Linear Algebra Quiz: Parametric Solutions
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Parametric SolutionsQuestion 1 of 20

The solution set for a system of linear equations Ax⃗=b⃗A\vec{x} = \vec{b} is given by the parametric vector form x⃗=(10−2)+t(3−14)\vec{x} = \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix} + t \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}, where t∈Rt \in \mathbb{R}. Which statement correctly describes a property of this system?

The corresponding homogeneous system Ax⃗=0⃗A\vec{x} = \vec{0} has a solution space spanned by the vector (3−14)\begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}.
The vector (10−2)\begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix} is the only particular solution to the non-homogeneous system Ax⃗=b⃗A\vec{x} = \vec{b}.
The system has a unique solution because there is only one free parameter, tt.
The null space of the coefficient matrix AA consists of the single vector (10−2)\begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}.
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Linear Algebra Quiz

Linear Algebra Quiz: Parametric Solutions

Practice Parametric Solutions in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Parametric Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

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Question 1

The solution set for a system of linear equations Ax⃗=b⃗A\vec{x} = \vec{b} is given by the parametric vector form x⃗=(10−2)+t(3−14)\vec{x} = \begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix} + t \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}, where t∈Rt \in \mathbb{R}. Which statement correctly describes a property of this system?

  1. The corresponding homogeneous system Ax⃗=0⃗A\vec{x} = \vec{0} has a solution space spanned by the vector (3−14)\begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}. (correct answer)
  2. The vector (10−2)\begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix} is the only particular solution to the non-homogeneous system Ax⃗=b⃗A\vec{x} = \vec{b}.
  3. The system has a unique solution because there is only one free parameter, tt.
  4. The null space of the coefficient matrix AA consists of the single vector (10−2)\begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}.
Explanation: The general solution to Ax⃗=b⃗A\vec{x} = \vec{b} is of the form x⃗=p⃗+x⃗h\vec{x} = \vec{p} + \vec{x}_h, where p⃗\vec{p} is a particular solution and x⃗h\vec{x}_h is the general solution to the corresponding homogeneous system Ax⃗=0⃗A\vec{x} = \vec{0}. In this case, x⃗h=t(3−14)\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=1t=1 giving x⃗=(4,−1,2)\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.

Question 2

The general solution to a linear system Ax⃗=b⃗A\vec{x} = \vec{b} is given by x⃗=s(210)+t(−103)\vec{x} = s\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} + t\begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix} for s,t∈Rs, t \in \mathbb{R}. What is the geometric representation of the solution set?

  1. A line through the origin in R3\mathbb{R}^3.
  2. A plane in R3\mathbb{R}^3 that does not pass through the origin.
  3. A plane through the origin in R3\mathbb{R}^3. (correct answer)
  4. Two intersecting lines in R3\mathbb{R}^3.
Explanation: The solution is expressed as a linear combination of two linearly independent vectors with two free parameters (ss and tt). 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=0s=0 and t=0t=0). Therefore, the solution set is a plane through the origin. This also implies that b⃗\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.

Question 3

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}

  1. A line in R4\mathbb{R}^4.
  2. A plane in R3\mathbb{R}^3.
  3. The system is inconsistent.
  4. A plane in R4\mathbb{R}^4. (correct answer)
Explanation: The variables are x1,x2,x3,x4x_1, x_2, x_3, x_4. The pivot columns are 1 and 3, corresponding to basic variables x1x_1 and x3x_3. The non-pivot columns are 2 and 4, corresponding to free variables x2x_2 and x4x_4. Let x2=sx_2 = s and x4=tx_4 = t. The equations are x1−2s+3t=5x_1 - 2s + 3t = 5 and x3−t=2x_3 - t = 2. This gives the parametric solution: x1=5+2s−3tx_1 = 5 + 2s - 3t, x2=sx_2 = s, x3=2+tx_3 = 2 + t, x4=tx_4 = t. The solution vector is x⃗=(5020)+s(2100)+t(−3011)\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 R4\mathbb{R}^4.

