Homogeneous systems and solutions in parametric vector form
A linear system has infinitely many solutions. How do you write the whole set down at once, and what shape does it have?
Where you are 7 in Row reduction
When you reach for it
Solving a system by row reduction: one, none or infinitely many solutions ends with a solution set in coordinates, such as (x, y, z) = (2 - t, 1 + 3t, t), one parameter per free variable. That is correct but hides the shape of the set, and says nothing about how it compares with the solutions for a different right side. This lesson writes the same answer as
where \mathbf{p} is one solution and s\mathbf{v}_1 + t\mathbf{v}_2 runs over every solution of the system with its right sides set to 0. Reach for it to describe, compare or picture a solution set, to decide uniqueness without solving, and to understand problems with many acceptable answers, such as mixing materials to a target. The tool underneath is Multiplying a matrix by a vector: A\mathbf{x} = x_1\mathbf{a}_1 + \dots + x_n\mathbf{a}_n, and A(c\mathbf{u} + d\mathbf{v}) = cA\mathbf{u} + dA\mathbf{v}.
The procedure
Day 1: homogeneous systems, and their solutions as a span
A system is homogeneous when every right side is 0: A\mathbf{x} = \mathbf{0}. Since A\mathbf{0} = \mathbf{0}, it is always consistent, with the trivial solution \mathbf{x} = \mathbf{0}. A solution \mathbf{x} \neq \mathbf{0} is nontrivial, and the question is whether any exist. Row reduce A alone: every row operation turns a column of zeros into a column of zeros, so the augmented column can be left off.
When nontrivial solutions exist. If every column of the reduced matrix has a pivot, the system says x_1 = 0, \dots, x_n = 0. If some column has none, set its free variable to 1 and the other free variables to 0, and solve for the basic variables: the result has a 1 in it, so it is not \mathbf{0}. So A\mathbf{x} = \mathbf{0} has a nontrivial solution exactly when there is a free variable, that is, when the rank of A is less than its number of columns. An m \times n matrix has at most one pivot per row, so if n > m some column has none: a homogeneous system with more unknowns than equations always has nontrivial solutions.
Parametric vector form. Write each basic variable in terms of the free ones, name the free variables s, t, and so on, and write out \mathbf{x} one entry per variable. Each entry is a sum of an s term, a t term, and so on. Vectors add and scale entry by entry, so \mathbf{x} splits into its s part plus its t part, each a free variable times a fixed vector: \mathbf{x} = s\mathbf{v}_1 + t\mathbf{v}_2 + \dots. Example 2 carries it out.
The solution set is a span. If the free variables give \mathbf{v}_1, \dots, \mathbf{v}_k, then
Why. The splitting shows every solution is a combination of the \mathbf{v}_i, with the free variables as weights. Conversely each \mathbf{v}_i is a solution (its own free variable 1, the others 0), and by linearity so is every combination:
With no free variable, k = 0 and the set is \{\mathbf{0}\}. Being a span, the set is a subspace of \mathbb{R}^n. The \mathbf{v}_i are linearly independent, since \mathbf{v}_i has a 1 in its own free entry and 0 in every other free entry, and there are n - \operatorname{rank} A of them. Later this set is named the kernel of A.
Day 2: one solution plus the homogeneous solutions
For \mathbf{b} \neq \mathbf{0}, the system A\mathbf{x} = \mathbf{b} is nonhomogeneous, and A\mathbf{x} = \mathbf{0} is its associated homogeneous system. Any one solution of A\mathbf{x} = \mathbf{b} is a particular solution.
Theorem. If A\mathbf{p} = \mathbf{b}, the solutions of A\mathbf{x} = \mathbf{b} are exactly the vectors
Proof. If \mathbf{x} is a solution, put \mathbf{v}_h = \mathbf{x} - \mathbf{p}; by linearity A\mathbf{v}_h = \mathbf{b} - \mathbf{b} = \mathbf{0}, and \mathbf{x} = \mathbf{p} + \mathbf{v}_h. Conversely, if A\mathbf{v}_h = \mathbf{0}, then A(\mathbf{p} + \mathbf{v}_h) = \mathbf{b} + \mathbf{0} = \mathbf{b}. With Day 1, every solution is \mathbf{x} = \mathbf{p} + c_1\mathbf{v}_1 + \dots + c_k\mathbf{v}_k.
