Linear Algebra Quiz: Parametric Solutions And Free Variables

19 questions · exam conditions

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Parametric Solutions And Free VariablesQuestion 1 of 19

The system $A\mathbf{x} = \mathbf{b}$ has the particular solution $\mathbf{x}_p = \begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix}$ and the associated homogeneous system $A\mathbf{x} = \mathbf{0}$ has general solution $\mathbf{x}_h = t\begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix}$ . If we want exactly 50% of the solutions to have $x_2 > 0$ , what constraint must be satisfied?

We must restrict

t > -1

to ensure

x_2 = t > -1

for half the solutions

We must restrict

t > 0

which gives exactly half the parameter space

No such constraint exists since

x_2 = t

takes all real values with equal probability

This is impossible since

x_2

starts at 0 and increases linearly with

t

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Linear Algebra Quiz

Linear Algebra Quiz: Parametric Solutions And Free Variables

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

What this quiz covers

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

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

We must restrict $t > -1$ to ensure $x_2 = t > -1$ for half the solutions
We must restrict $t > 0$ which gives exactly half the parameter space
No such constraint exists since $x_2 = t$ takes all real values with equal probability (correct answer)
This is impossible since $x_2$ starts at 0 and increases linearly with $t$

Explanation: When you encounter questions about linear systems with particular and homogeneous solutions, you're dealing with the fundamental structure of solution sets. The general solution to

A\mathbf{x} = \mathbf{b}

\mathbf{x} = \mathbf{x}_p + \mathbf{x}_h

, where every value of the parameter

t

gives a valid solution. Here, the general solution is

\mathbf{x} = \begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix} + t\begin{bmatrix} 2 \\ 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 + 2t \\ t \\ 3 - t \end{bmatrix}

. The second component is

x_2 = t

. The question asks about "exactly 50% of solutions" having

x_2 > 0

, but this reveals a conceptual issue. Since

t

can be any real number, there are infinitely many solutions. The concept of "50% of solutions" doesn't have mathematical meaning without specifying a probability distribution or finite constraint on

t

. Answer C correctly identifies this fundamental problem. Answer A incorrectly suggests restricting

t > -1

somehow gives "half the solutions," but this restriction would still yield infinitely many solutions, not a meaningful "half." Answer B makes the same error with

t > 0

, misunderstanding what "half the parameter space" means in this infinite context. Answer D incorrectly claims the situation is impossible due to the linear relationship, missing that the real issue is the undefined nature of "50%" over an infinite set. Remember: when dealing with parametric solutions to linear systems, be careful about probability language. Without additional constraints or distributions, fractions of infinite solution sets are meaningless.

Question 2

The reduced row echelon form of the augmented matrix for a system of linear equations is given by

\begin{pmatrix} 1 & 2 & 0 & -3 & | & 5 \\ 0 & 0 & 1 & 4 & | & -1 \\ 0 & 0 & 0 & 0 & | & 0 \end{pmatrix}

Letting the free variables be represented by parameters $s$ and $t$ , which of the following correctly expresses the general solution in parametric vector form $\vec{x} = \vec{p} + s\vec{u} + t\vec{v}$ ?

(correct answer)

Explanation: The pivot columns are 1 and 3, corresponding to variables

x_1

and

x_3

. The free variables are

x_2

and

x_4

. Let

x_2 = s

and

x_4 = t

. From the RREF, we write the system of equations:

x_1 + 2x_2 - 3x_4 = 5

x_3 + 4x_4 = -1

Express the basic variables in terms of the free variables:

x_1 = 5 - 2x_2 + 3x_4 = 5 - 2s + 3t

x_3 = -1 - 4x_4 = -1 - 4t

Now, write the solution vector:

This matches choice A. Distractor B has sign errors in the direction vectors for

s

and

t

. Distractor C incorrectly forms the particular solution vector

\vec{p}

and the direction vectors. Distractor D incorrectly forms the direction vectors by omitting the '1's for the free variables themselves.