Question 4

The solution set for a system of equations is described by x⃗=p⃗+s(121)+t(−2−4−2)\vec{x} = \vec{p} + s\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} + t\begin{pmatrix} -2 \\ -4 \\ -2 \end{pmatrix}, where s,t∈Rs, t \in \mathbb{R}. What is the dimension of this solution set?

  1. 0
  2. 1 (correct answer)
  3. 2
  4. 3
Explanation: 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 v⃗1=(121)\vec{v}_1 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} and v⃗2=(−2−4−2)\vec{v}_2 = \begin{pmatrix} -2 \\ -4 \\ -2 \end{pmatrix}. We can see that v⃗2=−2v⃗1\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.

Question 5

The solution to a system Ax⃗=b⃗A\vec{x}=\vec{b} is given by x⃗=(12)+t(3−1)\vec{x} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 3 \\ -1 \end{pmatrix}. If x⃗0=(70)\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?

  1. x⃗=(70)+s(3−1)\vec{x} = \begin{pmatrix} 7 \\ 0 \end{pmatrix} + s \begin{pmatrix} 3 \\ -1 \end{pmatrix} (correct answer)
  2. x⃗=(70)+s(12)\vec{x} = \begin{pmatrix} 7 \\ 0 \end{pmatrix} + s \begin{pmatrix} 1 \\ 2 \end{pmatrix}
  3. x⃗=(12)+s(70)\vec{x} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + s \begin{pmatrix} 7 \\ 0 \end{pmatrix}
  4. The representation is unique, so no other form is possible.
Explanation: 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 x⃗0=(70)\vec{x}_0 = \begin{pmatrix} 7 \\ 0 \end{pmatrix} is a particular solution. We can verify that it lies on the original line: (12)+t(3−1)=(70)\begin{pmatrix} 1 \\ 2 \end{pmatrix} + t \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 7 \\ 0 \end{pmatrix} yields 1+3t=7  ⟹  t=21+3t=7 \implies t=2 and 2−t=0  ⟹  t=22-t=0 \implies t=2. Since a consistent value of tt exists, the point is on the line. The solution set for the homogeneous system (the null space) is spanned by (3−1)\begin{pmatrix} 3 \\ -1 \end{pmatrix}. Therefore, we can write a new valid representation using the new particular solution and the same direction vector: x⃗=(70)+s(3−1)\vec{x} = \begin{pmatrix} 7 \\ 0 \end{pmatrix} + s \begin{pmatrix} 3 \\ -1 \end{pmatrix}.

Question 6

The solution set for a system Ax⃗=b⃗A\vec{x} = \vec{b} is a plane in R3\mathbb{R}^3 containing the points P=(1,0,0)P=(1,0,0), Q=(2,1,2)Q=(2,1,2), and R=(0,1,1)R=(0,1,1). Which of the following represents the general solution to the corresponding homogeneous system Ax⃗=0⃗A\vec{x} = \vec{0}?

  1. s(112)+t(−111)s\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} + t\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}, for s,t∈Rs, t \in \mathbb{R}. (correct answer)
  2. s(100)+t(212)s\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + t\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}, for s,t∈Rs, t \in \mathbb{R}.
  3. t(2−1−1)t\begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix}, for t∈Rt \in \mathbb{R}.
  4. s(112)+t(224)s\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} + t\begin{pmatrix} 2 \\ 2 \\ 4 \end{pmatrix}, for s,t∈Rs, t \in \mathbb{R}.
Explanation: The solution set to the homogeneous system, Nul(A)Nul(A), is a plane through the origin that is parallel to the solution plane of Ax⃗=b⃗A\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 u⃗=PQ⃗=Q−P=(2−11−02−0)=(112)\vec{u} = \vec{PQ} = Q-P = \begin{pmatrix} 2-1 \\ 1-0 \\ 2-0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} and v⃗=PR⃗=R−P=(0−11−01−0)=(−111)\vec{v} = \vec{PR} = R-P = \begin{pmatrix} 0-1 \\ 1-0 \\ 1-0 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}. Since u⃗\vec{u} and v⃗\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 Ax⃗=0⃗A\vec{x} = \vec{0} is the set of all linear combinations of these vectors, which is su⃗+tv⃗s\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.

