Linear Algebra Quiz: Rref And Solution Interpretation
20 questions · exam conditions
0:00
Rref And Solution InterpretationQuestion 1 of 20

Let AA be an n×nn \times n matrix. The augmented matrix [A∣In][A | I_n] is row-reduced to [R∣B][R | B], where RR is the reduced row echelon form of AA. If the system Ax⃗=b⃗A\vec{x}=\vec{b} has a unique solution for every b⃗∈Rn\vec{b} \in \mathbb{R}^n, what must be true about RR and BB?

RR is the zero matrix and B=A−1B = A^{-1}.
R=InR = I_n and B=InB = I_n.
RR has at least one row of zeros and BB is not unique.
R=InR = I_n and B=A−1B = A^{-1}.
← Back to quizzes

Linear Algebra Quiz

Linear Algebra Quiz: Rref And Solution Interpretation

Practice Rref And Solution Interpretation 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 Rref And Solution Interpretation, 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

Let AA be an n×nn \times n matrix. The augmented matrix [A∣In][A | I_n] is row-reduced to [R∣B][R | B], where RR is the reduced row echelon form of AA. If the system Ax⃗=b⃗A\vec{x}=\vec{b} has a unique solution for every b⃗∈Rn\vec{b} \in \mathbb{R}^n, what must be true about RR and BB?

  1. RR is the zero matrix and B=A−1B = A^{-1}.
  2. R=InR = I_n and B=InB = I_n.
  3. RR has at least one row of zeros and BB is not unique.
  4. R=InR = I_n and B=A−1B = A^{-1}. (correct answer)
Explanation: This question tests your understanding of the relationship between matrix invertibility, row reduction, and unique solutions to linear systems. When a system Ax⃗=b⃗A\vec{x} = \vec{b} has a unique solution for every b⃗∈Rn\vec{b} \in \mathbb{R}^n, this means AA is invertible. An n×nn \times n matrix is invertible if and only if its reduced row echelon form is the identity matrix InI_n. Here's why: If Ax⃗=b⃗A\vec{x} = \vec{b} has a unique solution for every b⃗\vec{b}, then AA must have full rank (rank nn). This means when you row-reduce AA, you get InI_n as the reduced row echelon form, so R=InR = I_n. Now, when you perform the same row operations on the augmented matrix [A∣In][A | I_n], whatever transforms AA into InI_n will transform InI_n into A−1A^{-1}. This is exactly how we compute matrix inverses! So B=A−1B = A^{-1}. Looking at the wrong answers: Choice A incorrectly states RR is the zero matrix, which would mean AA has rank 0 and the system has no solutions for most b⃗\vec{b}. Choice B claims B=InB = I_n, but this would only happen if A=InA = I_n originally. Choice C suggests RR 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 b⃗\vec{b}" is equivalent to "AA is invertible," and row-reducing [A∣In][A | I_n] gives you [In∣A−1][I_n | A^{-1}] when AA is invertible.

Question 2

An augmented matrix is row-reduced to the form shown below, where kk is a real number. For which value of kk will the corresponding linear system be inconsistent?

(103∣501−2∣400k2−4∣k−2)\begin{pmatrix} 1 & 0 & 3 & | & 5 \\ 0 & 1 & -2 & | & 4 \\ 0 & 0 & k^2 - 4 & | & k - 2 \end{pmatrix}
  1. k=2k = 2
  2. k=−2k = -2 (correct answer)
  3. k=±2k = \pm 2
  4. k=4k = 4
Explanation: A linear system is inconsistent if its augmented matrix has a row of the form [0 0 ... 0 ∣ c][0\ 0\ ...\ 0\ |\ c] where c≠0c \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 k2−4k^2 - 4 equals zero when k=2k=2 or k=−2k=-2. The expression k−2k-2 is the constant on the right. We need k2−4=0k^2 - 4 = 0 AND k−2≠0k - 2 \neq 0. If k=2k=2, the third row becomes [0 0 0 ∣ 0][0\ 0\ 0\ |\ 0], which indicates infinitely many solutions (a dependent system), not an inconsistent one. If k=−2k=-2, the third row becomes [0 0 0 ∣ −4][0\ 0\ 0\ |\ -4], which is a contradiction (0=−40=-4). Therefore, the system is inconsistent when k=−2k=-2.