Question 3

A linear system's solution is given in parametric form as $\vec{x} = \begin{pmatrix} 4 \\ 0 \\ -1 \\ 0 \end{pmatrix} + s \begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + t \begin{pmatrix} 0 \\ 0 \\ 3 \\ 1 \end{pmatrix}$ . Which of the following could be the reduced row echelon form of the system's augmented matrix?

$\begin{pmatrix} 1 & 2 & 0 & 0 & | & 4 \\ 0 & 0 & 1 & -3 & | & -1 \\ 0 & 0 & 0 & 0 & | & 0 \end{pmatrix}$ (correct answer)
$\begin{pmatrix} 1 & -2 & 0 & 0 & | & 4 \\ 0 & 0 & 1 & 3 & | & -1 \\ 0 & 0 & 0 & 0 & | & 0 \end{pmatrix}$
$\begin{pmatrix} 1 & 0 & 0 & 0 & | & 4 \\ 0 & 1 & 0 & 0 & | & 0 \\ 0 & 0 & 1 & 0 & | & -1 \end{pmatrix}$
$\begin{pmatrix} 1 & 2 & 0 & 0 & | & 4 \\ 0 & 0 & 1 & -3 & | & -1 \\ 0 & 0 & 0 & 0 & | & 1 \end{pmatrix}$

Explanation: From the parametric solution, we can identify the free and basic variables. The parameters

s

and

t

correspond to free variables. The vector for

s

has a 1 in the second position, so

x_2 = s

. The vector for

t

has a 1 in the fourth position, so

x_4 = t

. Thus,

x_2

and

x_4

are free variables, and

x_1

and

x_3

are basic variables. From the solution, we can write equations for the basic variables:

x_1 = 4 - 2s + 0t = 4 - 2x_2 \implies x_1 + 2x_2 = 4

x_3 = -1 + 0s + 3t = -1 + 3x_4 \implies x_3 - 3x_4 = -1

These two equations correspond to the rows of the RREF matrix. The first equation gives the row

[1 \ 2 \ 0 \ 0 \ | \ 4]

. The second equation gives the row

[0 \ 0 \ 1 \ -3 \ | \ -1]

. The system is consistent, so any additional rows must be zero rows. This matches matrix A. Distractor B has incorrect signs in the non-pivot entries. Distractor C implies a unique solution with no free variables, which contradicts the given parametric form. Distractor D contains the row

[0 \ 0 \ 0 \ 0 \ | \ 1]

, which means the system is inconsistent and has no solution.

Question 4

Let $\mathbf{A}$ be a $5 \times 7$ matrix. The solution set of the homogeneous equation $\mathbf{A}\vec{x} = \vec{0}$ is a 4-dimensional subspace of $\mathbb{R}^7$ . What is the rank of the matrix $\mathbf{A}$ ?

2
3 (correct answer)
4
5

Explanation: The Rank-Nullity Theorem states that for an

m \times n

matrix

\mathbf{A}

, the rank of

\mathbf{A}

plus the dimension of the null space of

\mathbf{A}

(its nullity) equals the number of columns,

n

. The dimension of the solution set of

\mathbf{A}\vec{x} = \vec{0}

is the nullity, which is also equal to the number of free variables. Here, the matrix is

5 \times 7

, so

n=7

. The dimension of the solution space (nullity) is given as 4. Using the theorem: rank(A) + nullity(A) =

n

. rank(A) + 4 = 7. rank(A) = 3. The rank is the number of pivot positions. Distractor A is an incorrect calculation. Distractor C incorrectly equates the rank with the nullity. Distractor D incorrectly equates the rank with the number of rows.

Question 5

The augmented matrix of a linear system with variables $x_1, x_2, x_3, x_4$ has been reduced to the following form:

\begin{pmatrix} 1 & 0 & -5 & 0 & | & 2 \\ 0 & 1 & 3 & 0 & | & 6 \\ 0 & 0 & 0 & 1 & | & -4 \end{pmatrix}

Which statement correctly describes a variable in this system?