Question 7

The set of all solutions to a system Ax⃗=b⃗A\vec{x} = \vec{b} is the line in R3\mathbb{R}^3 defined by x⃗=(411)+t(−201)\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 AA?

  1. The set containing only the zero vector, {0⃗}\{\vec{0}\}.
  2. The set of all scalar multiples of (411)\begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix}.
  3. The set of all scalar multiples of (−201)\begin{pmatrix} -2 \\ 0 \\ 1 \end{pmatrix}. (correct answer)
  4. The plane spanned by the vectors (411)\begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix} and (−201)\begin{pmatrix} -2 \\ 0 \\ 1 \end{pmatrix}.
Explanation: The solution set of Ax⃗=b⃗A\vec{x} = \vec{b} is a translation of the null space of AA (the solution set of Ax⃗=0⃗A\vec{x} = \vec{0}). The parametric form x⃗=p⃗+tv⃗\vec{x} = \vec{p} + t\vec{v} shows this translation, where p⃗\vec{p} is the translation vector (a particular solution) and tv⃗t\vec{v} is the null space. Here, the null space is given by all vectors of the form t(−201)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.

Question 8

A line in R3\mathbb{R}^3 is described by the parametric equation x⃗=(215)+t(1−12)\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?

  1. x⃗=(307)+s(−22−4)\vec{x} = \begin{pmatrix} 3 \\ 0 \\ 7 \end{pmatrix} + s\begin{pmatrix} -2 \\ 2 \\ -4 \end{pmatrix} (correct answer)
  2. x⃗=(215)+s(2−23)\vec{x} = \begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} + s\begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix}
  3. x⃗=(1−12)+s(215)\vec{x} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} + s\begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix}
  4. x⃗=(306)+s(1−12)\vec{x} = \begin{pmatrix} 3 \\ 0 \\ 6 \end{pmatrix} + s\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}
Explanation: 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 (−22−4)\begin{pmatrix} -2 \\ 2 \\ -4 \end{pmatrix} is −2-2 times the original direction vector (1−12)\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}, so they are parallel. We must check if the point (3,0,7)(3, 0, 7) is on the original line. Setting (215)+t(1−12)=(307)\begin{pmatrix} 2 \\ 1 \\ 5 \end{pmatrix} + t\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ 7 \end{pmatrix} gives t=1t=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)(3, 0, 6) is not on the original line.

Question 9

Let Nul(A)Nul(A) be the solution set for the homogeneous system Ax⃗=0⃗A\vec{x} = \vec{0}. A basis for Nul(A)Nul(A) is given by the set {(102),(01−1)}\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 Ax⃗=0⃗A\vec{x} = \vec{0}?

  1. (110)\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}
  2. (2−37)\begin{pmatrix} 2 \\ -3 \\ 7 \end{pmatrix} (correct answer)
  3. (10−1)\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}
  4. Any vector in R3\mathbb{R}^3 except the zero vector.
Explanation: A vector is a solution to Ax⃗=0⃗A\vec{x} = \vec{0} if and only if it is in the null space of AA. 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 c1,c2c_1, c_2 such that the vector equals c1(102)+c2(01−1)c_1\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + c_2\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}. For choice B, we check: c1(102)+c2(01−1)=(2−37)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 c1=2c_1 = 2. The second gives c2=−3c_2 = -3. The third component must also hold: 2c1−c2=2(2)−(−3)=4+3=72c_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.

Question 10

Let the solution set of a system Ax⃗=b⃗A\vec{x} = \vec{b} be given by x⃗=p⃗+tv⃗\vec{x} = \vec{p} + t\vec{v}, where tt is any real number, v⃗≠0⃗\vec{v} \neq \vec{0}, and b⃗≠0⃗\vec{b} \neq \vec{0}. Which of the following statements must be true?

