RREF & Solution Interpretation - Linear Algebra

1. Consider the augmented matrix $M$ for a linear system, where its row echelon form (not reduced) is given below. Without performing further row operations, which variables can be identified as free variables? $M_{REF} = \begin{pmatrix} 1 & 2 & 0 & 3 & 5 & | & 6 \\ 0 & 0 & 1 & 4 & -1 & | & 2 \\ 0 & 0 & 0 & 0 & 1 & | & 7 \\ 0 & 0 & 0 & 0 & 0 & | & 0 \end{pmatrix}$
   Answer: $x_2$ and $x_4$ Free variables correspond to the columns of the coefficient matrix that do not contain a pivot position (a leading 1 in a row of the REF). In the given row echelon form, the leading entries are in column 1 (for $x_1$), column 3 (for $x_3$), and column 5 (for $x_5$). These are the pivot columns. The columns that do not contain a pivot are columns 2 and 4, which correspond to variables $x_2$ and $x_4$. Therefore, $x_2$ and $x_4$ are the free variables. Note that there are only 5 variables, not 6.
2. A linear system has the RREF augmented matrix $\begin{bmatrix} 1 & 0 & 0 & 3 & | & -2 \\ 0 & 1 & 0 & -1 & | & 5 \\ 0 & 0 & 1 & 2 & | & 0 \\ 0 & 0 & 0 & 0 & | & 0 \end{bmatrix}$. If this system is modified by changing the last entry from 0 to 1, what happens to the solution set?
   Answer: The system becomes inconsistent because the last row represents $0 = 1$, which is impossible to satisfy Changing the last entry to 1 makes the last row $[0 \ 0 \ 0 \ 0 \ | \ 1]$, which represents the equation $0x_1 + 0x_2 + 0x_3 + 0x_4 = 1$, or $0 = 1$. This is a contradiction, making the system inconsistent (no solution). Choice B incorrectly assumes the system remains consistent. Choice C misunderstands how inconsistency works. Choice D incorrectly suggests the system is still consistent and introduces meaningless terminology about additional parameters.
3. A student row-reduces the augmented matrix for a system $A\vec{x} = \vec{b}$ to the form: $\begin{pmatrix} 1 & 2 & 0 & | & 3 \\ 0 & 0 & 1 & | & 4 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$ The student concludes the system has a unique solution given by $x_1=3$, $x_2=0$, and $x_3=4$. What is the error in the student's reasoning?
   Answer: The student incorrectly assumed that the variable $x_2$ must be zero; it is a free variable, leading to infinite solutions. The matrix is in reduced row echelon form. The pivot columns are the first and third columns, corresponding to variables $x_1$ and $x_3$. The second column, corresponding to $x_2$, is not a pivot column, which means $x_2$ is a free variable. The system of equations is $x_1 + 2x_2 = 3$ and $x_3 = 4$. The student's error was to assume the free variable $x_2$ must be 0. Since $x_2$ can be any real number, there are infinitely many solutions, not a unique one. The row of zeros indicates a dependent system, not an inconsistent one.
4. A consistent system of linear equations has a solution set described by $x_1 = 5 - 2t$, $x_2 = t$, $x_3 = -1$, where $t$ is any real number. Which of the following could be the reduced row echelon form of its augmented matrix?
   Answer: $\begin{pmatrix} 1 & 2 & 0 & | & 5 \\ 0 & 0 & 1 & | & -1 \end{pmatrix}$ From the parametric solution, we can derive the system's equations. $x_2=t$ indicates $x_2$ is the free variable. From $x_1 = 5 - 2t$, we get $x_1 = 5 - 2x_2$, which rearranges to $x_1 + 2x_2 = 5$. The equation for $x_3$ is simply $x_3 = -1$. These two equations, $x_1 + 2x_2 = 5$ and $x_3 = -1$, define the relationships between the variables. In RREF, the first equation corresponds to the row $[1\ 2\ 0\ |\ 5]$ and the second corresponds to $[0\ 0\ 1\ |\ -1]$. Combining these gives the correct matrix.
5. The reduced row echelon form of the augmented matrix for a system of three linear equations in three variables is given by: $\begin{pmatrix} 1 & 0 & -2 & | & 3 \\ 0 & 1 & 5 & | & -1 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$ What is the geometric interpretation of the solution set?