Question 3

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

(10−2∣3015∣−1000∣0)\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?

  1. A single point in R3\mathbb{R}^3
  2. A line in R3\mathbb{R}^3 (correct answer)
  3. A plane in R3\mathbb{R}^3
  4. The empty set (no solution)
Explanation: The given RREF corresponds to the system x1−2x3=3x_1 - 2x_3 = 3 and x2+5x3=−1x_2 + 5x_3 = -1. The third row, 0=00=0, indicates the system is consistent. The third column is not a pivot column, so x3x_3 is a free variable. Let x3=tx_3 = t. The solution can be written parametrically as x1=3+2tx_1 = 3 + 2t and x2=−1−5tx_2 = -1 - 5t. In vector form, the solution is x⃗=(3−10)+t(2−51)\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 R3\mathbb{R}^3 passing through the point (3,−1,0)(3, -1, 0) with direction vector (2,−5,1)(2, -5, 1).

Question 4

Let RR be the reduced row echelon form of a 3×43 \times 4 matrix AA. If the system Ax⃗=b⃗A\vec{x} = \vec{b} has a solution for all b⃗\vec{b} in R3\mathbb{R}^3, which of the following statements about RR must be true?

  1. RR has a column of zeros.
  2. RR has a row of zeros.
  3. RR has exactly three pivot columns. (correct answer)
  4. The last column of RR must be a pivot column.
Explanation: The statement that the system Ax⃗=b⃗A\vec{x} = \vec{b} has a solution for all b⃗\vec{b} in R3\mathbb{R}^3 means that the columns of AA span R3\mathbb{R}^3. This is equivalent to stating that the matrix AA (and its RREF, RR) must have a pivot position in every row. Since AA has 3 rows, its RREF RR must have 3 pivots. A pivot must be in a distinct column, so RR 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.

Question 5

A consistent system of linear equations has a solution set described by x1=5−2tx_1 = 5 - 2t, x2=tx_2 = t, x3=−1x_3 = -1, where tt is any real number. Which of the following could be the reduced row echelon form of its augmented matrix?

  1. (120∣5001∣−1)\begin{pmatrix} 1 & 2 & 0 & | & 5 \\ 0 & 0 & 1 & | & -1 \end{pmatrix} (correct answer)
  2. (1−20∣5001∣−1)\begin{pmatrix} 1 & -2 & 0 & | & 5 \\ 0 & 0 & 1 & | & -1 \end{pmatrix}
  3. (100∣5010∣0001∣−1)\begin{pmatrix} 1 & 0 & 0 & | & 5 \\ 0 & 1 & 0 & | & 0 \\ 0 & 0 & 1 & | & -1 \end{pmatrix}
  4. (102∣3010∣−1)\begin{pmatrix} 1 & 0 & 2 & | & 3 \\ 0 & 1 & 0 & | & -1 \end{pmatrix}
Explanation: From the parametric solution, we can derive the system's equations. x2=tx_2=t indicates x2x_2 is the free variable. From x1=5−2tx_1 = 5 - 2t, we get x1=5−2x2x_1 = 5 - 2x_2, which rearranges to x1+2x2=5x_1 + 2x_2 = 5. The equation for x3x_3 is simply x3=−1x_3 = -1. These two equations, x1+2x2=5x_1 + 2x_2 = 5 and x3=−1x_3 = -1, define the relationships between the variables. In RREF, the first equation corresponds to the row [1 2 0 ∣ 5][1\ 2\ 0\ |\ 5] and the second corresponds to [0 0 1 ∣ −1][0\ 0\ 1\ |\ -1]. Combining these gives the correct matrix.