  1. Ap⃗=0⃗A\vec{p} = \vec{0} and Av⃗=b⃗A\vec{v} = \vec{b}.
  2. Ap⃗=b⃗A\vec{p} = \vec{b} and Av⃗=0⃗A\vec{v} = \vec{0}. (correct answer)
  3. A(p⃗+v⃗)=0⃗A(\vec{p} + \vec{v}) = \vec{0}.
  4. Ap⃗=b⃗A\vec{p} = \vec{b} and Av⃗=b⃗A\vec{v} = \vec{b}.
Explanation: For x⃗=p⃗+tv⃗\vec{x} = \vec{p} + t\vec{v} to be a solution for all tt, we must have A(p⃗+tv⃗)=b⃗A(\vec{p} + t\vec{v}) = \vec{b}. Using linearity, this becomes Ap⃗+t(Av⃗)=b⃗A\vec{p} + t(A\vec{v}) = \vec{b}. This equation must hold for all tt. If we set t=0t=0, we get Ap⃗=b⃗A\vec{p} = \vec{b}. Substituting this back gives b⃗+t(Av⃗)=b⃗\vec{b} + t(A\vec{v}) = \vec{b}, which simplifies to t(Av⃗)=0⃗t(A\vec{v}) = \vec{0}. Since this must be true for any t≠0t \neq 0, it requires that Av⃗=0⃗A\vec{v} = \vec{0}. Thus, p⃗\vec{p} is a particular solution to the non-homogeneous system, and v⃗\vec{v} is a solution to the homogeneous system. Distractor A reverses these roles. Distractor C is incorrect because A(p⃗+v⃗)=Ap⃗+Av⃗=b⃗+0⃗=b⃗A(\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.

Question 11

The homogeneous system Bx=0Bx = 0 has the general solution x=r(3−210)+s(100−1)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=cBx = c has a particular solution xp=(41−32)x_p = \begin{pmatrix} 4 \\ 1 \\ -3 \\ 2 \end{pmatrix} , what is the complete solution to Bx=cBx = c?

  1. (correct answer)
Explanation: 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}

Question 12

Consider the matrix equation AX=BAX = B where AA is 3×43 \times 4, XX is 4×24 \times 2, and BB is 3×23 \times 2. If the first column of XX has the parametric form (1021)+s(210−1)\begin{pmatrix} 1 \\ 0 \\ 2 \\ 1 \end{pmatrix} + s\begin{pmatrix} 2 \\ 1 \\ 0 \\ -1 \end{pmatrix} and the second column has the form (03−12)+t(210−1)\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 AA?

  1. The null space of A cannot be determined without additional information about matrix A
  2. The null space of A is 2-dimensional and contains both (210−1)\begin{pmatrix} 2 \\ 1 \\ 0 \\ -1 \end{pmatrix} and $$ \begin{pmatrix} 1 \ 0 \ 2 \ 1 \end{pmatrix}
  3. The null space of A is empty since both systems have unique particular solutions
  4. The null space of A is 1-dimensional and is spanned by $$ \begin{pmatrix} 2 \ 1 \ 0 \ -1 \end{pmatrix} (correct answer)
Explanation: 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 XX have the same direction vector (210−1)\begin{pmatrix} 2 \\ 1 \\ 0 \\ -1 \end{pmatrix} in their parametric forms. This vector represents the homogeneous solution to Ax=0A\mathbf{x} = \mathbf{0}. Since both systems Ax=b1A\mathbf{x} = \mathbf{b_1} and Ax=b2A\mathbf{x} = \mathbf{b_2} (where b1\mathbf{b_1} and b2\mathbf{b_2} are the columns of BB) share this same direction vector, it must be in the null space of AA. 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 $$ \begin{pmatrix} 1 \ 0 \ 2 \ 1 \end{pmatrix} Remember: in parametric solutions to $$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.

