Linear Algebra Quiz: Common Errors And Pitfalls

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Common Errors And PitfallsQuestion 1 of 18

A data scientist is analyzing a system of linear equations represented by the augmented matrix $[A|\mathbf{b}]$ . After row reduction, they obtain the following reduced row-echelon form:

The variables are $x_1, x_2, x_3, x_4$ . The scientist concludes that the system has a unique solution. Which statement best explains the error in the scientist's conclusion?

The conclusion is incorrect because the row of zeros at the bottom indicates there is no solution to the system.

The conclusion is incorrect because the third column lacks a pivot, indicating a free variable and thus infinitely many solutions.

The conclusion is correct because every non-zero row contains a pivot (a leading 1), which guarantees a unique solution.

The conclusion is incorrect because there are more variables than equations, which always means there are no unique solutions.

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Linear Algebra Quiz

Linear Algebra Quiz: Common Errors And Pitfalls

Practice Common Errors And Pitfalls 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 Common Errors And Pitfalls, 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

A data scientist is analyzing a system of linear equations represented by the augmented matrix $[A|\mathbf{b}]$ . After row reduction, they obtain the following reduced row-echelon form:

The variables are $x_1, x_2, x_3, x_4$ . The scientist concludes that the system has a unique solution. Which statement best explains the error in the scientist's conclusion?

The conclusion is incorrect because the row of zeros at the bottom indicates there is no solution to the system.
The conclusion is incorrect because the third column lacks a pivot, indicating a free variable and thus infinitely many solutions. (correct answer)
The conclusion is correct because every non-zero row contains a pivot (a leading 1), which guarantees a unique solution.
The conclusion is incorrect because there are more variables than equations, which always means there are no unique solutions.

Explanation: The conclusion of a unique solution is incorrect. In the reduced row-echelon form, columns 1, 2, and 4 have pivots. Column 3 does not have a pivot, which means the corresponding variable,

x_3

, is a free variable. Because the system is consistent (no row of the form

[0, 0, 0, 0 | c]

where

c \neq 0

), the presence of a free variable implies that there are infinitely many solutions.

Question 2

An engineer calculates the product of two matrices, $C = AB$ , where $A = \begin{pmatrix} 2 & -1 \\ 3 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 4 \\ -2 & 5 \end{pmatrix}$ . The engineer's result is $C = \begin{pmatrix} 0 & 3 \\ 3 & 12 \end{pmatrix}$ . Which statement correctly identifies the error?

The entry $c_{11}$ is incorrect. The calculation should have been $(2)(1) + (-1)(-2) = 4$ . (correct answer)
The entry $c_{12}$ is incorrect. The calculation should have been $(2)(4) - (-1)(5) = 13$ .
The result is incorrect because the engineer computed $BA$ instead of $AB$ .
The entry $c_{22}$ is incorrect because it should be calculated as $(3)(4) + (0)(5)$ .

Explanation: The entry

c_{11}

is the dot product of the first row of

A

and the first column of

B

. The correct calculation is

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

. The engineer's result for

c_{11}

is 0, which is incorrect. A likely cause is a sign error, such as computing

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

. The other entries in the engineer's result (

c_{12}=3

c_{21}=3

c_{22}=12

) are all calculated correctly.

Question 3

A system of linear equations with variables $x_1, x_2, x_3, x_4$ is reduced to the augmented matrix:

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

A student attempts to write the general solution by setting the pivot variables $x_1$ and $x_3$ as free parameters: $x_1=t, x_3=s$ . They then try to solve for $x_2$ and $x_4$ . Why is this approach fundamentally flawed?

The approach is flawed because the system is inconsistent and has no solution.
This approach is flawed because a system with four variables requires four parameters for its solution.
Free variables must correspond to non-pivot columns; the student incorrectly assigned parameters to basic (pivot) variables. (correct answer)
The approach is correct, but the student should have set $x_1=t$ and $x_2=s$ as the parameters.

Explanation: In a parametric solution, parameters are assigned to the free variables. Free variables correspond to the columns in the coefficient matrix that do not contain a pivot. Here, the pivots are in columns 1 and 3, so

x_1

and

x_3

are basic variables. The non-pivot columns are 2 and 4, so

x_2

and

x_4

are the free variables. The student's approach of assigning parameters to the basic variables is fundamentally incorrect.