Question 6

Which of the following statements about the reduced row echelon form (RREF) of a matrix is always true?

  1. A matrix can have multiple RREFs depending on the sequence of row operations used.
  2. If the RREF of a square matrix has a row of zeros, the matrix is invertible.
  3. The RREF of a matrix AA is the same as the RREF of its transpose ATA^T.
  4. Every matrix has a unique reduced row echelon form. (correct answer)
Explanation: 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 rref(A)\text{rref}(A) and rref(AT)\text{rref}(A^T).

Question 7

Consider the augmented matrix MM 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?

  1. x2x_2 and x4x_4 (correct answer)
  2. x1x_1, x3x_3, and x5x_5
  3. x4x_4 and x6x_6
  4. x2x_2, x4x_4, and x6x_6
Explanation: 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 x1x_1), column 3 (for x3x_3), and column 5 (for x5x_5). These are the pivot columns. The columns that do not contain a pivot are columns 2 and 4, which correspond to variables x2x_2 and x4x_4. Therefore, x2x_2 and x4x_4 are the free variables. Note that there are only 5 variables, not 6.

Question 8

The augmented matrix of a linear system has been reduced to the following form:

(1−300∣20010∣−10001∣5)\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?

  1. The system has a unique solution where the free variable x2x_2 must be zero.
  2. The solution set is a plane in R4\mathbb{R}^4 because there are three pivot variables.
  3. The solution set is a line in R4\mathbb{R}^4 where x3x_3 and x4x_4 are fixed values. (correct answer)
  4. The system is inconsistent because the second column contains no pivot.
Explanation: The system is consistent. The pivot variables are x1x_1, x3x_3, and x4x_4. The variable x2x_2 corresponds to a non-pivot column, so it is a free variable. Let x2=tx_2 = t. The equations are x1−3x2=2x_1 - 3x_2 = 2, x3=−1x_3 = -1, and x4=5x_4 = 5. The general solution is x1=2+3tx_1 = 2 + 3t, x2=tx_2 = t, x3=−1x_3 = -1, x4=5x_4 = 5. This can be written in vector form as x⃗=(2,0,−1,5)+t(3,1,0,0)\vec{x} = (2, 0, -1, 5) + t(3, 1, 0, 0). This is the parametric equation of a line in R4\mathbb{R}^4. It is a line, not a plane, because there is only one free parameter.

Question 9

A student row-reduces the augmented matrix for a system Ax⃗=b⃗A\vec{x} = \vec{b} to the form:

(120∣3001∣4000∣0)\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 x1=3x_1=3, x2=0x_2=0, and x3=4x_3=4. What is the error in the student's reasoning?

  1. The student should have concluded there is no solution because of the row of zeros.
  2. The student incorrectly assumed that the variable x2x_2 must be zero; it is a free variable, leading to infinite solutions. (correct answer)
  3. The student did not complete the row reduction; the 2 in the first row should be a 0.
  4. The student misinterpreted the constant terms; the solution should be x1=−3x_1=-3 and x3=−4x_3=-4.
Explanation: The matrix is in reduced row echelon form. The pivot columns are the first and third columns, corresponding to variables x1x_1 and x3x_3. The second column, corresponding to x2x_2, is not a pivot column, which means x2x_2 is a free variable. The system of equations is x1+2x2=3x_1 + 2x_2 = 3 and x3=4x_3 = 4. The student's error was to assume the free variable x2x_2 must be 0. Since x2x_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.

Question 10

Consider the system of linear equations represented by the augmented matrix:

(112∣123a∣3123∣b)\begin{pmatrix} 1 & 1 & 2 & | & 1 \\ 2 & 3 & a & | & 3 \\ 1 & 2 & 3 & | & b \end{pmatrix}

For which values of aa and bb will the system have infinitely many solutions?