$x_4$ is a free variable because its corresponding column is the last one in the coefficient matrix.
$x_3$ is a free variable because its value can be chosen independently of the other variables.
$x_1$ , $x_2$ , and $x_4$ are basic variables, and their values are determined once a value for the free variable $x_3$ is chosen. (correct answer)
$x_2$ and $x_3$ are free variables because their columns in the coefficient matrix contain non-zero, non-pivot entries.

Explanation: The pivot positions are in columns 1, 2, and 4. Therefore,

x_1

x_2

, and

x_4

are basic (or pivot) variables. Column 3 does not have a pivot, so

x_3

is the free variable. The value of a free variable can be chosen arbitrarily. Once a value for

x_3

is set, the values for the basic variables

x_1

x_2

, and

x_4

are fixed by the equations:

x_1 = 2 + 5x_3

x_2 = 6 - 3x_3

, and

x_4 = -4

. Statement C correctly identifies the basic and free variables and describes their relationship. Distractor A is incorrect;

x_4

is a basic variable because its column contains a pivot. The position of the column does not determine if a variable is free. Distractor B is partially correct that

x_3

is free but the reasoning is incomplete; the key is that its value determines the others, as C states more fully. Distractor D is incorrect; only

x_3

is a free variable. The presence of non-zero entries does not determine if a variable is free; the absence of a pivot in its column does.

Question 6

A system of linear equations reduces to the single equation $x_1 - 3x_2 + 2x_3 = 5$ . Which of the following expressions represents the general solution to this system, using parameters $s$ and $t$ ?

(correct answer)

Explanation: With one equation and three variables, there will be one basic variable and two free variables. We can choose

x_1

as the basic variable and

x_2, x_3

as the free variables. Let

x_2 = s

and

x_3 = t

. Solve for the basic variable

x_1

x_1 = 5 + 3x_2 - 2x_3 = 5 + 3s - 2t

. Now, write the solution in vector form:

This matches choice A. Distractor B has sign errors in the direction vectors. It incorrectly moves the terms to the right side of the equation. Distractor C is the solution to the homogeneous equation

x_1 - 3x_2 + 2x_3 = 0

. It is missing the particular solution vector. Distractor D incorrectly constructs both the particular solution and the direction vectors.

Question 7

The solution set of a consistent system of linear equations in variables $x, y,$ and $z$ is parameterized by $x = 5t + 3$ , $y = -2t$ , and $z = t$ . What can be concluded about the reduced row echelon form (RREF) of the system's augmented matrix?

The RREF of the coefficient matrix has a pivot in every column.
The RREF of the augmented matrix contains exactly one row of all zeros.
The RREF of the augmented matrix contains a row of the form $[0 \ 0 \ 0 \ | \ 1]$ .
The RREF of the coefficient matrix has exactly two pivot positions. (correct answer)

Explanation: The solution is parameterized by a single variable,

t

. This means there is exactly one free variable. The system has 3 variables in total (

x, y, z

). The number of basic variables (pivots) is equal to the total number of variables minus the number of free variables. Number of pivots = 3 - 1 = 2. Therefore, the RREF of the coefficient matrix must have exactly two pivot positions. Distractor A is incorrect; a pivot in every column would mean zero free variables and a unique solution. Distractor B is not necessarily true. The matrix could have originated from two equations, e.g.,

x - 5z = 3

and

y + 2z = 0

. In this case, the

2 \times 4

RREF would be

\begin{pmatrix} 1 & 0 & -5 & | & 3 \\ 0 & 1 & 2 & | & 0 \end{pmatrix}

, which has no zero row. If the original system had three or more redundant equations, there would be zero rows. Distractor C is incorrect; this would indicate an inconsistent system with no solution, but the problem states a solution set exists.

Question 8

A consistent system of linear equations has 6 variables. It is known that the solution set forms a 2-dimensional plane in $\mathbb{R}^6$ . How many pivot columns must the reduced row echelon form of the system's coefficient matrix have?