Question 13

The solution set to a consistent linear system Cx=dCx = d is given by x=(2−1403)+α(1102−1)+β(02−111)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 CC has 5 columns, what is the minimum possible number of rows in CC?

  1. 2 rows, since the null space has dimension 2 and rank plus nullity equals 5
  2. 3 rows, since we need at least 3 constraints to determine a 2-dimensional solution space (correct answer)
  3. 4 rows, since the system must have at least as many equations as unknowns minus free variables
  4. 5 rows, since C must be square to have a consistent system with parameters
Explanation: 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.

Question 14

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}

  1. A line in R4\mathbb{R}^4 that passes through the origin since it contains the zero vector
  2. A hyperplane in R4\mathbb{R}^4 that does not pass through the origin since it has codimension 1
  3. A plane in R4\mathbb{R}^4 that passes through the origin since all linear combinations are included
  4. A line in R4\mathbb{R}^4 that does not pass through the origin since it's a translated copy of the null space (correct answer)
Explanation: 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=(20−13)+u(120−1)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 (20−13)\begin{pmatrix} 2 \\ 0 \\ -1 \\ 3 \end{pmatrix} and the null space spanned by (120−1)\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 R4\mathbb{R}^4. This line doesn't pass through the origin because when u=0u = 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 uu 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.

Question 15

Consider the system of linear equations Ax=bAx = b where AA is a 4×54 \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}

  1. The solution set is a 2-dimensional plane in R5\mathbb{R}^5 containing infinitely many solutions with exactly 2 free variables (correct answer)
  2. The solution set is a 3-dimensional hyperplane in R5\mathbb{R}^5 containing infinitely many solutions with exactly 3 free variables
  3. The solution set is a 1-dimensional line in R5\mathbb{R}^5 containing infinitely many solutions with exactly 1 free parameter
  4. The solution set is a 4-dimensional subspace in R5\mathbb{R}^5 containing infinitely many solutions with exactly 4 free variables
Explanation: The parametric solution shows 2 parameter variables (t and s), indicating a 2-dimensional affine subspace (plane) in R5\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.

Question 16

Two students obtain different parametric representations for the same linear system: Student A writes x=(120)+t(2−13)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}

  1. Student A's representation is correct while Student B's contains an error in the direction vector calculation
  2. The representations are different since they have different particular solutions, indicating at least one student made an error
  3. Both representations are equivalent since the direction vectors are scalar multiples and both particular solutions satisfy the original system (correct answer)
  4. The representations describe different solution sets since the direction vectors point in opposite directions
Explanation: 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 (2−13)\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} while Student B uses (−42−6)\begin{pmatrix} -4 \\ 2 \\ -6 \end{pmatrix}. Notice that Student B's direction vector equals −2-2 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=1t = 1 in Student A's equation gives $$ \begin{pmatrix} 3 \ 1 \ 3 \end{pmatrix} OptionAincorrectlyassumesStudentBmadeanerror,butscalarmultiplesofdirectionvectorsareperfectlyvalid.OptionBwronglysuggeststhatdifferentparticularsolutionsindicateanerror—multiplepointscanlieonthesamesolutionset.OptionDmisunderstandsthatoppositedirectionsstilldescribethesameline,justtraverseddifferentlyastheparametervaries.∗∗Studytip:∗∗Whencomparingparametricsolutions,alwayscheckifdirectionvectorsarescalarmultiplesandverifythatparticularsolutionscanbereachedfromeachotherbyadjustingparameters.Different−lookingrepresentationscandescribeidenticalsolutionsets. 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.

Question 17

Consider the parametric solution x=(41−32)+t(201−1)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 x1+x3=0x_1 + x_3 = 0, what values of the parameter tt satisfy this additional constraint?