  1. a=5a = 5 and b≠2b \neq 2
  2. a≠5a \neq 5 and bb is any real number
  3. a=5a = 5 and b=2b = 2 (correct answer)
  4. a=4a = 4 and b=1b = 1
Explanation: To find the conditions on aa and bb, we perform row reduction. R2→R2−2R1R_2 \to R_2 - 2R_1 and R3→R3−R1R_3 \to R_3 - R_1 gives: (112∣101a−4∣1011∣b−1)\begin{pmatrix} 1 & 1 & 2 & | & 1 \\ 0 & 1 & a-4 & | & 1 \\ 0 & 1 & 1 & | & b-1 \end{pmatrix} Next, R3→R3−R2R_3 \to R_3 - R_2 gives: (112∣101a−4∣1001−(a−4)∣(b−1)−1)=(112∣101a−4∣1005−a∣b−2)\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][0\ 0\ 0\ |\ 0]. This requires both 5−a=05-a = 0 and b−2=0b-2 = 0. Solving these gives a=5a=5 and b=2b=2.

Question 11

Let v⃗1=(10−1)\vec{v}_1 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, v⃗2=(210)\vec{v}_2 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}, v⃗3=(0−1−2)\vec{v}_3 = \begin{pmatrix} 0 \\ -1 \\ -2 \end{pmatrix}. To determine if these vectors span R3\mathbb{R}^3, a matrix AA is formed with these vectors as columns. The reduced row echelon form of AA is found to be:

What does this result imply about the span of v⃗1,v⃗2,v⃗3{\vec{v}_1, \vec{v}_2, \vec{v}_3}?

  1. The vectors span R3\mathbb{R}^3 because the system Ax⃗=b⃗A\vec{x} = \vec{b} is always consistent.
  2. The vectors do not span R3\mathbb{R}^3 because the vectors are linearly independent.
  3. The vectors span R3\mathbb{R}^3 because the RREF shows the vectors are linearly dependent.
  4. The vectors do not span R3\mathbb{R}^3 because the RREF has a row of zeros. (correct answer)
Explanation: 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 R3\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 AA. 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 R3\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 Ax⃗=b⃗A\vec{x} = \vec{b} is always consistent - but that row of zeros means some vectors b⃗\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×column 1−1×column 22 \times \text{column 1} - 1 \times \text{column 2}), and that's precisely why they don't span R3\mathbb{R}^3. Choice C makes two errors: claiming they span R3\mathbb{R}^3 and suggesting linear dependence would somehow help with spanning. Remember this pattern: for nn vectors to span Rn\mathbb{R}^n, you need exactly nn pivots in the RREF. Any row of zeros means insufficient rank and incomplete spanning.

Question 12

Consider the augmented matrix [12−1∣4001∣k000∣2k−1]\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 kk does the corresponding system have exactly one solution?

  1. k=12k = \frac{1}{2} only, because this makes the last equation 0=00 = 0 and eliminates the contradiction
  2. All real values of kk except k=12k = \frac{1}{2}, since the system always has a free variable x2x_2
  3. No values of kk, because the system always has infinitely many solutions due to the free variable x2x_2 (correct answer)
  4. k=0k = 0 only, because this creates the maximum number of pivot positions in the coefficient matrix
Explanation: The RREF shows pivot columns in positions 1 and 3, making x2x_2 a free variable regardless of kk. When k=12k = \frac{1}{2}, the last row becomes 0∣00|0 and the system is consistent with infinitely many solutions. When k≠12k \neq \frac{1}{2}, the last row becomes 0∣nonzero0|\text{nonzero} and the system is inconsistent. The system never has exactly one solution because x2x_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.