2
3
4 (correct answer)
6

Explanation: The dimension of the solution set corresponds to the number of free variables in the system. Since the solution set is a 2-dimensional plane, there are 2 free variables. The number of pivot columns (the rank of the matrix) is equal to the total number of variables minus the number of free variables. The system has 6 variables. Number of pivot columns = (Total variables) - (Free variables) Number of pivot columns = 6 - 2 = 4. Therefore, the reduced row echelon form of the coefficient matrix must have 4 pivot columns. Distractor A confuses the number of pivot columns with the number of free variables. Distractor B is an incorrect calculation. Distractor D incorrectly assumes there must be a pivot for every variable, which would imply a unique solution (a 0-dimensional solution set), contradicting the given information.

Question 9

Consider the system of linear equations represented by the augmented matrix below, where $k$ is a real constant.

\begin{pmatrix} 1 & 0 & -3 & | & -2 \\ 0 & 1 & 2 & | & 5 \\ 0 & 0 & k^2 - 4 & | & k + 2 \end{pmatrix}

For which value of $k$ does the system have infinitely many solutions characterized by one free variable?

$k = 2$
$k = -2$ (correct answer)
$k = 4$
For all $k \neq 2$ and $k \neq -2$ .

Explanation: For the system to have a free variable, the third column must not contain a pivot. This occurs when the entry

k^2 - 4 = 0

, which means

k=2

k=-2

. Case 1:

k = 2

. The third row becomes

[0 \ 0 \ 0 \ | \ 2+2]

, which is

[0 \ 0 \ 0 \ | \ 4]

. This equation,

0 = 4

, is a contradiction, so the system is inconsistent and has no solution. Case 2:

k = -2

. The third row becomes

[0 \ 0 \ 0 \ | \ -2+2]

, which is

[0 \ 0 \ 0 \ | \ 0]

. This is a trivial equation,

0 = 0

, and the system is consistent. The first two columns have pivots, but the third column does not. Therefore,

x_3

is a free variable. The system has infinitely many solutions parameterized by one free variable. Distractor A leads to an inconsistent system. Distractor C is a computational error, failing to take the square root of 4. Distractor D describes the condition for a unique solution, where there is a pivot in every column of the coefficient matrix.

Question 10

Consider the augmented matrix shown below, which is in row echelon form.

\begin{pmatrix} 1 & h & -4 & | & 1 \\ 0 & 2 & k & | & 6 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}

For what values of the real parameters $h$ and $k$ will the corresponding linear system have exactly one free variable?

Only when $h=0$ and $k=0$ .
Only when the product $hk=0$ .
The system cannot have exactly one free variable.
For all real values of $h$ and $k$ . (correct answer)

Explanation: The number of free variables is the number of columns in the coefficient matrix minus the number of pivot positions. The given matrix is in row echelon form. We can identify the pivot positions. The first pivot is in column 1 (the leading '1'). The second pivot is in column 2 (the leading '2'). There is no pivot in column 3. Therefore, the matrix has 2 pivot positions. The system has 3 variables (

x_1, x_2, x_3

). Number of free variables = (Number of variables) - (Number of pivots) = 3 - 2 = 1. The free variable is

x_3

. The values of

h

and

k

are in non-pivot positions and affect the specific values in the solution set, but they do not change the number of pivots or free variables. Thus, the system will have exactly one free variable for all real values of

h

and

k

. Distractor A and B are incorrect because

h

and

k

do not determine the number of free variables in this echelon form. Distractor C is incorrect because, as shown, the system always has one free variable.

Question 11

The general solution to a non-homogeneous system $\mathbf{A}\vec{x} = \vec{b}$ can be written as $\vec{x} = \vec{p} + \vec{x}_h$ , where $\vec{p}$ is one particular solution to $\mathbf{A}\vec{x} = \vec{b}$ and $\vec{x}_h$ is the general solution to the corresponding homogeneous system $\mathbf{A}\vec{x} = \vec{0}$ .

The solution set for a linear system $\mathbf{A}\vec{x} = \vec{b}$ is given by $\vec{x} = \begin{pmatrix} -3 \\ 1 \\ 0 \end{pmatrix} + t \begin{pmatrix} 4 \\ 0 \\ 2 \end{pmatrix}$ . Based on the passage, which statement must be true?

