Exercises — Pivot positions, free variables

WHAT to look for: the first nonzero entry of each nonzero row — each is a leading 1, and each sits in its own column. Row 1's leading 1 is in column 1. Row 2's is in column 2. Row 3's is in column 4.

• Pivot columns: $\{1,2,4\}$.
• Rank $r=3$ (three pivots).
• Column 3 has no leading 1 → $x_3$ is the only free variable.
• Check: $\#\text{free}=n-r=4-3=1$. ✓

• $A$: $n=2$ columns, one leading 1 (row 1, col 1) so $r=1$. $\#\text{free}=2-1=1$.
• $B$: $n=3$, three leading 1's on the diagonal so $r=3$. $\#\text{free}=3-3=0$.
• $C$: $n=3$, leading 1's in columns 1 and 2 so $r=2$; column 3 is free. $\#\text{free}=3-2=1$.

Step 1 — clear below the first pivot. $R_2\leftarrow R_2-2R_1$: $\left[\begin{array}{cccc}1& 3& 1& 2\\ 0& 0& 2& 2\end{array}\right].$ Why? We want zeros under the leading 1 in column 1 so the echelon staircase appears. Step 2 — normalise the second pivot. $R_2\leftarrow \frac{1}{2}{R}_2$: gives $x_3=1$. Pivots sit in columns 1 and 3; column 2 has none → $x_2$ is free ($r=2$, $n=3$, so $\#\text{free}=1$). Step 3 — back-substitute. Row 1: $x_1=2-3x_2-x_3=2-3x_2-1=1-3x_2$. With $x_2=t$: $\mathbf{x}=\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]+t\left[\begin{array}{c}-3\\ 1\\ 0\end{array}\right].$ One free variable ⇒ a line in 3D. ✓ (See Null Space and Solution Sets of Ax=0 for why the second vector is the null-space direction.)

Step 1 — get a leading 1. Swap rows so the tidy row is on top, $R_1\leftrightarrow R_2$: $\left[\begin{array}{ccc}1& 3& 5\\ 2& 4& 8\end{array}\right].$ Step 2 — clear below. $R_2\leftarrow R_2-2R_1$: $\left[\begin{array}{ccc}1& 3& 5\\ 0& -2& -2\end{array}\right]$. Step 3 — normalise. $R_2\leftarrow -\frac{1}{2}{R}_2\Rightarrow x_2=1$. Both columns are pivot columns → no free variables ($n-r=2-2=0$). Step 4 — back-substitute. $x_1=5-3x_2=5-3=2$. Unique solution $(2,1)$. ✓

$n=6$ variables, $r=3$ pivots. $\#\text{free}=n-r=6-3=3$. A consistent system with 3 free variables has a 3-dimensional flat (a "3-plane") of solutions. (This is exactly $\dim$ of the shifted null space; see Rank and the Rank-Nullity Theorem.)

Left matrix. The last row is all zeros — a harmless $0=0$, not a pivot in the augmented column. So it's consistent. Coefficient pivots sit in columns 1 and 3; column 2 is free ⇒ $\#\text{free}=1$. Solution set is a line. Writing $x_2=t$: $x_1=4-2t$, $x_3=-1$, so $\mathbf{x}=(4,0,-1)+t(-2,1,0)$. Right matrix. Bottom row reads $0=7$ — the augmented column is a pivot column. Inconsistent → no solution. Free-variable count is irrelevant here: freedom describes a solution set only when one exists.

$r=3=n$, so every column of $A$ is a pivot column → no free variables. Because $A$ has a pivot in every row too ($r=3=m$), the augmented column can never become a pivot column (there's no all-zero coefficient row to sit above a nonzero right side). So the system is consistent for every $\mathbf{b}$, and with $0$ free variables the solution is unique — exactly one point. (This is the invertible case; see Existence and Uniqueness of Solutions.)