  1. t=−1t = -1
  2. t=13t = \frac{1}{3}
  3. t=−13t = -\frac{1}{3} (correct answer)
  4. No values of tt satisfy the constraint since it's inconsistent with the original system
Explanation: 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 tt, substitute the parametric expressions into the constraint equation x1+x3=0x_1 + x_3 = 0. From the given solution, x1=4+2tx_1 = 4 + 2t and x3=−3+tx_3 = -3 + t. Substituting these into the constraint: (4+2t)+(−3+t)=0(4 + 2t) + (-3 + t) = 0 4+2t−3+t=04 + 2t - 3 + t = 0 1+3t=01 + 3t = 0 3t=−13t = -1 t=−13t = -\frac{1}{3} This confirms answer C is correct. Looking at the wrong answers: A) t=−1t = -1 would give x1+x3=(4+2(−1))+(−3+(−1))=2+(−4)=−2≠0x_1 + x_3 = (4 + 2(-1)) + (-3 + (-1)) = 2 + (-4) = -2 \neq 0. B) t=13t = \frac{1}{3} would yield x1+x3=(4+23)+(−3+13)=143−83=2≠0x_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 tt will give you the specific points on the solution line that satisfy your additional condition.

Question 18

Let SS be the solution set for a consistent linear system Ax⃗=b⃗A\vec{x}=\vec{b} where AA is an m×nm \times n matrix and b⃗≠0⃗\vec{b} \neq \vec{0}. Which statement about SS must be true?

  1. SS is a subspace of Rn\mathbb{R}^n.
  2. The dimension of SS is equal to the rank of AA.
  3. SS contains exactly one vector if m=nm=n.
  4. SS is a translation of the null space of AA. (correct answer)
Explanation: The general solution to a consistent system Ax⃗=b⃗A\vec{x}=\vec{b} can be written as x⃗=p⃗+x⃗h\vec{x} = \vec{p} + \vec{x}_h, where p⃗\vec{p} is any particular solution and x⃗h\vec{x}_h is the general solution to the homogeneous system Ax⃗=0⃗A\vec{x}=\vec{0} (i.e., the null space of A). Geometrically, this means the solution set SS is the null space of AA translated by the vector p⃗\vec{p}. Distractor A is false because for SS to be a subspace, it must contain the zero vector. But A0⃗=0⃗A\vec{0} = \vec{0}, and since b⃗≠0⃗\vec{b} \neq \vec{0}, the zero vector is not in SS. Distractor B is false; the dimension of SS (as a geometric object) is the nullity of A, not the rank. Distractor C is false; a unique solution requires the null space to be trivial, which is not guaranteed even if AA is square.

Question 19

The general solution to a consistent system Ax⃗=b⃗A\vec{x} = \vec{b} is a line in R5\mathbb{R}^5. What can be concluded about the matrix AA?

  1. The rank of AA is 1.
  2. The columns of AA are linearly independent.
  3. The nullity of AA is 1. (correct answer)
  4. AA is an invertible matrix.
Explanation: The geometric dimension of the solution set is equal to the dimension of the null space of AA, which is defined as the nullity of AA. Since the solution set is a line, its dimension is 1. Therefore, the nullity of AA must be 1. By the Rank-Nullity Theorem for a matrix with nn columns (here n=5n=5), we have rank(A)+nullity(A)=n\mathrm{rank}(A) + \mathrm{nullity}(A) = n. So, rank(A)+1=5\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.

Question 20

The coefficient matrix of a homogeneous system Ax=0Ax = 0 is 4×64 \times 6 with rank 4. If a related non-homogeneous system Ax=bAx = b is consistent, what is the dimension of its solution set?

  1. 4-dimensional, since the rank determines the dimension of the solution space
  2. 2-dimensional, since there are 6-4=2 free variables in the parametric representation (correct answer)
  3. 1-dimensional, since adding a particular solution reduces the dimension by 1
  4. 6-dimensional, since the solution vectors lie in R6\mathbb{R}^6
Explanation: For a consistent non-homogeneous system Ax=bAx = b, the solution set is an affine subspace of dimension equal to the dimension of the null space of A. Since A is 4×64 \times 6 with rank 4, by the rank-nullity theorem, the null space has dimension 6−4=26 - 4 = 2. Therefore, the solution set is a 2-dimensional affine subspace (plane) in R6\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.