The vector $\begin{pmatrix} 4 \\ 0 \\ 2 \end{pmatrix}$ is a particular solution to the system $\mathbf{A}\vec{x} = \vec{b}$ .
The solution set for the homogeneous system $\mathbf{A}\vec{x} = \vec{0}$ is a line passing through the point $(-3, 1, 0)$ .
The matrix-vector product $\mathbf{A} \begin{pmatrix} 4 \\ 0 \\ 2 \end{pmatrix}$ results in the zero vector. (correct answer)
The system $\mathbf{A}\vec{x} = \vec{b}$ has a unique solution given by the vector $\begin{pmatrix} -3 \\ 1 \\ 0 \end{pmatrix}$ .

Explanation: In the solution

\vec{x} = \vec{p} + \vec{x}_h

\vec{p} = \begin{pmatrix} -3 \\ 1 \\ 0 \end{pmatrix}

is the particular solution and

\vec{x}_h = t \begin{pmatrix} 4 \\ 0 \\ 2 \end{pmatrix}

is the solution to the homogeneous system

\mathbf{A}\vec{x} = \vec{0}

. This means that any vector of the form

t \begin{pmatrix} 4 \\ 0 \\ 2 \end{pmatrix}

must satisfy

\mathbf{A}\vec{x} = \vec{0}

. Specifically, if we take

t=1

, the vector

\vec{v} = \begin{pmatrix} 4 \\ 0 \\ 2 \end{pmatrix}

is a solution to

\mathbf{A}\vec{x} = \vec{0}

. Therefore,

\mathbf{A}\vec{v}

must be the zero vector. This matches choice C. Distractor A is incorrect; it confuses the direction vector of the homogeneous solution with the particular solution vector. Distractor B is incorrect; the solution to the homogeneous system is a line that must pass through the origin, not through the point given by the particular solution. Distractor D is incorrect; the presence of the parameter

t

indicates there are infinitely many solutions, not a unique one.

Question 12

Consider the augmented matrix $\begin{bmatrix} 1 & 2 & -1 & 3 & | & 4 \\ 0 & 0 & 1 & -2 & | & 1 \\ 0 & 0 & 0 & 0 & | & 0 \end{bmatrix}$ in reduced row echelon form. If the general solution is written as $\mathbf{x} = \mathbf{x}_p + s\mathbf{v}_1 + t\mathbf{v}_2$ where $s$ and $t$ are parameters, which of the following correctly identifies the relationship between the free variables and parameters?

$x_2 = s$ and $x_4 = t$ , where $s$ and $t$ can be any real numbers (correct answer)
$x_1 = s$ and $x_3 = t$ , where $s$ and $t$ can be any real numbers
$x_2 = s$ and $x_3 = t$ , where $s$ and $t$ must be non-negative real numbers
$x_1 = s$ and $x_4 = t$ , where $s$ and $t$ must be integers only

Explanation: From the RREF, we see that columns 1 and 3 contain leading 1's (pivot columns), making

x_1

and

x_3

basic variables. Columns 2 and 4 do not contain pivots, making

x_2

and

x_4

free variables. We can set

x_2 = s

and

x_4 = t

where

s, t ∈ ℝ

. Choice B incorrectly identifies basic variables as free variables. Choice C correctly identifies free variables but incorrectly restricts their domain. Choice D misidentifies the free variables and incorrectly restricts the parameter domain.

Question 13

Consider a system whose augmented matrix reduces to $\begin{bmatrix} 1 & a & 0 & | & 2 \\ 0 & 0 & 1 & | & b \\ 0 & 0 & 0 & | & 0 \end{bmatrix}$ where $a$ and $b$ are parameters. For the system to have a unique solution, which condition on $a$ and $b$ is necessary and sufficient?

