Parametric Solutions - Linear Algebra

1. Let the solution set of a system $A\vec{x} = \vec{b}$ be given by $\vec{x} = \vec{p} + t\vec{v}$, where $t$ is any real number, $\vec{v} \neq \vec{0}$, and $\vec{b} \neq \vec{0}$. Which of the following statements must be true?
   Answer: $A\vec{p} = \vec{b}$ and $A\vec{v} = \vec{0}$. For $\vec{x} = \vec{p} + t\vec{v}$ to be a solution for all $t$, we must have $A(\vec{p} + t\vec{v}) = \vec{b}$. Using linearity, this becomes $A\vec{p} + t(A\vec{v}) = \vec{b}$. This equation must hold for all $t$. If we set $t=0$, we get $A\vec{p} = \vec{b}$. Substituting this back gives $\vec{b} + t(A\vec{v}) = \vec{b}$, which simplifies to $t(A\vec{v}) = \vec{0}$. Since this must be true for any $t \neq 0$, it requires that $A\vec{v} = \vec{0}$. Thus, $\vec{p}$ is a particular solution to the non-homogeneous system, and $\vec{v}$ is a solution to the homogeneous system. Distractor A reverses these roles. Distractor C is incorrect because $A(\vec{p} + \vec{v}) = A\vec{p} + A\vec{v} = \vec{b} + \vec{0} = \vec{b}$. Distractor D incorrectly claims the direction vector also solves the non-homogeneous system.
2. Consider the system of linear equations $Ax = b$ where $A$ is a $4 \times 5$ matrix with rank 3. If the system is consistent and has the parametric solution $x = \begin{pmatrix} 2 \\ -1 \\ 0 \\ 3 \\ 0 \end{pmatrix} + t\begin{pmatrix} 1 \\ 2 \\ -1 \\ 0 \\ 1 \end{pmatrix} + s\begin{pmatrix} 0 \\ 1 \\ 1 \\ -2 \\ 3 \end{pmatrix}$, which statement best describes the solution set?
   Answer: The solution set is a 2-dimensional plane in $\mathbb{R}^5$ containing infinitely many solutions with exactly 2 free variables The parametric solution shows 2 parameter variables (t and s), indicating a 2-dimensional affine subspace (plane) in $\mathbb{R}^5$. Since the matrix A has rank 3 and 5 columns, the nullspace has dimension 5-3=2, confirming 2 free variables. Choice B incorrectly identifies dimension as 3. Choice C incorrectly counts only 1 parameter. Choice D incorrectly assumes 4 free variables, which would require rank 1.
3. The general solution to a consistent system $A\vec{x} = \vec{b}$ is a line in $\mathbb{R}^5$. What can be concluded about the matrix $A$?
   Answer: The nullity of $A$ is 1. The geometric dimension of the solution set is equal to the dimension of the null space of $A$, which is defined as the nullity of $A$. Since the solution set is a line, its dimension is 1. Therefore, the nullity of $A$ must be 1. By the Rank-Nullity Theorem for a matrix with $n$ columns (here $n=5$), we have $\mathrm{rank}(A) + \mathrm{nullity}(A) = n$. So, $\mathrm{rank}(A) + 1 = 5$, which means the rank must be 4. Thus, distractor A is incorrect. If the columns of A were linearly independent (distractor B) or if A were invertible (distractor D), the nullity would be 0, leading to a unique solution (a point), not a line.
4. The solution set to a consistent linear system $Cx = d$ is given by $x = \begin{pmatrix} 2 \\ -1 \\ 4 \\ 0 \\ 3 \end{pmatrix} + \alpha\begin{pmatrix} 1 \\ 1 \\ 0 \\ 2 \\ -1 \end{pmatrix} + \beta\begin{pmatrix} 0 \\ 2 \\ -1 \\ 1 \\ 1 \end{pmatrix}$. If $C$ has 5 columns, what is the minimum possible number of rows in $C$?