   Answer: A line in $\mathbb{R}^3$ The given RREF corresponds to the system $x_1 - 2x_3 = 3$ and $x_2 + 5x_3 = -1$. The third row, $0=0$, indicates the system is consistent. The third column is not a pivot column, so $x_3$ is a free variable. Let $x_3 = t$. The solution can be written parametrically as $x_1 = 3 + 2t$ and $x_2 = -1 - 5t$. In vector form, the solution is $\vec{x} = \begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix} + t \begin{pmatrix} 2 \\ -5 \\ 1 \end{pmatrix}$. This is the vector equation of a line in $\mathbb{R}^3$ passing through the point $(3, -1, 0)$ with direction vector $(2, -5, 1)$.
6. Consider the system of linear equations represented by the augmented matrix: $\begin{pmatrix} 1 & 1 & 2 & | & 1 \\ 2 & 3 & a & | & 3 \\ 1 & 2 & 3 & | & b \end{pmatrix}$ For which values of $a$ and $b$ will the system have infinitely many solutions?
   Answer: $a = 5$ and $b = 2$ To find the conditions on $a$ and $b$, we perform row reduction. $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - R_1$ gives: $\begin{pmatrix} 1 & 1 & 2 & | & 1 \\ 0 & 1 & a-4 & | & 1 \\ 0 & 1 & 1 & | & b-1 \end{pmatrix}$ Next, $R_3 \to R_3 - R_2$ gives: $\begin{pmatrix} 1 & 1 & 2 & | & 1 \\ 0 & 1 & a-4 & | & 1 \\ 0 & 0 & 1-(a-4) & | & (b-1)-1 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 2 & | & 1 \\ 0 & 1 & a-4 & | & 1 \\ 0 & 0 & 5-a & | & b-2 \end{pmatrix}$ For the system to have infinitely many solutions, the last row must be $[0\ 0\ 0\ |\ 0]$. This requires both $5-a = 0$ and $b-2 = 0$. Solving these gives $a=5$ and $b=2$.
7. Consider the augmented matrix $\begin{bmatrix} 1 & 2 & -1 & | & 4 \\ 0 & 0 & 1 & | & k \\ 0 & 0 & 0 & | & 2k-1 \end{bmatrix}$ which is already in RREF. For which value(s) of $k$ does the corresponding system have exactly one solution?
   Answer: No values of $k$, because the system always has infinitely many solutions due to the free variable $x_2$ The RREF shows pivot columns in positions 1 and 3, making $x_2$ a free variable regardless of $k$. When $k = \frac{1}{2}$, the last row becomes $0|0$ and the system is consistent with infinitely many solutions. When $k \neq \frac{1}{2}$, the last row becomes $0|\text{nonzero}$ and the system is inconsistent. The system never has exactly one solution because $x_2$ is always free when consistent. Choice A and B incorrectly assume one solution is possible. Choice D picks an arbitrary value without proper justification.
8. Which of the following statements about the reduced row echelon form (RREF) of a matrix is always true?
   Answer: Every matrix has a unique reduced row echelon form. A fundamental theorem in linear algebra states that every matrix is row equivalent to one and only one reduced row echelon matrix. This means the RREF of any given matrix is unique, regardless of the specific sequence of elementary row operations used to obtain it. Choice A is true for row echelon form (REF), but not for RREF. Choice B is false; a row of zeros in the RREF of a square matrix implies it is not invertible. Choice C is false; transposition and row reduction do not commute, and there is no general relationship between $\text{rref}(A)$ and $\text{rref}(A^T)$.
9. Let $\vec{v}_1 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$, $\vec{v}_2 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$, $\vec{v}_3 = \begin{pmatrix} 0 \\ -1 \\ -2 \end{pmatrix}$. To determine if these vectors span $\mathbb{R}^3$, a matrix $A$ is formed with these vectors as columns. The reduced row echelon form of $A$ is found to be: $\text{rref}(A) = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}$ What does this result imply about the span of ${\vec{v}_1, \vec{v}_2, \vec{v}_3}$?