Question 4

While trying to find the intersection of three planes, a student sets up a system of equations and arrives at the following augmented matrix after several row operations:

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

The student concludes that an arithmetic error must have occurred because a row of the form $[0, 0, 0 | c]$ was produced. Which statement provides the correct interpretation of this result?

The student is correct; a row like $[0, 0, 0 | 3]$ can only result from a calculation mistake and the reduction should be redone.
The result indicates infinitely many solutions, with one free variable corresponding to the third column.
The result does not imply an arithmetic error; the final row corresponds to the equation $0=3$ , meaning the system is inconsistent. (correct answer)
The result indicates a unique solution exists, but it can only be found by using Cramer's rule, not row reduction.

Explanation: The final row of the augmented matrix,

[0, 0, 0 | 3]

, translates to the equation

0x_1 + 0x_2 + 0x_3 = 3

, which simplifies to

0=3

. This is a contradiction. This is a valid outcome of row reduction and indicates that the original system of equations has no solution. Such a system is called inconsistent. Geometrically, the three planes have no point of common intersection.

Question 5

A student computes the determinant of matrix $B$ using cofactor expansion along the second row.

The student's calculation is: $\det(B) = (2) \det \begin{pmatrix} 0 & 3 \\ 1 & 0 \end{pmatrix} + (-1) \det \begin{pmatrix} 1 & 3 \\ 5 & 0 \end{pmatrix} + (4) \det \begin{pmatrix} 1 & 0 \\ 5 & 1 \end{pmatrix}$ $= 2(0-3) - 1(0-15) + 4(1-0) = -6 + 15 + 4 = 13$ What is the error in the student's calculation?

The signs for the cofactor expansion are incorrect; the checkerboard pattern was not applied. (correct answer)
The minor for the element $b_{22} = -1$ was calculated incorrectly.
The arithmetic in the final step is incorrect; $-6 + 15 + 4$ is not equal to 13.
The student should have expanded along the first row because it contains a zero.

Explanation: The cofactor expansion requires a sign pattern based on the position of the element, given by

(-1)^{i+j}

. For the second row (

i=2

), the signs should alternate starting with a negative:

(b_{21}, b_{22}, b_{23})

should have signs

(-, +, -)

. The student used positive signs for all terms. The correct calculation is

\det(B) = -(2)(-3) + (-1)(-15) - (4)(1) = 6 + 15 - 4 = 17

Question 6

A student is tasked with finding a basis for the column space of matrix $A$ .

The student correctly row reduces $A$ to its reduced row-echelon form, $R$ .

The student states: "The pivot columns in $R$ are columns 1 and 2. Therefore, a basis for the column space of $A$ is $\{ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \}$ ." What is the flaw in the student's reasoning?

The row reduction is incorrect; the RREF of $A$ should be the identity matrix.
While the pivot columns of $R$ identify the correct columns, the basis vectors must be taken from the original matrix $A$ . (correct answer)
The basis for the column space should be the non-zero rows of the RREF matrix $R$ .
The student identified the wrong pivot columns; the pivots should be in columns 1 and 3.

Explanation: When finding a basis for the column space of a matrix, you need to understand the relationship between row operations and column spaces. Row operations preserve the relationships between columns but change the actual column vectors themselves. The student correctly identified that columns 1 and 2 are the pivot columns in the reduced row-echelon form

R

. This tells us that the first two columns of the original matrix

A

are linearly independent and span the column space. However, the student made a critical error by taking the basis vectors from

R

instead of

A

. The correct basis for the column space of

A

should be

\left\{ \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix} \right\}

— the first two columns of the original matrix

A

. Row operations change the column space, so while

R

helps us identify which columns form a basis, we must use the corresponding columns from

A

. Looking at the wrong answers: (A) is incorrect because the row reduction is performed correctly — you can verify this step by step. (C) confuses column space with row space; non-zero rows of the RREF give a basis for the row space, not the column space. (D) is wrong because columns 1 and 2 clearly contain the leading 1's (pivots) in

R

. Key takeaway: Use the RREF to identify pivot columns, but always take your basis vectors from the original matrix. The RREF tells you which columns to choose, not what those basis vectors are.