Both parts from one reduction. Reducing [\,A \mid \mathbf{b}\,] does to the columns of A what it would do with \mathbf{0} as the last column, so only the constants differ from the homogeneous reduction. In parametric vector form the constants make up \mathbf{p} (the solution with every free variable 0), and the free-variable terms are the same \mathbf{v}_i.
The picture. Adding \mathbf{p} to every vector of a set slides the set without turning or stretching it. So the solutions of A\mathbf{x} = \mathbf{b} are those of A\mathbf{x} = \mathbf{0} translated by \mathbf{p}: a line through the origin and a parallel line through \mathbf{p}, or a plane and a parallel plane. The translate misses \mathbf{0}, since A\mathbf{0} \neq \mathbf{b}, so it is not a span. The proof used nothing about \mathbf{p} but A\mathbf{p} = \mathbf{b}, so any solution can serve as \mathbf{p} and gives the same set.
Uniqueness without solving. Different \mathbf{v}_h give different solutions \mathbf{p} + \mathbf{v}_h, so a consistent system has exactly one solution when A\mathbf{x} = \mathbf{0} has only the trivial solution, and infinitely many otherwise. Every consistent right side gives a set of the same shape, with n - \operatorname{rank} A parameters; only its position changes. Whether A\mathbf{x} = \mathbf{b} is consistent at all is a separate question: whether \mathbf{b} is in the span of the columns.
Worked examples
Day 1: one free variable, or none
Find every solution of A\mathbf{x} = \mathbf{0} and of B\mathbf{x} = \mathbf{0}, where
For A, subtract 2 times row 1 from row 2, and row 1 from row 3. Both new rows are (0, 1, 3), so row 3 minus row 2 is zero. Subtracting 2 times row 2 from row 1 gives
Column 3 has no pivot, so x_3 = t is free, and x_1 = 7t, x_2 = -3t:
The solution set is \operatorname{span}\{(7, -3, 1)\}, a line through the origin in \mathbb{R}^3. Check: the rows of A times (7, -3, 1) give 7 - 6 - 1, 14 - 15 + 1 and 7 - 9 + 2, all 0.
For B the same steps give rows (0, 1, 3) and (0, 1, 4), whose difference (0, 0, 1) puts a pivot in every column. There is no free variable, and \mathbf{x} = \mathbf{0} is the only solution. One entry decided between a line and a point.
Day 1: two free variables, and the set as a span
Write the solutions of A\mathbf{x} = \mathbf{0} in parametric vector form, where
Subtract 2 times row 1 from row 2, and add row 1 to row 3. Both give (0, 0, 1, -3), so row 3 then becomes zero:
This is reduced, with pivots in columns 1 and 3. The rank is 2, so 4 - 2 = 2 variables are free, x_2 = s and x_4 = t, and the rows give x_1 = 3s - 2t, x_3 = 3t. Split by free variable:
So the solution set is \operatorname{span}\{\mathbf{v}_1, \mathbf{v}_2\} with \mathbf{v}_1 = (3, 1, 0, 0) and \mathbf{v}_2 = (-2, 0, 3, 1), a plane through the origin in \mathbb{R}^4.
Check by columns: A\mathbf{v}_1 = 3\mathbf{a}_1 + \mathbf{a}_2 = \mathbf{0} because \mathbf{a}_2 = -3\mathbf{a}_1, and A\mathbf{v}_2 = -2\mathbf{a}_1 + 3\mathbf{a}_3 + \mathbf{a}_4 has entries -2 + 0 + 2, -4 + 3 + 1 and 2 + 3 - 5, all 0. Each homogeneous solution records a way the columns combine to \mathbf{0}.
Day 1: one equation, a plane
Write the solutions of x_1 - 2x_2 + 3x_3 = 0 as a span, and find the weights that give the solution (1, 2, 1).
SolutionOne equation in three unknowns has one pivot, so two variables are free. With x_2 = s, x_3 = t and x_1 = 2s - 3t,
The solution set is the plane \operatorname{span}\{(2, 1, 0), (-3, 0, 1)\}: the vectors whose dot product with (1, -2, 3) is 0. Since 1 - 4 + 3 = 0, (1, 2, 1) is a solution, and its weights are its free entries, s = x_2 = 2 and t = x_3 = 1: indeed 2(2, 1, 0) + (-3, 0, 1) = (1, 2, 1). Reading weights off the free entries always works, because each \mathbf{v}_i has a 1 in its own free entry and 0 in the others.