   Answer: The vectors do not span $\mathbb{R}^3$ because the RREF has a row of zeros. When you're asked whether vectors span a particular vector space, you're essentially asking: "Can these vectors be combined to create any vector in that space?" The key insight is that vectors span $\mathbb{R}^3$ if and only if the matrix formed by these vectors has full rank (rank 3). The reduced row echelon form reveals the rank of matrix $A$. Since the RREF has only two pivot positions (leading 1's in rows 1 and 2) and a complete row of zeros in the third row, the rank is 2, not 3. This means the three vectors actually span only a 2-dimensional subspace of $\mathbb{R}^3$ - essentially a plane passing through the origin. You cannot reach every point in 3-dimensional space using just these three vectors. Let's examine why the other answers miss the mark. Choice A incorrectly assumes spanning means the system $A\vec{x} = \vec{b}$ is always consistent - but that row of zeros means some vectors $\vec{b}$ will make the system inconsistent. Choice B gets the conclusion right but the reasoning backwards - the vectors are actually linearly dependent (notice column 3 equals $2 \times \text{column 1} - 1 \times \text{column 2}$), and that's precisely why they don't span $\mathbb{R}^3$. Choice C makes two errors: claiming they span $\mathbb{R}^3$ and suggesting linear dependence would somehow help with spanning. Remember this pattern: for $n$ vectors to span $\mathbb{R}^n$, you need exactly $n$ pivots in the RREF. Any row of zeros means insufficient rank and incomplete spanning.
10. Let $R$ be the reduced row echelon form of a $3 \times 4$ matrix $A$. If the system $A\vec{x} = \vec{b}$ has a solution for all $\vec{b}$ in $\mathbb{R}^3$, which of the following statements about $R$ must be true?
    Answer: $R$ has exactly three pivot columns. The statement that the system $A\vec{x} = \vec{b}$ has a solution for all $\vec{b}$ in $\mathbb{R}^3$ means that the columns of $A$ span $\mathbb{R}^3$. This is equivalent to stating that the matrix $A$ (and its RREF, $R$) must have a pivot position in every row. Since $A$ has 3 rows, its RREF $R$ must have 3 pivots. A pivot must be in a distinct column, so $R$ must have exactly three pivot columns. A row of zeros would contradict the condition. A column of zeros is possible for the free variable but not required. The pivots can be in any three of the four columns.
11. An augmented matrix is row-reduced to the form shown below, where $k$ is a real number. For which value of $k$ will the corresponding linear system be inconsistent? $\begin{pmatrix} 1 & 0 & 3 & | & 5 \\ 0 & 1 & -2 & | & 4 \\ 0 & 0 & k^2 - 4 & | & k - 2 \end{pmatrix}$
    Answer: $k = -2$ A linear system is inconsistent if its augmented matrix has a row of the form $[0\ 0\ ...\ 0\ |\ c]$ where $c \neq 0$. In the given matrix, this occurs if the third row's left side is zero while the right side is non-zero. The expression $k^2 - 4$ equals zero when $k=2$ or $k=-2$. The expression $k-2$ is the constant on the right. We need $k^2 - 4 = 0$ AND $k - 2 \neq 0$. If $k=2$, the third row becomes $[0\ 0\ 0\ |\ 0]$, which indicates infinitely many solutions (a dependent system), not an inconsistent one. If $k=-2$, the third row becomes $[0\ 0\ 0\ |\ -4]$, which is a contradiction ($0=-4$). Therefore, the system is inconsistent when $k=-2$.
12. A student reduces an augmented matrix and claims the final RREF is $\begin{bmatrix} 1 & 0 & 2 & | & 3 \\ 0 & 2 & -4 & | & 6 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}$. What should you conclude about this student's work?
    Answer: The work is incorrect because the second row should be divided by 2 to make the leading entry equal to 1 When evaluating whether a matrix is in reduced row echelon form (RREF), you need to check three key requirements: leading entries must be 1, all entries above and below leading entries must be zero, and each leading entry must be to the right of the leading entry in the row above it. Looking at this matrix, the first row satisfies RREF requirements with a leading 1 in the first column. However, the second row has a leading entry of 2 in the second column, which violates the fundamental RREF rule that all leading entries must equal 1. To correct this, the student should divide the entire second row by 2, giving $\begin{bmatrix} 0 & 1 & -2 & | & 3 \end{bmatrix}$. Answer A correctly identifies this error. Answer B is wrong because having zeros above and below pivots isn't sufficient—the leading entries themselves must be 1. Answer C misunderstands RREF requirements; the 2 in position (1,3) doesn't need elimination since it's not above or below a leading entry. Answer D incorrectly claims the second and third rows are proportional, but $\begin{bmatrix} 0 & 2 & -4 & | & 6 \end{bmatrix}$ and $\begin{bmatrix} 0 & 0 & 0 & | & 0 \end{bmatrix}$ are not proportional—you cannot multiply the zero row by any constant to get the second row. Study tip: When checking RREF, always verify the "leading 1" requirement first. Many students focus on the zero patterns but forget that pivot entries must actually equal 1, not just be the leftmost non-zero entry in their row.