Question 7

A student is finding the determinant of a $4 \times 4$ matrix $A$ and performs the following row operation:

The student then calculates $\det(B) = 12$ . Based on this, the student concludes that $\det(A) = 12$ . What is the logical error in this conclusion?

The determinant of a matrix does not change when one of its rows is multiplied by a non-zero scalar.
The relationship is $\det(A) = 3 \det(B)$ , so $\det(A) = 36$ .
The student should have performed the operation $R_2 \to R_2 - 2R_1$ , which would not change the determinant.
Scaling a single row by a factor $c$ scales the determinant by $c$ . Therefore, $\det(A) = \det(B)/3 = 4$ . (correct answer)

Explanation: A property of determinants states that if a matrix

B

is obtained from a matrix

A

by multiplying one row by a scalar

c

, then

\det(B) = c \cdot \det(A)

. In this case,

B

was obtained by multiplying row 2 of

A

c=3

. Thus,

\det(B) = 3 \det(A)

. Given that

\det(B) = 12

, we have

12 = 3 \det(A)

, which implies

\det(A) = 4

. The student incorrectly assumed the determinant was unchanged by this type of row operation.

Question 8

A student is asked to find the determinant of $C = 2A$ , where $A$ is a $3 \times 3$ matrix with $\det(A) = 5$ . The student's reasoning is as follows: "Since $C = 2A$ , I can use the property $\det(kA) = k \cdot \det(A)$ . Therefore, $\det(C) = 2 \cdot \det(A) = 2 \cdot 5 = 10$ ." Why is the student's reasoning flawed?

The property $\det(kA) = k \cdot \det(A)$ is only true if $A$ is an invertible matrix.
The student should have used the property $\det(A+B) = \det(A) + \det(B)$ .
The determinant of a matrix cannot be found without knowing the specific entries of the matrix.
The property for an $n \times n$ matrix is $\det(kA) = k^n \cdot \det(A)$ , which was applied incorrectly. (correct answer)

Explanation: The correct property for scalar multiplication of a determinant is

\det(kA) = k^n \det(A)

for an

n \times n

matrix

A

. Since

A

is a

3 \times 3

matrix, the scalar 2 should be raised to the power of 3. The correct calculation is

\det(2A) = 2^3 \det(A) = 8 \cdot 5 = 40

. The student's reasoning is flawed because they used

n=1

instead of

n=3

Question 9

A student tries to find the inverse of $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ using Gauss-Jordan elimination. Their work is shown below. Step 0: $\left( \begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 3 & 4 & 0 & 1 \end{array} \right)$ Step 1: $R_2 \to R_2 - 3R_1$ $\left( \begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & -2 & -3 & 1 \end{array} \right)$ Step 2: $R_2 \to -\frac{1}{2}R_2$ $\left( \begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & 1 & \frac{3}{2} & -\frac{1}{2} \end{array} \right)$ Step 3: $R_1 \to R_1 - 2R_2$ $\left( \begin{array}{cc|cc} 1 & 0 & 2 & -1 \\ 0 & 1 & \frac{3}{2} & -\frac{1}{2} \end{array} \right)$ Where did the student first make an error?

In the calculation for Step 1.
In the calculation for Step 2.
In the calculation for Step 3. (correct answer)
The student made no error; the resulting matrix is the correct inverse.

Explanation: Steps 1 and 2 are calculated correctly. The error occurs in Step 3. The operation is

R_1 \to R_1 - 2R_2

. The new row 1 should be

[1, 2, | 1, 0] - 2[0, 1, | \frac{3}{2}, -\frac{1}{2}]

. The left side is correct:

1-2(0)=1

and

2-2(1)=0

. The right side is incorrect. The calculation should be

1 - 2(\frac{3}{2}) = 1 - 3 = -2

and

0 - 2(-\frac{1}{2}) = 0 - (-1) = 1

. The student's result of

[2, -1]

for these entries is wrong, likely due to sign errors.

Question 10

A student is solving a system of linear equations by row reducing the augmented matrix $A$ . Their first two steps are shown below.

Step 1: $R_3 \to R_3 - 2R_1$

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

Step 2: $R_3 \to R_3 + R_2$

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

What error, if any, did the student make?

The notation for the operation in Step 1 is incorrect.
The calculation in Step 1 contains an arithmetic error.
The calculation in Step 2 contains an arithmetic error. (correct answer)
There are no errors in the student's work shown.