Day 2: the same matrix with a right side
With A from Example 2 and \mathbf{b} = (2, 5, -1), solve A\mathbf{x} = \mathbf{b} in parametric vector form.
SolutionThe operations of Example 2 turn the constants into 2, then 5 - 4 = 1 and -1 + 2 = 1, then 1 - 1 = 0:
The last row says 0 = 0, so the system is consistent, and x_1 = 2 + 3s - 2t, x_3 = 1 + 3t:
The constant vector \mathbf{p} = (2, 0, 1, 0) is the solution with s = t = 0; check A\mathbf{p} = 2\mathbf{a}_1 + \mathbf{a}_3 = (2, 4, -2) + (0, 1, 1) = (2, 5, -1). The rest is exactly the solution set of Example 2, so the solutions of A\mathbf{x} = \mathbf{b} form that plane slid by \mathbf{p}, a plane in \mathbb{R}^4 missing the origin. Any member could replace \mathbf{p}: with s = t = 1 it is (3, 1, 4, 1), and A(3, 1, 4, 1) = (2, 5, -1) as well.
Day 2: two parallel lines in the plane
Solve A\mathbf{x} = \mathbf{0} and A\mathbf{x} = \mathbf{b}, and draw both solution sets, for
Adding 2 times row 1 to row 2 of [\,A \mid \mathbf{b}\,] gives (0, 0 \mid -6 + 6), all zeros, so the system is consistent and comes down to x_1 - x_2 = 3. With x_2 = t free,
Without \mathbf{p} = (3, 0) this is t(1, 1), the homogeneous solutions: the line x_2 = x_1 through the origin. The solutions of A\mathbf{x} = \mathbf{b} form the line x_2 = x_1 - 3, the same direction pushed over by \mathbf{p}.
Adding the homogeneous solution \mathbf{v}_h = (1, 1) to \mathbf{p} gives (4, 1), still on the line A\mathbf{x} = \mathbf{b}: every homogeneous solution points along that line, so adding one slides along it. And (4, 1) serves as a particular solution just as well, since (4, 1) + t(1, 1) = (3, 0) + (t + 1)(1, 1).
Day 2: mixing cullet to a target
Binta, Mamadou and Lamine run a glassblowing studio that melts three kinds of cullet (crushed glass). In simplified round figures, a bag of kind 1 holds 7 kilograms of silica, 2 of soda and 1 of lime; kind 2 holds 8, 1 and 1; kind 3 holds 6, 3 and 1. A batch needs 36 kilograms of silica, 9 of soda and 5 of lime. Find every mix, in bags with fractions allowed, and the mixes that can actually be weighed out.
SolutionWith x_j bags of kind j, each bag adds its column of contents, so the batch is A\mathbf{x} = \mathbf{b} with
Swap rows 1 and 3 to put the lime equation on top, then subtract 2 times it from the soda row and 7 times it from the silica row:
Row 3 plus row 2 is zero with constant 0, so the system is consistent. Scaling row 2 by -1 gives x_2 - x_3 = 1, and subtracting that from row 1 gives x_1 + 2x_3 = 4. With x_3 = t,
The particular solution is 4 bags of kind 1 and 1 of kind 2: 4(7, 2, 1) + (8, 1, 1) = (36, 9, 5). The homogeneous vector (-2, 1, 1) is a trade: one bag each of kinds 2 and 3 holds (14, 4, 2), exactly what two bags of kind 1 hold, so swapping them leaves the batch unchanged.
Amounts cannot be negative, so 4 - 2t \ge 0, 1 + t \ge 0 and t \ge 0, which give 0 \le t \le 2. The workable mixes form the segment from (4, 1, 0) to (0, 3, 2), whose far end uses no kind 1: 3(8, 1, 1) + 2(6, 3, 1) = (36, 9, 5). The algebra gives a line; the studio keeps a piece of it.
Day 2: every solution from two of them
A 3 \times 4 matrix A has rank 3, and \mathbf{u} = (1, 2, 0, 3) and \mathbf{w} = (3, 1, 1, 2) both solve A\mathbf{x} = \mathbf{b}. Without knowing A or \mathbf{b}, find every solution, decide whether (5, 0, 2, 1) is one, and decide whether \mathbf{b} could be \mathbf{0}.