A plane = $2$ free variables. With $n=4$, we need $\#\text{free}=n-r=2$, so $r=2$. Build any RREF with exactly two leading 1's, e.g. $A=\left[\begin{array}{cccc}1& 0& 5& 3\\ 0& 1& -2& 4\\ 0& 0& 0& 0\end{array}\right].$ Pivots in columns 1, 2 → $r=2$. Columns 3, 4 free → $\#\text{free}=4-2=2$. A homogeneous system is always consistent ($0$ works), so the solution set is a genuine 2-plane through the origin. ✓ (Its two spanning directions form a basis of the null space — see Basis and Dimension.)

Columns are independent $⟺$ every column is a pivot column $⟺r=2$ (no free variables in $A\mathbf{x}=0$; see Linear Independence). Reduce: $R_2\leftarrow R_2-2R_1$ gives $\left[\begin{array}{cc}1& 2\\ 0& k-4\end{array}\right]$.

• If $k-4\ne 0$ (e.g. $k=0$): second pivot exists, $r=2$ → independent.
• If $k-4=0$, i.e. $k=4$: second row is zero, only $r=1$, column 2 is free → dependent. Answer: $k=4$ makes them dependent; $k=0$ (any $k\ne 4$) keeps them independent.

Make the two coefficient rows proportional but the right-hand sides not matching. E.g. $\left[\begin{array}{cccc}1& 1& 1& 1\\ 2& 2& 2& 5\end{array}\right]\stackrel{R_2\leftarrow R_2-2R_1}{\to }\left[\begin{array}{cccc}1& 1& 1& 1\\ 0& 0& 0& 3\end{array}\right].$ The bottom row is $0=3$: the augmented column earns a pivot, so the system is inconsistent. ✓

(a) Reduce. $R_2\leftarrow R_2-R_1$ and $R_3\leftarrow R_3-2R_1$: $\left[\begin{array}{cccc}1& 2& 1& 1\\ 0& 0& 1& 2\\ 0& 0& 1& h-2\end{array}\right].$ Now $R_3\leftarrow R_3-R_2$: $\left[\begin{array}{cccc}1& 2& 1& 1\\ 0& 0& 1& 2\\ 0& 0& 0& h-4\end{array}\right].$ (b) Consistency. The bottom row reads $0=h-4$. This is a true statement (no augmented pivot) exactly when $h-4=0$, i.e. $h=4$. For any $h\ne 4$ the augmented column takes a pivot → inconsistent. (c) At $h=4$. Coefficient pivots sit in columns 1 and 3; column 2 is free ($r=2$, $n=3$, $\#\text{free}=1$). Back-substitute with $x_2=t$: row 2 gives $x_3=2$; row 1 gives $x_1=1-2x_2-x_3=1-2t-2=-1-2t$. So $\mathbf{x}=\left[\begin{array}{c}-1\\ 0\\ 2\end{array}\right]+t\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right].$ (d) Shape. One free variable ⇒ a line in 3D, passing through $(-1,0,2)$ in the direction $(-2,1,0)$.

The coefficient reduction is identical, giving pivots in columns 1, 3 and free column 2 — so $\#\text{free}=1$ regardless of the right-hand side (pivots depend only on $A$). Setting the constants to zero: $x_3=0$, $x_1=-2t$, $x_2=t$: ${\mathbf{x}}_{\text{hom}}=t\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right].$ This is the line through the origin in the same direction $(-2,1,0)$. The inhomogeneous solution set from 5.1 is this line shifted by the particular point $(-1,0,2)$ — geometrically, two parallel lines. That "particular + null-space" split is the general structure of solution sets (Null Space and Solution Sets of Ax=0).

Free variables count non-pivot columns ($n-r$), the right-hand side controls only consistency, and pivot positions belong to $A$ alone. Give the master formula and the two independent questions it splits into ::: $\#\text{free}=n-r$; Question 1 (existence) depends on $\mathbf{b}$/consistency, Question 2 (size of solution set) depends only on $A$'s pivots.