R1←R1+3R2R_1 \leftarrow R_1 + 3R_2 (kills the −3-3 in column 2):

∼[1−300920100−41000194000000]∼[1000−350100−41000194000000]\sim \left[\begin{array}{ccccc|c} 1 & -3 & 0 & 0 & 9 & 2 \\ 0 & 1 & 0 & 0 & -4 & 1 \\ 0 & 0 & 0 & 1 & 9 & 4 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] \sim \left[\begin{array}{ccccc|c} 1 & 0 & 0 & 0 & -3 & 5 \\ 0 & 1 & 0 & 0 & -4 & 1 \\ 0 & 0 & 0 & 1 & 9 & 4 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right]

(Row 1: 9+3(−4)=−39 + 3(-4) = -3 and 2+3(1)=52 + 3(1) = 5.) Basic x1,x2,x4x_1, x_2, x_4; free x3,x5x_3, x_5. Equations: x1−3x5=5x_1 - 3x_5 = 5, x2−4x5=1x_2 - 4x_5 = 1, x4+9x5=4x_4 + 9x_5 = 4:

{x1=5+3x5x2=1+4x5x3 is freex4=4−9x5x5 is free\begin{cases} x_1 = 5 + 3x_5 \\ x_2 = 1 + 4x_5 \\ x_3 \text{ is free} \\ x_4 = 4 - 9x_5 \\ x_5 \text{ is free} \end{cases}

(Again: x3x_3 is free, not 00 — column 3 is all zeros, so it has no pivot.)

(b) Exercise 14. Pivots in columns 1, 2, 5. Column 5 already has zeros above its pivot, so only R1←R1−2R2R_1 \leftarrow R_1 - 2R_2 is needed:

∼[10700−901−6−302000010000000]\sim \left[\begin{array}{ccccc|c} 1 & 0 & 7 & 0 & 0 & -9 \\ 0 & 1 & -6 & -3 & 0 & 2 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right]

(Row 1: −5−2(−6)=7-5 - 2(-6) = 7, −6−2(−3)=0-6 - 2(-3) = 0, −5−2(2)=−9-5 - 2(2) = -9.) Basic x1,x2,x5x_1, x_2, x_5; free x3,x4x_3, x_4. Equations: x1+7x3=−9x_1 + 7x_3 = -9, x2−6x3−3x4=2x_2 - 6x_3 - 3x_4 = 2, x5=0x_5 = 0:

{x1=−9−7x3x2=2+6x3+3x4x3 is freex4 is freex5=0\begin{cases} x_1 = -9 - 7x_3 \\ x_2 = 2 + 6x_3 + 3x_4 \\ x_3 \text{ is free} \\ x_4 \text{ is free} \\ x_5 = 0 \end{cases}

(Here x5=0x_5 = 0 is correct: x5x_5 is a basic variable whose equation happens to say x5=0x_5 = 0.)

Exercise 5

(Adapted from Lay §1.2, Exercises 15 & 16)

Each matrix below is the augmented matrix of a linear system, in echelon form (■\blacksquare = nonzero leading entry, ∗\ast = any value, possibly zero). In each case determine whether the system is consistent, and if so, whether the solution is unique.

15(a) [■∗∗∗0■∗∗00■0]15(b) [0■∗∗∗00■∗∗0000■]\text{15(a)}\ \begin{bmatrix} \blacksquare & \ast & \ast & \ast \\ 0 & \blacksquare & \ast & \ast \\ 0 & 0 & \blacksquare & 0 \end{bmatrix} \qquad \text{15(b)}\ \begin{bmatrix} 0 & \blacksquare & \ast & \ast & \ast \\ 0 & 0 & \blacksquare & \ast & \ast \\ 0 & 0 & 0 & 0 & \blacksquare \end{bmatrix}16(a) [■∗∗0■∗000]16(b) [■∗∗∗∗00■∗∗000■∗]\text{16(a)}\ \begin{bmatrix} \blacksquare & \ast & \ast \\ 0 & \blacksquare & \ast \\ 0 & 0 & 0 \end{bmatrix} \qquad \text{16(b)}\ \begin{bmatrix} \blacksquare & \ast & \ast & \ast & \ast \\ 0 & 0 & \blacksquare & \ast & \ast \\ 0 & 0 & 0 & \blacksquare & \ast \end{bmatrix}
Show solution ↓
Solution

Apply Theorem 2: (1) is the last column a pivot column? (2) if not, is every variable column a pivot column?

15(a) 3 variables (columns 1–3), last column = augmented. Pivots in columns 1, 2, 3; the augmented column has no pivot (its bottom entry is 0, below no leading entry) → consistent. Every variable column has a pivot → no free variables → unique solution.

15(b) 4 variables (columns 1–4). The last row is [0 0 0 0 ■][0\ 0\ 0\ 0\ \blacksquare], i.e. 0=nonzero0 = \text{nonzero}. The rightmost column is a pivot column → inconsistent.

16(a) 2 variables (columns 1–2). Pivots in columns 1 and 2; the last row is all zeros (0=00 = 0, harmless), so the augmented column has no pivot → consistent. Both variable columns are pivot columns → unique solution.

16(b) 4 variables (columns 1–4). Pivots in columns 1, 3, 4; the augmented column (5) has no pivot → consistent. Column 2 has no pivot, so x2x_2 is free → infinitely many solutions.

These questions never need numbers — only the shape (where the pivots are). Count variable columns carefully: the last column is always the augmented column, not a variable.

Exercise 6

(Adapted from Lay §1.2, Exercises 17 & 18)

Determine the value(s) of hh such that the matrix is the augmented matrix of a consistent linear system.

(a) [23h467](b) [1−3−25h−7]\text{(a)}\ \left[\begin{array}{cc|c} 2 & 3 & h \\ 4 & 6 & 7 \end{array}\right] \qquad \text{(b)}\ \left[\begin{array}{cc|c} 1 & -3 & -2 \\ 5 & h & -7 \end{array}\right]
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Solution

(a) Exercise 17. R2←R2−2R1R_2 \leftarrow R_2 - 2R_1:

[23h467]∼[23h007−2h]\left[\begin{array}{cc|c} 2 & 3 & h \\ 4 & 6 & 7 \end{array}\right] \sim \left[\begin{array}{cc|c} 2 & 3 & h \\ 0 & 0 & 7 - 2h \end{array}\right]

Row 2 says 0=7−2h0 = 7 - 2h. The system has a solution only if 7−2h=07 - 2h = 0, i.e. h=72h = \tfrac72. (Then the two equations describe the same line → infinitely many solutions. For any other hh the lines are parallel → no solution.)

(b) Exercise 18. R2←R2−5R1R_2 \leftarrow R_2 - 5R_1:

[1−3−25h−7]∼[1−3−20h+153]\left[\begin{array}{cc|c} 1 & -3 & -2 \\ 5 & h & -7 \end{array}\right] \sim \left[\begin{array}{cc|c} 1 & -3 & -2 \\ 0 & h + 15 & 3 \end{array}\right]
  • If h+15=0h + 15 = 0, i.e. h=−15h = -15, row 2 reads 0=30 = 3 → no solution.
  • If h≠−15h \ne -15, h+15h + 15 is a pivot and the augmented column is not a pivot column → consistent (unique solution).

Answer: consistent for all h≠−15h \ne -15.

Notice the two different shapes of answer: in (a) the hh sits in the augmented column, so consistency requires one specific value; in (b) the hh sits in a coefficient position, so consistency fails for one specific value. Always reduce first, then ask “can the last row become [0 0∣nonzero][0\ 0 \mid \text{nonzero}]?”