SolutionRank 3 leaves 4 - 3 = 1 free variable, so the homogeneous solutions are the span of one nonzero vector. By the theorem, \mathbf{w} - \mathbf{u} = (2, -1, 1, -1) is a homogeneous solution, and it is not \mathbf{0}, so it spans them. With \mathbf{u} as particular solution,
For (5, 0, 2, 1) the third entry forces t = 2, and t = 2 gives (1 + 4, 2 - 2, 0 + 2, 3 - 2) = (5, 0, 2, 1), so it is a solution.
If \mathbf{b} were \mathbf{0}, then \mathbf{u} would be a homogeneous solution, c(2, -1, 1, -1) for some c. The third entries force c = 0, but \mathbf{u} \neq \mathbf{0}. So \mathbf{b} \neq \mathbf{0}, and the line of solutions misses the origin.
Boundaries
- The theorem needs one solution to exist. If A\mathbf{x} = \mathbf{b} is inconsistent there is no \mathbf{p} and the solution set is empty, though A\mathbf{x} = \mathbf{0} always has solutions. The homogeneous system gives the shape of the solutions if there are any, never whether there are any.
- Only for linear equations. The proof used A(\mathbf{x} - \mathbf{p}) = A\mathbf{x} - A\mathbf{p}. The solutions 2 and -2 of x^2 = 4 differ by 4, which does not solve x^2 = 0.
- With \mathbf{b} \neq \mathbf{0} the solution set is not a span. It misses \mathbf{0}, and the sum of two solutions solves A\mathbf{x} = 2\mathbf{b}.
- The description is not unique. Any solution can be \mathbf{p}, and the \mathbf{v}_i can be traded for other vectors with the same span. Different-looking descriptions can give the same set (Exercise 14), so compare sets, not lists of numbers. The count n - \operatorname{rank} A is for the vectors read from the free variables; a spanning list found another way may hold redundant vectors.
- Conditions from the setting are extra. Nonnegative amounts or whole numbers of items cut the set down, as in Example 6; they are not linear equations.
Where this goes next
The solution set of A\mathbf{x} = \mathbf{0} is the kernel of A, and its count n - \operatorname{rank} A is half of Rank and nullity. “Exactly one solution when A\mathbf{x} = \mathbf{0} has only the trivial one” becomes a condition in the The invertible matrix theorem. In Linear systems in use: networks, chemical equations, fitting curves, balancing a chemical equation is a homogeneous system, and a network of flows with free flows is the cullet problem on a larger scale. The pattern of a particular solution plus every solution of the zero problem returns for linear differential equations.
Exercises
Day 1: homogeneous systems in parametric vector form
In Exercises 1 to 6, find every solution of A\mathbf{x} = \mathbf{0} in parametric vector form, and say whether the solution set is \{\mathbf{0}\}, a line, a plane, or larger.
- A = \begin{bmatrix} 1 & -1 & 2 \\ 3 & -2 & 7 \end{bmatrix}
- A = \begin{bmatrix} 1 & -2 & 3 \\ 2 & -3 & 8 \\ 3 & -5 & 10 \end{bmatrix}
- The single equation x_1 - 5x_2 + 2x_3 = 0
- A = \begin{bmatrix} 1 & 2 & 0 & -1 \\ 0 & 0 & 1 & 4 \end{bmatrix}
- A = \begin{bmatrix} 1 & -2 & 1 & 3 \\ 3 & -6 & 4 & 7 \end{bmatrix}
- A = \begin{bmatrix} 1 & -1 & 2 & 0 & 3 \\ 0 & 0 & 0 & 1 & -2 \end{bmatrix}
- Show that \mathbf{w} = (1, 3, 2, 1) solves the homogeneous system of Exercise 5, and find its weights in your parametric vector form.
Day 1: trivial or nontrivial
- Without row reducing, explain why every homogeneous system of 3 equations in 5 unknowns has a nontrivial solution.
- For which k does \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ 1 & 4 & k \end{bmatrix}\mathbf{x} = \mathbf{0} have nontrivial solutions? For that k, write the solution set as a span.
- Let A = \begin{bmatrix} 2 & 1 & 4 \\ -1 & 3 & 5 \\ 0 & 2 & 4 \end{bmatrix}. Check that column 3 is column 1 plus 2 times column 2, and use it to give a nontrivial solution of A\mathbf{x} = \mathbf{0} without row reducing.
Day 2: particular plus homogeneous
- With A from Exercise 1, solve A\mathbf{x} = (1, 5) in parametric vector form, and name the particular and homogeneous parts.