$a \neq 0$ and $b$ can be any real number, since we need 3 pivot columns
$a = 0$ and $b \neq 0$ , since this makes the system have exactly one free variable
$a \neq 0$ and $b = 0$ , since we need full column rank and consistency
No values of $a$ and $b$ work since we have 3 variables but only 2 pivot positions (correct answer)

Explanation: For a unique solution, we need the number of pivot columns to equal the number of variables. Here we have 3 variables (

x_1, x_2, x_3

) but only 2 pivot positions (columns 1 and 3), regardless of the values of

a

and

b

. The parameter

a

affects the coefficient of

x_2

but doesn't create a new pivot since

x_2

corresponds to column 2, which has no leading 1. Choice A incorrectly thinks

a \neq 0

creates a pivot. Choice B misunderstands the pivot structure. Choice C incorrectly believes the values of

a

and

b

can change the fundamental structure.

Question 14

A linear system has the parametric solution $\mathbf{x} = \begin{bmatrix} 2 \\ 1 \\ 0 \\ -1 \end{bmatrix} + r\begin{bmatrix} 1 \\ 0 \\ -2 \\ 1 \end{bmatrix} + s\begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix}$ . If we form a new system by adding the equation $x_1 + x_2 + x_3 + x_4 = 5$ to the original system, what happens to the solution set?

The solution set becomes a single point since both parameters become fixed
The solution set becomes a line since one parameter becomes fixed while the other remains free (correct answer)
The solution set remains a plane since the new equation is satisfied by all original solutions
The solution set becomes empty since the new equation is incompatible with the parametric solution

Explanation: Substituting the parametric solution into

x_1 + x_2 + x_3 + x_4 = 5

(2+r) + (1+s) + (0-2r+s) + (-1+r) = 5

. Simplifying:

2 + r + 1 + s - 2r + s - 1 + r = 2 + 2s = 5

, so

s = \frac{3}{2}

. With

s

fixed but

r

free, the solution set becomes a line. Choice A incorrectly assumes both parameters are fixed. Choice C would be true if the equation was automatically satisfied. Choice D is wrong since we found a valid value for

s

Question 15

The solution set of a homogeneous system $\mathbf{A}\vec{x} = \vec{0}$ is given by all linear combinations of the vectors $\vec{u} = \begin{pmatrix} 1 \\ 0 \\ -2 \\ 1 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 0 \\ 1 \\ 3 \\ 0 \end{pmatrix}$ . Which of the following vectors is NOT a solution to this system?

$\begin{pmatrix} 2 \\ -1 \\ -7 \\ 2 \end{pmatrix}$
$\begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}$
$\begin{pmatrix} -1 \\ -2 \\ -4 \\ -1 \end{pmatrix}$ (correct answer)
$\begin{pmatrix} 3 \\ 2 \\ 0 \\ 3 \end{pmatrix}$

Explanation: A vector

\vec{w}

is a solution if and only if it can be written as a linear combination

\vec{w} = s\vec{u} + t\vec{v}

for some scalars

s

and

t

. We must check which option does not satisfy this condition. A) For

\begin{pmatrix} 2 \\ -1 \\ -7 \\ 2 \end{pmatrix}

: From row 1,

s=2

. From row 2,

t=-1

. Check other rows: Row 3:

2(-2) + (-1)(3) = -4 - 3 = -7

. Correct. Row 4:

2(1) + (-1)(0) = 2

. Correct. This is a solution. B) For

\begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}

: From row 1,

s=1

. From row 2,

t=1

. Check other rows: Row 3:

1(-2) + 1(3) = 1

. Correct. Row 4:

1(1) + 1(0) = 1

. Correct. This is a solution. D) For

\begin{pmatrix} 3 \\ 2 \\ 0 \\ 3 \end{pmatrix}

: From row 1,

s=3

. From row 2,

t=2

. Check other rows: Row 3:

3(-2) + 2(3) = -6 + 6 = 0

. Correct. Row 4:

3(1) + 2(0) = 3

. Correct. This is a solution. C) For

\begin{pmatrix} -1 \\ -2 \\ -4 \\ -1 \end{pmatrix}

: From row 1,

s=-1

. From row 2,

t=-2

. Check other rows: Row 3:

