9.1 Systems of Linear Equations, Matrix Rank, and Row Reduction

9.1 Systems of Linear Equations, Matrix Rank, and Row Reduction

Linear systems of equations constitute the foundational algebraic machinery tested on the GRE Mathematics Subject Test. A system of $m$ linear equations in $n$ unknowns is represented compactly in matrix form as $A\mathbf{x} = \mathbf{b}$, where $A \in \mathbb{R}^{m \times n}$ is the coefficient matrix, $\mathbf{x} \in \mathbb{R}^n$ is the column vector of unknowns, and $\mathbf{b} \in \mathbb{R}^m$ is the right-hand side vector. Evaluating the existence, uniqueness, and structural dimension of the solution set requires systematic row reduction, matrix rank theorems, and the geometric duality of the four fundamental subspaces.

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Augmented Matrices and Elementary Row Operations

The entire algebraic data of the nonhomogeneous system $A\mathbf{x} = \mathbf{b}$ is encoded within the augmented matrix $[A \mid \mathbf{b}] \in \mathbb{R}^{m \times (n+1)}$. Solving the system corresponds to executing a sequence of elementary row operations, each of which corresponds to left-multiplication by an invertible elementary matrix $E$:

1. Row Swap ($R_i \leftrightarrow R_j$): Transposes row $i$ and row $j$. The corresponding elementary matrix $E_{ij}$ has determinant $-1$.
2. Row Scaling ($c R_i \to R_i$, $c \neq 0$): Multiplies every entry in row $i$ by a non-zero scalar $c$. The corresponding elementary matrix has determinant $c$.
3. Row Addition ($R_i + c R_j \to R_i$, $j \neq i$): Adds $c$ times row $j$ to row $i$. The corresponding shear matrix has determinant $1$.

Because every elementary matrix is invertible, row equivalence preserves the solution set: $A\mathbf{x} = \mathbf{b}$ and $(E_k \cdots E_1 A)\mathbf{x} = (E_k \cdots E_1 \mathbf{b})$ share identical solutions.

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Row Echelon Form (REF) and Reduced Row Echelon Form (RREF)

Gaussian elimination systematically simplifies an augmented matrix into Row Echelon Form (REF):

• All all-zero rows reside strictly at the bottom of the matrix.
• In each non-zero row, the leading non-zero coefficient (the pivot) appears strictly to the right of the leading coefficient of the row directly above it.

Gauss-Jordan elimination continues the reduction to produce the unique Reduced Row Echelon Form (RREF):

• Every pivot entry is scaled to $1$.
• Each pivot column contains zeros in every entry except the pivot itself.

While an arbitrary matrix possesses infinitely many valid REFs depending on the choice of row operations, its RREF is strictly unique. Columns containing pivots correspond to basic (pivot) variables, whereas non-pivot columns correspond to free variables that parametrize the solution manifold.

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The Rank Theorem and Solvability Criteria

The rank of a matrix $A$, denoted $\operatorname{rank}(A)$, is the number of pivots in its echelon form.

The Rank Theorem

For any $m \times n$ matrix $A$, the dimension of the vector space spanned by its rows (row rank) equals the dimension of the vector space spanned by its columns (column rank): $\operatorname{dim}(\operatorname{row}(A)) = \operatorname{dim}(\operatorname{col}(A)) = \operatorname{rank}(A) = r \le \min(m, n).$ Elementary row operations preserve the row space $\operatorname{row}(A)$ and preserve all linear dependence relations between columns, proving that row rank equals column rank.

The Rouché-Capelli Consistency Criterion

A linear system $A\mathbf{x} = \mathbf{b}$ is consistent if and only if $\mathbf{b} \in \operatorname{col}(A)$, which holds if and only if: $\operatorname{rank}(A) = \operatorname{rank}([A \mid \mathbf{b}]).$ If $\operatorname{rank}([A \mid \mathbf{b}]) = \operatorname{rank}(A) + 1$, the augmented column contains a pivot, producing the contradiction $0 = 1$ (no solution).

When consistent ($\operatorname{rank}(A) = \operatorname{rank}([A \mid \mathbf{b}]) = r$):

• Unique Solution: If $r = n$ (full column rank), there are no free variables. If $A$ is square ($n \times n$), $A$ is invertible and $\mathbf{x} = A^{-1}\mathbf{b}$.
• Infinitely Many Solutions: If $r < n$, there are $n - r$ free variables. The complete solution set is the affine subspace $\mathbf{x} = \mathbf{x}_p + \mathbf{x}_h$, where $\mathbf{x}_p$ is any particular solution and $\mathbf{x}_h \in \operatorname{null}(A)$.

For homogeneous systems $A\mathbf{x} = \mathbf{0}$, $\mathbf{x} = \mathbf{0}$ is always a solution. Non-trivial solutions exist if and only if $\operatorname{rank}(A) < n$.