- With A from Exercise 5, solve A\mathbf{x} = (3, 10) in parametric vector form.
- Let A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ 1 & 4 & 5 \end{bmatrix}. Solve A\mathbf{x} = \mathbf{b} in parametric vector form, or show there is no solution, for (a) \mathbf{b} = (1, 3, 2) and (b) \mathbf{b} = (1, 3, 3).
- With A from Exercise 1, check that (1, 1, 1) solves A\mathbf{x} = (2, 8) and write every solution with it as \mathbf{p}. Then solve by row reduction, and show the two descriptions give the same set.
- A 3 \times 3 matrix A has rank 2, and \mathbf{u} = (2, 0, 1) and \mathbf{w} = (1, 1, 3) both solve A\mathbf{x} = \mathbf{b}. Find every solution, and the solution whose first entry is 0.
Day 2: in context
- Mamadou needs a batch with 41 kilograms of silica, 13 of soda and 6 of lime, from the cullet of Example 6. Find every mix, the values of the free amount that keep every amount nonnegative, and the workable mix with no kind 2.
- Lamine asks for 40 kilograms of silica, 10 of soda and 5 of lime from the same three kinds. Show that no mix works.
Reasoning
- Prove that a consistent system A\mathbf{x} = \mathbf{b} has exactly one solution exactly when A\mathbf{x} = \mathbf{0} has only the trivial solution.
- Suppose \mathbf{p} and \mathbf{q} both solve A\mathbf{x} = \mathbf{b}, where \mathbf{b} \neq \mathbf{0}. Which of \mathbf{p} + \mathbf{q}, \mathbf{p} - \mathbf{q}, 2\mathbf{p} - \mathbf{q} and \displaystyle \frac{1}{2}(\mathbf{p} + \mathbf{q}) also solve it? Explain.
- Give a counterexample to the claim “if A\mathbf{x} = \mathbf{0} has a nontrivial solution, then A\mathbf{x} = \mathbf{b} has infinitely many solutions for every \mathbf{b}.”
- A 4 \times 6 matrix A has rank 4. Explain why A\mathbf{x} = \mathbf{b} is consistent for every \mathbf{b} in \mathbb{R}^4, and say how many parameters its solution set needs.
Answers
- \mathbf{x} = t(-3, -1, 1), a line. Row 2 minus 3 times row 1 is (0, 1, 1); adding it to row 1 gives (1, 0, 3).
- Only \mathbf{x} = \mathbf{0}. Elimination gives rows (0, 1, 2) and (0, 1, 1), whose difference leaves a pivot in every column.
- \mathbf{x} = s(5, 1, 0) + t(-2, 0, 1), a plane, from x_1 = 5x_2 - 2x_3.
- \mathbf{x} = s(-2, 1, 0, 0) + t(1, 0, -4, 1), a plane in \mathbb{R}^4: x_1 = -2x_2 + x_4 and x_3 = -4x_4.
- \mathbf{x} = s(2, 1, 0, 0) + t(-5, 0, 2, 1), a plane in \mathbb{R}^4. Row 2 minus 3 times row 1 is (0, 0, 1, -2), and row 1 minus that is (1, -2, 0, 5).
- \mathbf{x} = r(1, 1, 0, 0, 0) + s(-2, 0, 1, 0, 0) +\, t(-3, 0, 0, 2, 1), larger than a plane, with x_2, x_3, x_5 free: x_1 = x_2 - 2x_3 - 3x_5 and x_4 = 2x_5.
- A\mathbf{w} = (1 - 6 + 2 + 3,\ 3 - 18 + 8 + 7) = (0, 0). The weights are the free entries, s = x_2 = 3 and t = x_4 = 1: 3(2, 1, 0, 0) + (-5, 0, 2, 1) = (1, 3, 2, 1).
- A 3 \times 5 matrix has at most 3 pivots among 5 columns, so at least 2 variables are free, and setting one to 1 gives a solution that is not \mathbf{0}.
- k = 5, with solution set \operatorname{span}\{(-1, -1, 1)\}. Elimination gives rows (0, 1, 1) and (0, 2, k - 3), then (0, 0, k - 5). For k = 5, x_2 = -x_3 and x_1 = -2x_2 - 3x_3 = -x_3.