13. The RREF of the augmented matrix for a system of linear equations is $\begin{bmatrix} 1 & 0 & 0 & 2 & | & 5 \\ 0 & 1 & 0 & -3 & | & 1 \\ 0 & 0 & 1 & 4 & | & -2 \\ 0 & 0 & 0 & 0 & | & 0 \end{bmatrix}$. If we set the free variable equal to 2, what is the resulting particular solution?
    Answer: $(x_1, x_2, x_3, x_4) = (1, 7, -10, 2)$ because each basic variable adjusts according to its coefficient of the free variable From the RREF, $x_4$ is the free variable. Setting $x_4 = 2$: $x_1 = 5 - 2x_4 = 5 - 2(2) = 1$, $x_2 = 1 - (-3)x_4 = 1 + 3(2) = 7$, $x_3 = -2 - 4x_4 = -2 - 4(2) = -10$. So the solution is $(1, 7, -10, 2)$. Choice A ignores the coefficients of $x_4$ in the equations. Choice C incorrectly adds instead of using the proper relationships. Choice D uses an incorrect computation method.
14. A $4 \times 6$ matrix $A$ is row-reduced to RREF and has exactly 3 pivot columns. If the system $A\mathbf{x} = \mathbf{b}$ is consistent, what can be concluded about the solution set?
    Answer: The solution set is a 3-dimensional affine subspace of $\mathbb{R}^6$ translated from the origin by a particular solution With 6 variables and 3 pivot columns, there are $6-3=3$ free variables, making the solution set 3-dimensional. Since the system is $A\mathbf{x} = \mathbf{b}$ (not homogeneous), the solution set is an affine subspace (translated from origin), not a vector subspace through the origin. Choice A incorrectly calls it a subspace rather than affine subspace. Choice C miscalculates the dimension as $6-3-1$ instead of $6-3$. Choice D incorrectly computes the dimension and misunderstands the role of the constraint.
15. A homogeneous system $A\mathbf{x} = \mathbf{0}$ has coefficient matrix $A$ that reduces to RREF with 2 pivot columns. If $A$ is a $3 \times 5$ matrix, which statement about the null space of $A$ is correct?
    Answer: The null space has dimension 3 and is spanned by 3 linearly independent vectors corresponding to the free variables When you encounter questions about homogeneous systems and null spaces, focus on the relationship between pivot columns, free variables, and the dimension of the null space. This connection is fundamental to understanding solution spaces. The key insight is that the dimension of the null space equals the number of free variables, which equals the total number of variables minus the number of pivot columns. Since $A$ is $3 \times 5$ with 2 pivot columns, you have $5 - 2 = 3$ free variables. By the rank-nullity theorem, $\text{dim}(\text{null space}) = n - \text{rank}(A) = 5 - 2 = 3$. The null space is spanned by 3 linearly independent vectors, each corresponding to one of the free variables (the non-pivot columns). Choice A incorrectly states the dimension is 2. This confuses the number of pivot columns with the dimension of the null space—these are different quantities related by subtraction from the total variables. Choice B drastically overestimates the dimension as 5, wrongly claiming the null space includes all of $\mathbb{R}^5$. The null space is a proper subspace when there are pivot columns, not the entire space. Choice C reflects a fundamental misunderstanding, claiming dimension 1 and only the trivial solution. Homogeneous systems always have the trivial solution, but when there are free variables (as here), infinitely many non-trivial solutions exist. Remember this pattern: for any matrix, the dimension of the null space always equals the number of columns minus the number of pivot columns. This gives you the number of linearly independent vectors needed to span the null space.
16. Let $A$ be a $4 \times 5$ matrix. What can be definitively concluded about the number of solutions to the homogeneous system $A\vec{x} = \vec{0}$?
    Answer: The system must have infinitely many solutions. The system $A\vec{x} = \vec{0}$ is a homogeneous system with 5 variables and 4 equations. A homogeneous system is always consistent because the trivial solution $\vec{x} = \vec{0}$ is always a solution. The number of free variables is given by (number of variables) - rank(A). Since $A$ is a $4 \times 5$ matrix, its rank (the number of pivots in its RREF) can be at most 4. Therefore, the number of free variables is at least $5 - 4 = 1$. Because there is at least one free variable, the system must have infinitely many solutions.
17. Two students obtain different RREF forms for the same augmented matrix: Student A gets $\begin{bmatrix} 1 & 0 & 2 & | & 3 \\ 0 & 1 & -1 & | & 4 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}$ and Student B gets $\begin{bmatrix} 1 & 2 & 0 & | & 11 \\ 0 & 0 & 1 & | & 4 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}$. What can be concluded?