Question 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}

  1. (x1,x2,x3,x4)=(5,1,−2,2)(x_1, x_2, x_3, x_4) = (5, 1, -2, 2) because we substitute the free variable directly into the constant terms
  2. (x1,x2,x3,x4)=(1,7,−10,2)(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 (correct answer)
  3. (x1,x2,x3,x4)=(9,−5,6,2)(x_1, x_2, x_3, x_4) = (9, -5, 6, 2) because we must add the free variable contribution to each constant term
  4. (x1,x2,x3,x4)=(3,4,2,2)(x_1, x_2, x_3, x_4) = (3, 4, 2, 2) because the free variable value modifies the original constant vector uniformly
Explanation: From the RREF, x4x_4 is the free variable. Setting x4=2x_4 = 2: x1=5−2x4=5−2(2)=1x_1 = 5 - 2x_4 = 5 - 2(2) = 1, x2=1−(−3)x4=1+3(2)=7x_2 = 1 - (-3)x_4 = 1 + 3(2) = 7, x3=−2−4x4=−2−4(2)=−10x_3 = -2 - 4x_4 = -2 - 4(2) = -10. So the solution is (1,7,−10,2)(1, 7, -10, 2). Choice A ignores the coefficients of x4x_4 in the equations. Choice C incorrectly adds instead of using the proper relationships. Choice D uses an incorrect computation method.

Question 14

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}

  1. The system becomes inconsistent because the last row represents 0=10 = 1, which is impossible to satisfy (correct answer)
  2. The system still has infinitely many solutions, but the particular solution vector changes while keeping the same free variable
  3. The system becomes consistent with a unique solution because the additional constraint eliminates the free variable
  4. The system becomes overdetermined but still consistent, requiring an additional parameter in the general solution
Explanation: Changing the last entry to 1 makes the last row [0 0 0 0 ∣ 1][0 \ 0 \ 0 \ 0 \ | \ 1], which represents the equation 0x1+0x2+0x3+0x4=10x_1 + 0x_2 + 0x_3 + 0x_4 = 1, or 0=10 = 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.

Question 15

Consider a consistent system whose RREF augmented matrix has the form [1a0b∣c001d∣e0000∣0]\begin{bmatrix} 1 & a & 0 & b & | & c \\ 0 & 0 & 1 & d & | & e \\ 0 & 0 & 0 & 0 & | & 0 \end{bmatrix} where a,b,c,d,ea, b, c, d, e are constants. How many parameters are needed to express the general solution?

  1. One parameter, because there are 4 variables and 3 constraints, leaving 1 degree of freedom
  2. Three parameters, because the last row of zeros indicates an additional degree of freedom beyond the free variables
  3. Two parameters, because variables x2x_2 and x4x_4 correspond to non-pivot columns and are free (correct answer)
  4. No parameters, because the system has a unique solution when it's consistent and in RREF
Explanation: When you encounter a system in reduced row echelon form (RREF), the key to finding the number of parameters is identifying which variables are free variables. Free variables correspond to columns that don't contain pivot positions (leading 1's). Looking at this RREF matrix, you can see that the pivot columns are columns 1 and 3 (containing the leading 1's). This means variables x1x_1 and x3x_3 are basic variables, while x2x_2 and x4x_4 are free variables since columns 2 and 4 have no pivots. Each free variable becomes a parameter in your general solution, so you need exactly two parameters to express all solutions. Option A incorrectly applies a simple "variables minus equations" formula (4 - 3 = 1), but this doesn't account for the fact that one row is all zeros, meaning you effectively have only 2 constraints, not 3. The correct count is 4 variables minus 2 constraints equals 2 free variables. Option B misunderstands what the zero row represents. The bottom row of zeros simply indicates that the original system had a redundant equation or that row operations eliminated an equation. It doesn't create additional degrees of freedom. Option D confuses this situation with systems that have unique solutions. Unique solutions only occur when the number of pivot columns equals the number of variables, which isn't the case here. Study tip: Always count pivot columns in RREF to find basic variables. The remaining variables are free, and each free variable contributes one parameter to your general solution.