Explanation: The calculation in Step 2 is incorrect. The operation is

R_3 \to R_3 + R_2

. Using the matrices from the end of Step 1 and the beginning of Step 2, this means adding row 2 to row 3:

[0, -1, 3, | -2] + [0, 1, 3, | 5]

. The correct result is

[0, 0, 6, | 3]

. The student incorrectly calculated

-2 + 5 = 2

. Step 1 was calculated correctly:

[2, 3, 1, | 4] - 2[1, 2, -1, | 3] = [0, -1, 3, | -2]

. The notation is standard.

Question 11

A student reduces $\begin{bmatrix} 1 & 2 & 3 & | & 6 \\ 2 & 4 & 6 & | & 12 \\ 1 & 2 & 5 & | & 10 \end{bmatrix}$ and gets $\begin{bmatrix} 1 & 2 & 3 & | & 6 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 2 & | & 4 \end{bmatrix}$ . They conclude the solution is $x_1 = 0$ , $x_2 = 0$ , $x_3 = 2$ . What is the most significant error in this interpretation?

The student failed to express $x_2$ as a free parameter in the general solution
The student made arithmetic errors during the row reduction process
The student didn't complete the back-substitution to find $x_1$ correctly
The student misinterpreted the zero row as meaning $x_2 = 0$ when $x_2$ is actually free (correct answer)

Explanation: From the reduced matrix,

x_3 = 2

from the third row, but

x_2

is a free parameter since it doesn't correspond to a pivot column. The first equation gives

x_1 + 2x_2 + 3(2) = 6

, so

x_1 = -2x_2

. The correct solution is

x_1 = -2t

x_2 = t

x_3 = 2

where

t

is any real number. The student's main error was assuming

x_2 = 0

instead of recognizing it as free.

Question 12

While solving a system, a student applies $R_1 \leftarrow \frac{1}{2}R_1$ to $\begin{bmatrix} 2 & -4 & 6 & | & 8 \\ 1 & -1 & 2 & | & 3 \\ 3 & -5 & 7 & | & 10 \end{bmatrix}$ and gets $\begin{bmatrix} 1 & -2 & 3 & | & 4 \\ 1 & -1 & 2 & | & 3 \\ 3 & -5 & 7 & | & 10 \end{bmatrix}$ . Next, they perform $R_2 \leftarrow R_2 - R_1$ and obtain $$ \begin{bmatrix} 1 & -2 & 3 & | & 4 \ 0 & 1 & -1 & | & -1 \ 3 & -5 & 7 & | & 10 \end{bmatrix}

The row scaling operation in the first step may have introduced computational errors
The elimination in the second step was performed in the wrong direction
The student needs to continue eliminating to complete the row reduction process
The student should verify the arithmetic in $R_2 \leftarrow R_2 - R_1$ before proceeding further (correct answer)

Explanation: When working through row operations in Gaussian elimination, accuracy at each step is crucial because errors compound throughout the process. Let's examine what happened in the student's work. The first operation

R_1 \leftarrow \frac{1}{2}R_1

correctly transforms

[2, -4, 6, |, 8]

into

[1, -2, 3, |, 4]

. However, let's check the second operation

R_2 \leftarrow R_2 - R_1

R_2 - R_1 = [1, -1, 2, |, 3] - [1, -2, 3, |, 4] = [0, 1, -1, |, -1]

The arithmetic here is:

1-1=0

-1-(-2)=1

2-3=-1

, and

3-4=-1

. This matches the student's result, so the calculation is actually correct. Looking at the answer choices: (A) suggests the scaling introduced errors, but

\frac{1}{2} \times [2, -4, 6, 8] = [1, -2, 3, 4]

is correct. (B) claims wrong direction, but eliminating below the pivot (making the entry below the leading 1 into a zero) is the standard approach. (C) suggests continuing the process, but while true eventually, it's not the most pressing concern at this stage. The correct answer is (D). Even though the arithmetic happens to be correct in this case, the student should always verify their row operations before proceeding. One arithmetic mistake early in the process will propagate through all subsequent steps, making the entire solution incorrect. Study tip: Always double-check your arithmetic after each row operation. It's much easier to catch and fix errors immediately than to restart the entire elimination process later.