    Answer: One student made an error because RREF is unique, though both represent systems with the same solution set The RREF of any matrix is unique. Both forms cannot be correct RREF for the same original matrix, so one student made an error in row operations. However, it's possible both represent equivalent systems (same solution set) if they started from equivalent augmented matrices. Choice A incorrectly states RREF is not unique. Choice B incorrectly assumes pivot columns must be consecutive from the left. Choice C misunderstands that pivot columns can be non-consecutive in RREF.
18. The augmented matrix of a linear system has been reduced to the following form: $\begin{pmatrix} 1 & -3 & 0 & 0 & | & 2 \\ 0 & 0 & 1 & 0 & | & -1 \\ 0 & 0 & 0 & 1 & | & 5 \end{pmatrix}$ Which of the following statements correctly describes the solution set?
    Answer: The solution set is a line in $\mathbb{R}^4$ where $x_3$ and $x_4$ are fixed values. The system is consistent. The pivot variables are $x_1$, $x_3$, and $x_4$. The variable $x_2$ corresponds to a non-pivot column, so it is a free variable. Let $x_2 = t$. The equations are $x_1 - 3x_2 = 2$, $x_3 = -1$, and $x_4 = 5$. The general solution is $x_1 = 2 + 3t$, $x_2 = t$, $x_3 = -1$, $x_4 = 5$. This can be written in vector form as $\vec{x} = (2, 0, -1, 5) + t(3, 1, 0, 0)$. This is the parametric equation of a line in $\mathbb{R}^4$. It is a line, not a plane, because there is only one free parameter.
19. Let $A$ be an $n \times n$ matrix. The augmented matrix $[A | I_n]$ is row-reduced to $[R | B]$, where $R$ is the reduced row echelon form of $A$. If the system $A\vec{x}=\vec{b}$ has a unique solution for every $\vec{b} \in \mathbb{R}^n$, what must be true about $R$ and $B$?
    Answer: $R = I_n$ and $B = A^{-1}$. This question tests your understanding of the relationship between matrix invertibility, row reduction, and unique solutions to linear systems. When a system $A\vec{x} = \vec{b}$ has a unique solution for every $\vec{b} \in \mathbb{R}^n$, this means $A$ is invertible. An $n \times n$ matrix is invertible if and only if its reduced row echelon form is the identity matrix $I_n$. Here's why: If $A\vec{x} = \vec{b}$ has a unique solution for every $\vec{b}$, then $A$ must have full rank (rank $n$). This means when you row-reduce $A$, you get $I_n$ as the reduced row echelon form, so $R = I_n$. Now, when you perform the same row operations on the augmented matrix $[A | I_n]$, whatever transforms $A$ into $I_n$ will transform $I_n$ into $A^{-1}$. This is exactly how we compute matrix inverses! So $B = A^{-1}$. Looking at the wrong answers: Choice A incorrectly states $R$ is the zero matrix, which would mean $A$ has rank 0 and the system has no solutions for most $\vec{b}$. Choice B claims $B = I_n$, but this would only happen if $A = I_n$ originally. Choice C suggests $R$ has zero rows, which again contradicts the uniqueness condition since this would mean infinitely many solutions or no solutions. Study tip: Remember that "unique solution for every $\vec{b}$" is equivalent to "$A$ is invertible," and row-reducing $[A | I_n]$ gives you $[I_n | A^{-1}]$ when $A$ is invertible.
20. A system of linear equations is reduced to the following RREF matrix: $\begin{bmatrix} 1 & 0 & 3 & 0 & 7 \\ 0 & 1 & -2 & 0 & 4 \\ 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$. If the original system had variables $x_1, x_2, x_3, x_4$, which statement correctly describes the solution set?
    Answer: The system has infinitely many solutions with $x_3$ as a free variable and $x_1 = 7 - 3t, x_2 = 4 + 2t, x_4 = -1$ where $t$ is arbitrary In RREF, the pivot columns are 1, 2, and 4, making $x_1, x_2, x_4$ basic variables and $x_3$ a free variable. Setting $x_3 = t$, we get $x_1 = 7 - 3t$, $x_2 = 4 + 2t$, and $x_4 = -1$. Choice A is wrong because $x_3$ is free, not fixed at 0. Choice C misinterprets the zero row ($0=0$ means consistent, not contradictory). Choice D incorrectly identifies $x_4$ as free when column 4 is a pivot column.

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