   Answer: 3 rows, since we need at least 3 constraints to determine a 2-dimensional solution space The parametric solution has 2 parameters (α and β), indicating the null space of C has dimension 2. By the rank-nullity theorem, rank(C) + nullity(C) = 5, so rank(C) = 3. The rank of C cannot exceed the number of rows, so C must have at least 3 rows. Choice A incorrectly assumes minimum rank equals nullity. Choice C applies an incorrect rule about equations vs unknowns. Choice D incorrectly assumes square matrices are required for parametric solutions.
5. Consider the matrix equation $AX = B$ where $A$ is $3 \times 4$, $X$ is $4 \times 2$, and $B$ is $3 \times 2$. If the first column of $X$ has the parametric form $\begin{pmatrix} 1 \\ 0 \\ 2 \\ 1 \end{pmatrix} + s\begin{pmatrix} 2 \\ 1 \\ 0 \\ -1 \end{pmatrix}$ and the second column has the form $\begin{pmatrix} 0 \\ 3 \\ -1 \\ 2 \end{pmatrix} + t\begin{pmatrix} 2 \\ 1 \\ 0 \\ -1 \end{pmatrix}$, what can be concluded about the null space of $A$?
   Answer: The null space of A is 1-dimensional and is spanned by $\begin{pmatrix} 2 \\ 1 \\ 0 \\ -1 \end{pmatrix}$ When you encounter parametric solutions to matrix equations, you're seeing the general solution structure that reveals crucial information about the null space. The key insight is recognizing what the parametric form tells us about the underlying linear system. The correct answer is D. Notice that both columns of $X$ have the same direction vector $\begin{pmatrix} 2 \\ 1 \\ 0 \\ -1 \end{pmatrix}$ in their parametric forms. This vector represents the homogeneous solution to $A\mathbf{x} = \mathbf{0}$. Since both systems $A\mathbf{x} = \mathbf{b_1}$ and $A\mathbf{x} = \mathbf{b_2}$ (where $\mathbf{b_1}$ and $\mathbf{b_2}$ are the columns of $B$) share this same direction vector, it must be in the null space of $A$. The parametric form shows exactly one free variable in each case, indicating the null space is 1-dimensional. A is wrong because the parametric structure gives us complete information about the null space dimension and basis. B incorrectly identifies the dimension and confuses particular solutions with null space vectors. The vector $\begin{pmatrix} 1 \\ 0 \\ 2 \\ 1 \end{pmatrix}$ is a particular solution, not a null space vector. C misunderstands parametric solutions—the presence of free parameters means infinitely many solutions, not unique ones. Remember: in parametric solutions to $A\mathbf{x} = \mathbf{b}$, the direction vectors of the parameters always span the null space of $A$. The number of free parameters equals the dimension of the null space.
6. The solution to a system $A\vec{x}=\vec{b}$ is given by $\vec{x} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 3 \\ -1 \end{pmatrix}$. If $\vec{x}_0 = \begin{pmatrix} 7 \\ 0 \end{pmatrix}$ is a particular solution to this same system, which of the following could be another valid parametric representation of the solution set?
   Answer: $\vec{x} = \begin{pmatrix} 7 \\ 0 \end{pmatrix} + s \begin{pmatrix} 3 \\ -1 \end{pmatrix}$ A parametric representation of a solution set is defined by any particular solution (any point on the line) and a direction vector for the corresponding homogeneous system. We are given that $\vec{x}_0 = \begin{pmatrix} 7 \\ 0 \end{pmatrix}$ is a particular solution. We can verify that it lies on the original line: $\begin{pmatrix} 1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 7 \\ 0 \end{pmatrix}$ yields $1+3t=7 \implies t=2$ and $2-t=0 \implies t=2$. Since a consistent value of $t$ exists, the point is on the line. The solution set for the homogeneous system (the null space) is spanned by $\begin{pmatrix} 3 \\ -1 \end{pmatrix}$. Therefore, we can write a new valid representation using the new particular solution and the same direction vector: $\vec{x} = \begin{pmatrix} 7 \\ 0 \end{pmatrix} + s \begin{pmatrix} 3 \\ -1 \end{pmatrix}$.
7. The solution set for a system of equations is described by $\vec{x} = \vec{p} + s\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} + t\begin{pmatrix} -2 \\ -4 \\ -2 \end{pmatrix}$, where $s, t \in \mathbb{R}$. What is the dimension of this solution set?