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The Four Fundamental Subspaces

An $m \times n$ matrix $A$ of rank $r$ defines four canonical vector subspaces:

Fundamental Theorem of Linear Algebra

The row space and null space are orthogonal complements in $\mathbb{R}^n$: $\operatorname{row}(A)^\perp = \operatorname{null}(A), \quad \mathbb{R}^n = \operatorname{row}(A) \oplus \operatorname{null}(A).$ The column space and left null space are orthogonal complements in $\mathbb{R}^m$: $\operatorname{col}(A)^\perp = \operatorname{null}(A^T), \quad \mathbb{R}^m = \operatorname{col}(A) \oplus \operatorname{null}(A^T).$

The Rank-Nullity Theorem asserts that $\operatorname{dim}(\operatorname{col}(A)) + \operatorname{dim}(\operatorname{null}(A)) = n$.

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Step-by-Step Worked Problems

Problem 1: Parameter-Dependent System Solvability

Determine all values of $k \in \mathbb{R}$ for which the system has: (a) no solution, (b) a unique solution, and (c) infinitely many solutions. $\begin{aligned} x + y + kz &= 2 \\ 3x + 4y + 2z &= k \\ 2x + 3y - z &= 1 \end{aligned}$

Solution:

1. Form the augmented matrix and apply row operations: $\begin{pmatrix} 1 & 1 & k & 2 \\ 3 & 4 & 2 & k \\ 2 & 3 & -1 & 1 \end{pmatrix} \xrightarrow{R_2 - 3R_1, \; R_3 - 2R_1} \begin{pmatrix} 1 & 1 & k & 2 \\ 0 & 1 & 2 - 3k & k - 6 \\ 0 & 1 & -1 - 2k & -3 \end{pmatrix}$
2. Subtract $R_2$ from $R_3$: $\xrightarrow{R_3 - R_2} \begin{pmatrix} 1 & 1 & k & 2 \\ 0 & 1 & 2 - 3k & k - 6 \\ 0 & 0 & k - 3 & 3 - k \end{pmatrix}$
3. Analyze the bottom row $[0 \quad 0 \quad k - 3 \mid 3 - k]$:
   • If $k \neq 3$: The pivot entry is $k - 3 \neq 0$. Thus $\operatorname{rank}(A) = \operatorname{rank}([A \mid \mathbf{b}]) = 3 = n$. The system has a unique solution.
   • If $k = 3$: The bottom row becomes $[0 \quad 0 \quad 0 \mid 0]$. Thus $\operatorname{rank}(A) = \operatorname{rank}([A \mid \mathbf{b}]) = 2 < 3$. The system has infinitely many solutions parameterized by $3 - 2 = 1$ free variable.
   • There is no value of $k$ for which the system has no solution.

Problem 2: Basis for the Left Null Space

Find a basis for $\operatorname{null}(A^T)$ where $A = \begin{pmatrix} 1 & 2 & 0 \ 0 & 1 & 1 \ 1 & 3 & 1 \end{pmatrix}$.

Solution:

1. Transpose $A$: $A^T = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{pmatrix}$
2. Row-reduce $[A^T \mid \mathbf{0}]$: $\begin{pmatrix} 1 & 0 & 1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{pmatrix} \xrightarrow{R_2 - 2R_1} \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix} \xrightarrow{R_3 - R_2} \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$
3. Express basic variables in terms of the free variable $y_3 = t$: $y_1 = -t, \quad y_2 = -t, \quad y_3 = t \implies \mathbf{y} = t \begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix}.$
4. A basis for $\operatorname{null}(A^T)$ is \{(-1, -1, 1)^T\}. Notice that this vector is orthogonal to every column of $A$: $(-1)(1) + (-1)(0) + 1(1) = 0$, $(-1)(2) + (-1)(1) + 1(3) = 0$, and $(-1)(0) + (-1)(1) + 1(1) = 0$.

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GRE Exam Traps & Pitfalls

Trap 1: Extracting Column Space Basis from RREF Columns Elementary row operations change the column space! The pivot columns of the RREF identify which columns are linearly independent, but the basis vectors for $\operatorname{col}(A)$ must be extracted from the original matrix $A$, not from the RREF.

Trap 2: Mismatching Ambient Dimensions in Fundamental Subspaces For an $m \times n$ matrix $A$, $\operatorname{null}(A)$ and $\operatorname{row}(A)$ live in $\mathbb{R}^n$, whereas $\operatorname{col}(A)$ and $\operatorname{null}(A^T)$ live in $\mathbb{R}^m$. Never take inner products between vectors from $\operatorname{null}(A)$ and $\operatorname{col}(A)$ when $m \neq n$.

Trap 3: Overlooking Inconsistency in Parameter Problems A system is inconsistent only if a pivot lands in the augmented column ($[0 \dots 0 \mid c]$ with $c \neq 0$). Do not confuse a zero row in $A$ with inconsistency unless the corresponding right-hand entry is non-zero.