- \mathbf{a}_1 + 2\mathbf{a}_2 is (2 + 2,\ -1 + 6,\ 0 + 4), which is (4, 5, 4) = \mathbf{a}_3, so \mathbf{a}_1 + 2\mathbf{a}_2 - \mathbf{a}_3 = \mathbf{0}, which says A(1, 2, -1) = \mathbf{0}.
- \mathbf{x} = (3, 2, 0) + t(-3, -1, 1): particular part (3, 2, 0), homogeneous part t(-3, -1, 1) from Exercise 1. The reduced rows are x_1 + 3x_3 = 3 and x_2 + x_3 = 2.
- \mathbf{x} = (2, 0, 1, 0) + s(2, 1, 0, 0) + t(-5, 0, 2, 1). The constants reduce to 10 - 9 = 1 and 3 - 1 = 2, so x_1 = 2 + 2x_2 - 5x_4 and x_3 = 1 + 2x_4.
- (a) No solution: elimination leaves (0, 1, 1 \mid 1) and (0, 2, 2 \mid 1), and the second minus twice the first is (0, 0, 0 \mid -1). (b) \mathbf{x} = (-1, 1, 0) + t(-1, -1, 1): the last row is now zero, x_2 = 1 - x_3 and x_1 = 1 - 2x_2 - 3x_3 = -1 - x_3.
- A(1, 1, 1) = (1 - 1 + 2,\ 3 - 2 + 7) = (2, 8), so \mathbf{x} = (1, 1, 1) + t(-3, -1, 1). Row reduction gives x_1 + 3x_3 = 4 and x_2 + x_3 = 2, so \mathbf{x} = (4, 2, 0) + t(-3, -1, 1). Since (1, 1, 1) = (4, 2, 0) + (-3, -1, 1), the first description at t is the second at t + 1: the same line.
- \mathbf{x} = (2, 0, 1) + t(-1, 1, 2), and the solution with first entry 0 is (0, 2, 5), at t = 2. Rank 2 leaves one free variable, and \mathbf{w} - \mathbf{u} = (-1, 1, 2) is a nonzero homogeneous solution.
- \mathbf{x} = (7, -1, 0) + t(-2, 1, 1), workable for \displaystyle 1 \le t \le \frac{7}{2}; the mix with no kind 2 is 5 bags of kind 1 and 1 of kind 3. Elimination gives x_2 - x_3 = -1 and x_1 + 2x_3 = 7. Nonnegative amounts need 7 - 2t \ge 0 and t - 1 \ge 0. No kind 2 means t = 1; check 5(7, 2, 1) + (6, 3, 1) = (41, 13, 6).
- With the lime row on top, subtracting 2 and 7 times it from the soda and silica rows gives (0, -1, 1 \mid 0) and (0, 1, -1 \mid 5). Their sum is (0, 0, 0 \mid 5), which says 0 = 5.
- Take a solution \mathbf{p}. Every solution is \mathbf{p} + \mathbf{v}_h with A\mathbf{v}_h = \mathbf{0}. If \mathbf{0} is the only \mathbf{v}_h, then \mathbf{p} is the only solution; if some \mathbf{v}_h \neq \mathbf{0}, then \mathbf{p} and \mathbf{p} + \mathbf{v}_h are two different solutions.
- 2\mathbf{p} - \mathbf{q} and \displaystyle \frac{1}{2}(\mathbf{p} + \mathbf{q}) do; the others do not. By linearity A(\alpha\mathbf{p} + \beta\mathbf{q}) = (\alpha + \beta)\mathbf{b}, which is \mathbf{b} exactly when \alpha + \beta = 1, as \mathbf{b} \neq \mathbf{0}. So \mathbf{p} + \mathbf{q} solves A\mathbf{x} = 2\mathbf{b} and \mathbf{p} - \mathbf{q} solves A\mathbf{x} = \mathbf{0}.
- A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} and \mathbf{b} = (1, 0). A(1, -1) = \mathbf{0} is nontrivial, but x_1 + x_2 = 1 and x_1 + x_2 = 0 cannot both hold, so A\mathbf{x} = \mathbf{b} has no solution.
- Rank 4 puts a pivot in each of the 4 rows, so reduction never produces a row (0, \dots, 0 \mid d) with d \neq 0, and every \mathbf{b} gives a consistent system. There are 6 - 4 = 2 free variables, so two parameters: \mathbf{x} = \mathbf{p} + s\mathbf{v}_1 + t\mathbf{v}_2.