   Answer: 1 The dimension of the solution set is the dimension of the corresponding homogeneous solution space, which is spanned by the direction vectors. The dimension is equal to the number of linearly independent direction vectors. In this case, the direction vectors are $\vec{v}_1 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$ and $\vec{v}_2 = \begin{pmatrix} -2 \\ -4 \\ -2 \end{pmatrix}$. We can see that $\vec{v}_2 = -2\vec{v}_1$. Since one vector is a scalar multiple of the other, they are linearly dependent and do not span a plane. They both lie on the same line through the origin. Therefore, the solution set is a line (a 1-dimensional object). The common mistake is to assume the dimension equals the number of parameters, which is only true if the direction vectors are linearly independent.
8. A linear system has the parametric solution $x = \begin{pmatrix} 5 \\ 0 \\ -2 \end{pmatrix} + t\begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix}$. Which of the following vectors is NOT in the solution set?
   Answer: $\begin{pmatrix} 3 \\ 2 \\ -8 \end{pmatrix}$ To check if a vector is in the solution set, we substitute into the parametric form and solve for t. For choice A: $\begin{pmatrix} 7 \\ -2 \\ 4 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \\ -2 \end{pmatrix} + t\begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix}$ gives t=2 (consistent). For choice B: $\begin{pmatrix} 3 \\ 2 \\ -8 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \\ -2 \end{pmatrix} + t\begin{pmatrix} 1 \\ -1 \\ 3 \end{pmatrix}$ requires t=-2 from first component, but t=-2 from second component, and t=-2 from third component gives inconsistent results. Choice C is the particular solution (t=0). Choice D gives t=-1 consistently.
9. The homogeneous system $Bx = 0$ has the general solution $x = r\begin{pmatrix} 3 \\ -2 \\ 1 \\ 0 \end{pmatrix} + s\begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix}$. If the related non-homogeneous system $Bx = c$ has a particular solution $x_p = \begin{pmatrix} 4 \\ 1 \\ -3 \\ 2 \end{pmatrix}$, what is the complete solution to $Bx = c$?
   Answer: $x = \begin{pmatrix} 4 \\ 1 \\ -3 \\ 2 \end{pmatrix} + r\begin{pmatrix} 3 \\ -2 \\ 1 \\ 0 \end{pmatrix} + s\begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix}$ The general solution to a non-homogeneous system is the sum of any particular solution and the general solution to the associated homogeneous system. Thus: $x = x_p + x_h = \begin{pmatrix} 4 \\ 1 \\ -3 \\ 2 \end{pmatrix} + r\begin{pmatrix} 3 \\ -2 \\ 1 \\ 0 \end{pmatrix} + s\begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix}$. Choice B incorrectly adds the particular solution to one basis vector. Choice C incorrectly combines the basis vectors. Choice D incorrectly treats both the particular solution and a basis vector as parameters.
10. The general solution to a linear system $A\vec{x} = \vec{b}$ is given by $\vec{x} = s\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} + t\begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix}$ for $s, t \in \mathbb{R}$. What is the geometric representation of the solution set?
    Answer: A plane through the origin in $\mathbb{R}^3$. The solution is expressed as a linear combination of two linearly independent vectors with two free parameters ($s$ and $t$). This structure defines a plane. Since there is no particular solution vector (or it is the zero vector), the solution set includes the origin (when $s=0$ and $t=0$). Therefore, the solution set is a plane through the origin. This also implies that $\vec{b}$ must be the zero vector. Distractor A is incorrect because two parameters define a plane, not a line. Distractor B is incorrect because the solution set contains the origin. Distractor D misinterprets the linear combination of two vectors.
11. The solution set for a system of linear equations $A\vec{x} = \vec{b}$ is given by the parametric vector form $\vec{x} = \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix} + t \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}$, where $t \in \mathbb{R}$. Which statement correctly describes a property of this system?