Question 16

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}

  1. The work is incorrect because the second row should be divided by 2 to make the leading entry equal to 1 (correct answer)
  2. The work is correct since all entries below and above pivot positions are zero, satisfying RREF requirements
  3. The work is correct but incomplete since the student should continue reducing to eliminate the 2 in position (1,3)
  4. The work is incorrect because the second and third rows are proportional, indicating an error in the row operations
Explanation: 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.

Question 17

Two students obtain different RREF forms for the same augmented matrix: Student A gets [102∣301−1∣4000∣0]\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}

  1. Both forms are correct since RREF is not unique when there are free variables in the system
  2. Student A's form is correct because the pivot columns should be the first kk columns where kk is the rank
  3. Student B's form is incorrect because column 2 cannot be skipped when column 3 is a pivot column
  4. One student made an error because RREF is unique, though both represent systems with the same solution set (correct answer)
Explanation: 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.

Question 18

A homogeneous system Ax=0A\mathbf{x} = \mathbf{0} has coefficient matrix AA that reduces to RREF with 2 pivot columns. If AA is a 3×53 \times 5 matrix, which statement about the null space of AA is correct?

  1. The null space has dimension 2 and is spanned by 2 linearly independent vectors corresponding to the non-pivot columns
  2. The null space has dimension 5 and includes all vectors in R5\mathbb{R}^5 since there are more variables than constraints
  3. The null space has dimension 1 and contains only the trivial solution since the system is homogeneous
  4. The null space has dimension 3 and is spanned by 3 linearly independent vectors corresponding to the free variables (correct answer)
Explanation: 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 AA is 3×53 \times 5 with 2 pivot columns, you have 5−2=35 - 2 = 3 free variables. By the rank-nullity theorem, dim(null space)=n−rank(A)=5−2=3\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 R5\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.

Question 19

A system of linear equations is reduced to the following RREF matrix: [1030701−2040001−100000]\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 x1,x2,x3,x4x_1, x_2, x_3, x_4, which statement correctly describes the solution set?

  1. The system has a unique solution with x3=0x_3 = 0 and all other variables determined uniquely
  2. The system has infinitely many solutions with x3x_3 as a free variable and x1=7−3t,x2=4+2t,x4=−1x_1 = 7 - 3t, x_2 = 4 + 2t, x_4 = -1 where tt is arbitrary (correct answer)
  3. The system has no solution because the last row represents 0=00 = 0, which is a contradiction
  4. The system has infinitely many solutions with both x3x_3 and x4x_4 as free variables since there are fewer pivot columns than variables
Explanation: In RREF, the pivot columns are 1, 2, and 4, making x1,x2,x4x_1, x_2, x_4 basic variables and x3x_3 a free variable. Setting x3=tx_3 = t, we get x1=7−3tx_1 = 7 - 3t, x2=4+2tx_2 = 4 + 2t, and x4=−1x_4 = -1. Choice A is wrong because x3x_3 is free, not fixed at 0. Choice C misinterprets the zero row (0=00=0 means consistent, not contradictory). Choice D incorrectly identifies x4x_4 as free when column 4 is a pivot column.

Question 20

Let AA be a 4×54 \times 5 matrix. What can be definitively concluded about the number of solutions to the homogeneous system Ax⃗=0⃗A\vec{x} = \vec{0}?

  1. The system must have a unique solution, the trivial solution.
  2. The system must have infinitely many solutions. (correct answer)
  3. The system must be inconsistent.
  4. The system has either a unique solution or infinitely many solutions.
Explanation: The system Ax⃗=0⃗A\vec{x} = \vec{0} is a homogeneous system with 5 variables and 4 equations. A homogeneous system is always consistent because the trivial solution x⃗=0⃗\vec{x} = \vec{0} is always a solution. The number of free variables is given by (number of variables) - rank(A). Since AA is a 4×54 \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=15 - 4 = 1. Because there is at least one free variable, the system must have infinitely many solutions.