(-1)(-2) + (-2)(3) = 2 - 6 = -4

. Correct. Row 4:

(-1)(1) + (-2)(0) = -1

. Correct. Oh, wait, I made a mistake in my scratchpad. Let me re-verify all of them. Re-check B:

s=1, t=1

. Row 3:

1(-2) + 1(3) = 1

. Correct. Row 4:

1(1) + 1(0) = 1

. Correct. Let me re-check C:

s=-1, t=-2

. Row 3:

(-1)(-2) + (-2)(3) = 2-6 = -4

. Correct. Row 4:

(-1)(1) = -1

. Correct. My scratchpad was wrong, C is also a solution. Let me check the question and my vectors again. My distractors were all solutions. I need one that isn't. Let's make a new distractor. Let's change C. New C:

\begin{pmatrix} 1 \\ -2 \\ 8 \\ 1 \end{pmatrix}

. Test:

s=1, t=-2

. Row 3:

1(-2) + (-2)(3) = -2-6 = -8

. The vector has 8, not -8. So this is not a solution. Okay, let me re-write the options. Correct answer choice will be a vector not in the span. Let's test

\begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}

again.

s=1, t=1

s(-2) + t(3) = -2+3 = 1

. Correct. So B is a solution. Let's test

\begin{pmatrix} 2 \\ -1 \\ -7 \\ 2 \end{pmatrix}

s=2, t=-1

s(-2) + t(3) = 2(-2)+(-1)(3) = -4-3 = -7

. Correct. So A is a solution. Let's test

\begin{pmatrix} 3 \\ 2 \\ 0 \\ 3 \end{pmatrix}

s=3, t=2

s(-2)+t(3) = 3(-2)+2(3) = -6+6=0

. Correct. So D is a solution. Let me construct one that fails. Take

s=1, t=1

which gives

(1,1,1,1)

. Let's change the third component. How about

(1,1,2,1)

? Is this a solution?

s=1, t=1

. Third component should be 1, not 2. So

(1,1,2,1)

is not a solution. I will use this as the correct answer. Let's change the options in the final json. Let's swap B for the new non-solution. Final check: For B)

\begin{pmatrix} 1 \\ 1 \\ 2 \\ 1 \end{pmatrix}

: from row 1,

s=1

; from row 2,

t=1

. We test row 3:

s(-2) + t(3) = (1)(-2) + (1)(3) = 1

. However, the third component of the vector is 2. Since

1 \neq 2

, this vector is not a linear combination of

\vec{u}

and

\vec{v}

and thus is not a solution. The others are verified solutions.

Question 16

Consider the parametric solution $\mathbf{x} = \begin{bmatrix} 2 \\ 0 \\ -1 \end{bmatrix} + s\begin{bmatrix} -1 \\ 1 \\ 2 \end{bmatrix} + t\begin{bmatrix} 3 \\ -2 \\ 1 \end{bmatrix}$ to a linear system. If we require that $x_1 + x_2 = 1$ , how many solutions satisfy this additional constraint?

Exactly one solution exists since the constraint eliminates one degree of freedom completely
Infinitely many solutions exist since the constraint defines a line in the solution plane (correct answer)
No solutions exist since the constraint is incompatible with the original system
Exactly two solutions exist since the constraint equation is quadratic in the parameters

Explanation: Substituting the parametric solution into

x_1 + x_2 = 1

(2 - s + 3t) + (0 + s - 2t) = 1

, which simplifies to

2 + t = 1

, so

t = -1

. With

t = -1

fixed,

s

remains free, giving infinitely many solutions forming a line. Choice A incorrectly assumes both parameters become fixed. Choice C is wrong since

t = -1

provides a valid constraint. Choice D incorrectly assumes the constraint equation is quadratic when it's actually linear in the parameters.

Question 17

A homogeneous system $A\mathbf{x} = \mathbf{0}$ has a $4 \times 6$ coefficient matrix $A$ with rank 3. When the general solution is expressed in parametric form, what is the minimum number of linearly independent vectors needed in the spanning set for the solution space?