    Answer: The corresponding homogeneous system $A\vec{x} = \vec{0}$ has a solution space spanned by the vector $\begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}$. The general solution to $A\vec{x} = \vec{b}$ is of the form $\vec{x} = \vec{p} + \vec{x}_h$, where $\vec{p}$ is a particular solution and $\vec{x}_h$ is the general solution to the corresponding homogeneous system $A\vec{x} = \vec{0}$. In this case, $\vec{x}_h = t \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}$, which describes all scalar multiples of the vector. This is the span of the vector, which constitutes the null space. Distractor B is incorrect because any point on the line, such as for $t=1$ giving $\vec{x} = (4, -1, 2)$, is also a valid particular solution. Distractor C is incorrect because the presence of a free parameter implies infinitely many solutions, not a unique one. Distractor D incorrectly identifies the particular solution vector as the null space, rather than the direction vector.
12. Consider the parametric solution $x = \begin{pmatrix} 4 \\ 1 \\ -3 \\ 2 \end{pmatrix} + t\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}$ to a linear system. If we require that $x_1 + x_3 = 0$, what values of the parameter $t$ satisfy this additional constraint?
    Answer: $t = -\frac{1}{3}$ When you encounter a parametric solution with an additional constraint, you're essentially finding where a line (represented by the parametric equations) intersects with a plane or hyperplane (represented by the constraint equation). To find the correct value of $t$, substitute the parametric expressions into the constraint equation $x_1 + x_3 = 0$. From the given solution, $x_1 = 4 + 2t$ and $x_3 = -3 + t$. Substituting these into the constraint: $(4 + 2t) + (-3 + t) = 0$ $4 + 2t - 3 + t = 0$ $1 + 3t = 0$ $3t = -1$ $t = -\frac{1}{3}$ This confirms answer C is correct. Looking at the wrong answers: A) $t = -1$ would give $x_1 + x_3 = (4 + 2(-1)) + (-3 + (-1)) = 2 + (-4) = -2 \neq 0$. B) $t = \frac{1}{3}$ would yield $x_1 + x_3 = (4 + \frac{2}{3}) + (-3 + \frac{1}{3}) = \frac{14}{3} - \frac{8}{3} = 2 \neq 0$. D) incorrectly assumes inconsistency—additional linear constraints on parametric solutions are typically consistent unless they contradict the solution space's geometry. Study tip: When adding constraints to parametric solutions, always substitute the parametric expressions directly into the new constraint equation. The resulting equation in $t$ will give you the specific points on the solution line that satisfy your additional condition.
13. The reduced row echelon form of the augmented matrix for a linear system is $\begin{pmatrix} 1 & -2 & 0 & 3 & | & 5 \\ 0 & 0 & 1 & -1 & | & 2 \\ 0 & 0 & 0 & 0 & | & 0 \end{pmatrix}$. What is the geometric nature of the solution set?
    Answer: A plane in $\mathbb{R}^4$. The variables are $x_1, x_2, x_3, x_4$. The pivot columns are 1 and 3, corresponding to basic variables $x_1$ and $x_3$. The non-pivot columns are 2 and 4, corresponding to free variables $x_2$ and $x_4$. Let $x_2 = s$ and $x_4 = t$. The equations are $x_1 - 2s + 3t = 5$ and $x_3 - t = 2$. This gives the parametric solution: $x_1 = 5 + 2s - 3t$, $x_2 = s$, $x_3 = 2 + t$, $x_4 = t$. The solution vector is $\vec{x} = \begin{pmatrix} 5 \\ 0 \\ 2 \\ 0 \end{pmatrix} + s\begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + t\begin{pmatrix} -3 \\ 0 \\ 1 \\ 1 \end{pmatrix}$. Since there are two free variables and the direction vectors are linearly independent, the solution set is a plane. The system has 4 variables, so the solution set is a plane in $\mathbb{R}^4$.
14. Two students obtain different parametric representations for the same linear system: Student A writes $x = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} + t\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ while Student B writes $x = \begin{pmatrix} 3 \\ 1 \\ 3 \end{pmatrix} + s\begin{pmatrix} -4 \\ 2 \\ -6 \end{pmatrix}$. Which statement correctly analyzes these representations?
