Parametric Solutions & Free Variables - Linear Algebra

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?
Answer: $x_2 = s$ and $x_4 = t$ , where $s$ and $t$ can be any real numbers 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.
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?
Answer: For all real values of $h$ and $k$ . 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.
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?
Answer: $x_1$ , $x_2$ , and $x_4$ are basic variables, and their values are determined once a value for the free variable $x_3$ is chosen. 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.
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?
Answer: $\begin{pmatrix} -1 \\ -2 \\ -4 \\ -1 \end{pmatrix}$ 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.
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?
Answer: The RREF of the coefficient matrix has exactly two pivot positions. 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.
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?
Answer: No such constraint exists since $x_2 = t$ takes all real values with equal probability 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 $A\mathbf{x} = \mathbf{b}$ is $\mathbf{x} = \mathbf{x}_p + \mathbf{x}_h$ , 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.
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?
Answer: No values of $a$ and $b$ work since we have 3 variables but only 2 pivot positions 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.
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?
Answer: $\begin{pmatrix} 1 & 2 & 0 & 0 & | & 4 \\ 0 & 0 & 1 & -3 & | & -1 \\ 0 & 0 & 0 & 0 & | & 0 \end{pmatrix}$ 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.
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?
Answer: $k = -2$ 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 $k=2$ or $k=-2$ . 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.
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?
Answer: 4 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.
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?
Answer: A line through the origin since one parameter becomes dependent on the other 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.
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 $\vec{x} = \vec{p} + s\vec{u} + t\vec{v}$ ?
Answer: $\vec{x} = \begin{pmatrix} 5 \\ 0 \\ -1 \\ 0 \end{pmatrix} + s \begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + t \begin{pmatrix} 3 \\ 0 \\ -4 \\ 1 \end{pmatrix}$ 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: $\vec{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} 5 - 2s + 3t \\ s \\ -1 - 4t \\ t \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \\ -1 \\ 0 \end{pmatrix} + s \begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + t \begin{pmatrix} 3 \\ 0 \\ -4 \\ 1 \end{pmatrix}$ 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.
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?
Answer: The spanning set requires exactly 3 linearly independent vectors since nullity(A) = 3 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.
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?
Answer: The matrix-vector product $\mathbf{A} \begin{pmatrix} 4 \\ 0 \\ 2 \end{pmatrix}$ results in the zero vector. 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.
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 $\mathbf{A}$ ?
Answer: 3 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.
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?
Answer: Infinitely many solutions exist since the constraint defines a line in the solution plane 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.
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 $t$ ?
Answer: $\vec{x} = \begin{pmatrix} 5 \\ 0 \\ 0 \end{pmatrix} + s \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} + t \begin{pmatrix} -2 \\ 0 \\ 1 \end{pmatrix}$ 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: $\vec{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 5 + 3s - 2t \\ s \\ t \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \\ 0 \end{pmatrix} + s \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} + t \begin{pmatrix} -2 \\ 0 \\ 1 \end{pmatrix}$ 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.
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 $x_1$ ?
Answer: The coefficient of $t$ in $x_1$ is $-2$ since $x_1 = 3 - 2x_3$ 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 $t$ is $-2$ . 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.
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?
Answer: The solution set becomes a line since one parameter becomes fixed while the other remains free 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 $s$ .

Start this quiz