The spanning set requires exactly 3 linearly independent vectors since rank(A) = 3
The spanning set requires exactly 4 linearly independent vectors since A has 4 rows
The spanning set requires exactly 6 linearly independent vectors since A has 6 columns
The spanning set requires exactly 3 linearly independent vectors since nullity(A) = 3 (correct answer)

Explanation: By the Rank-Nullity Theorem, nullity(A) = number of columns - rank(A) = 6 - 3 = 3. The dimension of the null space equals the nullity, so we need exactly 3 linearly independent vectors to span the solution space. Choice A confuses rank with nullity. Choice B incorrectly uses the number of rows. Choice C incorrectly uses the total number of columns. The key insight is that the dimension of the solution space equals the number of free variables, which equals the nullity.

Question 18

For a homogeneous system with general solution $\mathbf{x} = s\begin{bmatrix} 1 \\ -2 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 0 \\ 1 \\ 1 \\ -1 \end{bmatrix}$ , consider the subset of solutions where $x_1 - x_3 = 0$ . This subset forms which type of geometric object?

A line through the origin since one parameter becomes dependent on the other (correct answer)
A plane through the origin since we eliminate one degree of freedom from a 2D space
A single point at the origin since both parameters become fixed at zero
The empty set since the constraint is incompatible with the parametric solution

Explanation: Substituting into

x_1 - x_3 = 0

(s \cdot 1 + t \cdot 0) - (s \cdot 1 + t \cdot 1) = 0

, which gives

s - s - t = 0

, so

t = 0

. With

t = 0

, the solution becomes

\mathbf{x} = s\begin{bmatrix} 1 \\ -2 \\ 1 \\ 0 \end{bmatrix}

, which is a line through the origin. Choice B would be correct if we started with a 3D space. Choice C incorrectly concludes both parameters are fixed. Choice D is wrong since

t = 0

is a valid solution.

Question 19

Given the augmented matrix $\begin{bmatrix} 1 & 0 & 2 & 0 & | & 3 \\ 0 & 1 & -1 & 0 & | & 2 \\ 0 & 0 & 0 & 1 & | & -1 \end{bmatrix}$ in RREF, if the parametric solution is expressed with $x_3 = t$ , what is the coefficient of $t$ in the expression for $x_1$ ?

The coefficient of $t$ in $x_1$ is $0$ since $x_1$ is a basic variable
The coefficient of $t$ in $x_1$ is $2$ since $x_1 = 3 + 2x_3$
The coefficient of $t$ in $x_1$ is $-2$ since $x_1 = 3 - 2x_3$ (correct answer)
The coefficient of $t$ in $x_1$ is $3$ since that's the constant term

Explanation: When you encounter an augmented matrix in reduced row echelon form (RREF), you're looking at a system where you can directly read off the relationships between variables. The key is identifying which variables are basic (have leading 1s in their columns) and which are free (can take any value). From this RREF matrix, you can see that columns 1, 2, and 4 contain leading 1s, making

x_1

x_2

, and

x_4

basic variables. Column 3 has no leading 1, so

x_3

is a free variable that we set equal to parameter

t

. To find

x_1

in terms of

t

, read the first row:

1 \cdot x_1 + 0 \cdot x_2 + 2 \cdot x_3 + 0 \cdot x_4 = 3

. This gives us

x_1 + 2x_3 = 3

. Solving for

x_1

x_1 = 3 - 2x_3 = 3 - 2t

. The coefficient of

t

-2

. Choice A incorrectly assumes basic variables can't depend on free variables. While

x_1

is basic, it still depends on the free variable

x_3

. Choice B makes a sign error, writing

x_1 = 3 + 2x_3

instead of correctly isolating

x_1

on the left side. Choice D confuses the coefficient of

t

with the constant term. Study tip: When working with RREF matrices, always solve for basic variables by moving free variable terms to the right side of the equation. The coefficients of free variables in the matrix become negative coefficients in your parametric solution.