    Answer: Both representations are equivalent since the direction vectors are scalar multiples and both particular solutions satisfy the original system When you encounter different parametric representations of the same solution set, remember that multiple correct forms can exist. The key is checking whether the direction vectors are scalar multiples and whether both particular solutions satisfy the original system. Let's verify these representations are equivalent. First, examine the direction vectors: Student A uses $\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ while Student B uses $\begin{pmatrix} -4 \\ 2 \\ -6 \end{pmatrix}$. Notice that Student B's direction vector equals $-2$ times Student A's direction vector, making them scalar multiples that span the same line direction. Next, check if both particular solutions lie on the same solution set. Substituting $t = 1$ in Student A's equation gives $\begin{pmatrix} 3 \\ 1 \\ 3 \end{pmatrix}$, which is exactly Student B's particular solution. This confirms both points satisfy the same system. Therefore, both representations describe identical solution sets, just with different parameterizations. Option A incorrectly assumes Student B made an error, but scalar multiples of direction vectors are perfectly valid. Option B wrongly suggests that different particular solutions indicate an error—multiple points can lie on the same solution set. Option D misunderstands that opposite directions still describe the same line, just traversed differently as the parameter varies. Study tip: When comparing parametric solutions, always check if direction vectors are scalar multiples and verify that particular solutions can be reached from each other by adjusting parameters. Different-looking representations can describe identical solution sets.
15. The set of all solutions to a system $A\vec{x} = \vec{b}$ is the line in $\mathbb{R}^3$ defined by $\vec{x} = \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix} + t\begin{pmatrix} -2 \\ 0 \\ 1 \end{pmatrix}$. What is the null space of the matrix $A$?
    Answer: The set of all scalar multiples of $\begin{pmatrix} -2 \\ 0 \\ 1 \end{pmatrix}$. The solution set of $A\vec{x} = \vec{b}$ is a translation of the null space of $A$ (the solution set of $A\vec{x} = \vec{0}$). The parametric form $\vec{x} = \vec{p} + t\vec{v}$ shows this translation, where $\vec{p}$ is the translation vector (a particular solution) and $t\vec{v}$ is the null space. Here, the null space is given by all vectors of the form $t\begin{pmatrix} -2 \\ 0 \\ 1 \end{pmatrix}$, which is the set of all scalar multiples of that direction vector. Distractor A would imply a unique solution, not a line. Distractor B confuses the particular solution with the null space basis. Distractor D overestimates the dimension of the null space.
16. A linear system has the general solution $x = \begin{pmatrix} 2 \\ 0 \\ -1 \\ 3 \end{pmatrix} + u\begin{pmatrix} 1 \\ 2 \\ 0 \\ -1 \end{pmatrix}$. Which geometric object best describes the solution set, and what is its relationship to the origin?
    Answer: A line in $\mathbb{R}^4$ that does not pass through the origin since it's a translated copy of the null space When you encounter a linear system's general solution in parametric form, you're looking at the structure of an affine subspace - essentially a linear subspace that's been translated away from the origin. The given solution $x = \begin{pmatrix} 2 \\ 0 \\ -1 \\ 3 \end{pmatrix} + u\begin{pmatrix} 1 \\ 2 \\ 0 \\ -1 \end{pmatrix}$ has two key components: a particular solution $\begin{pmatrix} 2 \\ 0 \\ -1 \\ 3 \end{pmatrix}$ and the null space spanned by $\begin{pmatrix} 1 \\ 2 \\ 0 \\ -1 \end{pmatrix}$. Since the null space is one-dimensional (spanned by a single vector), the solution set forms a line in $\mathbb{R}^4$. This line doesn't pass through the origin because when $u = 0$, you get the particular solution, not the zero vector. Choice A incorrectly claims the line passes through the origin. To check this, set the parametric equation equal to zero - there's no value of $u$ that makes this true. Choice B misidentifies the geometric object as a hyperplane, which would require the null space to have dimension 3, not 1. Choice C incorrectly calls it a plane (which needs a 2D null space) and wrongly claims it passes through the origin. The correct answer is D: this is indeed a line that doesn't pass through the origin, representing a translated copy of the one-dimensional null space. Study tip: The dimension of the null space determines the geometric object's type (line = 1D, plane = 2D, etc.), while the particular solution determines whether it passes through the origin.
17. A line in $\mathbb{R}^3$ is described by the parametric equation $\vec{x} = \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} + t\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$. Which of the following parametric equations represents the same line?
    Answer: $\vec{x} = \begin{pmatrix} 3 \\ 0 \\ 7 \end{pmatrix} + s\begin{pmatrix} -2 \\ 2 \\ -4 \end{pmatrix}$ Two parametric equations represent the same line if their direction vectors are parallel (non-zero scalar multiples of each other) and if a point from one line lies on the other. In choice A, the direction vector $\begin{pmatrix} -2 \\ 2 \\ -4 \end{pmatrix}$ is $-2$ times the original direction vector $\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}$, so they are parallel. We must check if the point $(3, 0, 7)$ is on the original line. Setting $\begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} + t\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ 7 \end{pmatrix}$ gives $t=1$ for all three components, so the point is on the line. Thus, it represents the same line. Distractor B has a non-parallel direction vector. Distractor C incorrectly swaps the point and direction vector. Distractor D has a parallel direction vector, but its point $(3, 0, 6)$ is not on the original line.
18. The solution set for a system $A\vec{x} = \vec{b}$ is a plane in $\mathbb{R}^3$ containing the points $P=(1,0,0)$, $Q=(2,1,2)$, and $R=(0,1,1)$. Which of the following represents the general solution to the corresponding homogeneous system $A\vec{x} = \vec{0}$?
    Answer: $s\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} + t\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$, for $s, t \in \mathbb{R}$. The solution set to the homogeneous system, $Nul(A)$, is a plane through the origin that is parallel to the solution plane of $A\vec{x} = \vec{b}$. We can find two non-parallel direction vectors that lie in the solution plane by taking differences between the given points. Let $\vec{u} = \vec{PQ} = Q-P = \begin{pmatrix} 2-1 \\ 1-0 \\ 2-0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ and $\vec{v} = \vec{PR} = R-P = \begin{pmatrix} 0-1 \\ 1-0 \\ 1-0 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$. Since $\vec{u}$ and $\vec{v}$ are not scalar multiples of each other, they are linearly independent and form a basis for the direction of the plane. Thus, the general solution to $A\vec{x} = \vec{0}$ is the set of all linear combinations of these vectors, which is $s\vec{u} + t\vec{v}$. Distractor B incorrectly uses the position vectors of points as direction vectors. Distractor C represents a line, not a plane. Distractor D uses two linearly dependent vectors, which would also only span a line.
19. The coefficient matrix of a homogeneous system $Ax = 0$ is $4 \times 6$ with rank 4. If a related non-homogeneous system $Ax = b$ is consistent, what is the dimension of its solution set?
    Answer: 2-dimensional, since there are 6-4=2 free variables in the parametric representation For a consistent non-homogeneous system $Ax = b$, the solution set is an affine subspace of dimension equal to the dimension of the null space of A. Since A is $4 \times 6$ with rank 4, by the rank-nullity theorem, the null space has dimension $6 - 4 = 2$. Therefore, the solution set is a 2-dimensional affine subspace (plane) in $\mathbb{R}^6$. Choice A confuses rank with solution dimension. Choice C incorrectly thinks non-homogeneous systems have lower dimension than their homogeneous counterparts. Choice D confuses the ambient space dimension with the solution set dimension.
20. Let $Nul(A)$ be the solution set for the homogeneous system $A\vec{x} = \vec{0}$. A basis for $Nul(A)$ is given by the set $\left\{ \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} \right\}$. Which of the following vectors is a solution to $A\vec{x} = \vec{0}$?
    Answer: $\begin{pmatrix} 2 \\ -3 \\ 7 \end{pmatrix}$ A vector is a solution to $A\vec{x} = \vec{0}$ if and only if it is in the null space of $A$. This means the vector must be a linear combination of the basis vectors of the null space. We need to find if there exist scalars $c_1, c_2$ such that the vector equals $c_1\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + c_2\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$. For choice B, we check: $c_1\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + c_2\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 7 \end{pmatrix}$. The first component gives $c_1 = 2$. The second gives $c_2 = -3$. The third component must also hold: $2c_1 - c_2 = 2(2) - (-3) = 4+3=7$. Since all components match, this vector is in the null space. The other vector choices cannot be formed as linear combinations of the basis vectors.

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