Question 13

While performing row operations, a student transforms $\begin{bmatrix} 3 & -6 & 9 & | & 12 \\ 1 & -2 & 3 & | & 4 \\ 2 & -4 & 6 & | & 7 \end{bmatrix}$ into $\begin{bmatrix} 1 & -2 & 3 & | & 4 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & | & -1 \end{bmatrix}$ and states that since the last row gives $0 = -1$ , the system has no solution. However, a classmate argues this conclusion is premature. What should the student verify?

Whether the row operations were performed correctly, since the arithmetic may contain errors (correct answer)
Whether the original system was already in reduced form and didn't require row operations
Whether additional row operations could eliminate the inconsistency in the third row
Whether the coefficient matrix has full rank before concluding about consistency

Explanation: The student should verify the row operations. Notice that the first row is exactly 3 times the second row in the coefficient part (3×[1,-2,3] = [3,-6,9]), and the augmented entry follows this pattern (3×4 = 12). However, the third row [2,-4,6] = 2×[1,-2,3], so its augmented entry should be 2×4 = 8 for consistency, not 7. The apparent inconsistency in the reduced form suggests an arithmetic error occurred during row operations rather than true inconsistency in the original system.

Question 14

A student performs the row operation $R_2 \leftarrow R_2 + 3R_1$ on $\begin{bmatrix} 2 & -1 & 4 & | & 5 \\ -6 & 3 & -12 & | & -15 \\ 1 & 0 & 2 & | & 3 \end{bmatrix}$ and obtains $$ \begin{bmatrix} 2 & -1 & 4 & | & 5 \ 0 & 0 & 0 & | & 0 \ 1 & 0 & 2 & | & 3 \end{bmatrix}

Continue row reduction to eliminate the leading coefficient in the third row using the first row (correct answer)
Verify that the second row elimination was correct and examine the relationship between remaining equations
Immediately write the general solution using the two remaining non-zero equations
Check if the zero row indicates that one of the original equations was redundant or inconsistent

Explanation: While the student correctly eliminated the second row, they haven't completed the row reduction process. They should continue by using

R_3 \leftarrow 2R_3 - R_1

to eliminate the leading coefficient in the third row, which would give

[0, 1, 0, 1]

. Only after completing the full reduction can they properly determine the solution structure. The zero row alone doesn't guarantee infinitely many solutions - they need to see the final reduced form.

Question 15

A student attempts to solve $\begin{bmatrix} 2 & 4 & 6 \\ 1 & 2 & 3 \\ 3 & 6 & 9 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 8 \\ 4 \\ 12 \end{bmatrix}$ and after row reduction obtains $\begin{bmatrix} 1 & 2 & 3 & | & 4 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}$ . The student concludes there is exactly one solution: $x = 4$ , $y = 0$ , $z = 0$ . What is wrong with this conclusion?

The student made computational errors during row reduction that led to an incorrect reduced matrix
The student failed to recognize that variables $y$ and $z$ are free parameters, giving infinitely many solutions (correct answer)
The student incorrectly assumed that zero rows in the coefficient matrix mean those variables equal zero
The student should have concluded the system is inconsistent due to the presence of zero rows

Explanation: The reduced matrix represents the single equation

x + 2y + 3z = 4

. Since there are 3 variables but only 1 equation,

y

and

z

are free parameters. The general solution is

x = 4 - 2s - 3t

y = s

z = t

where

s, t

are any real numbers. The student's error was setting the free variables to zero and claiming a unique solution when infinitely many solutions exist.

Question 16

A student reduces the augmented matrix $$ \begin{bmatrix} 1 & 2 & -1 & | & 3 \ 2 & 4 & -2 & | & 6 \ 1 & 2 & -1 & | & 5 \end{bmatrix}

The student failed to complete the row reduction process and stopped too early in the elimination
The student misapplied row operations, creating artificial dependencies between the original independent equations
The student failed to recognize that the third row leads to $0 = 2$ , making the system inconsistent (correct answer)
The student incorrectly concluded that identical first and third rows automatically create infinite solutions

Explanation: When reducing this matrix,

R_2 \leftarrow R_2 - 2R_1

gives

[0, 0, 0, 0]

and

R_3 \leftarrow R_3 - R_1

gives

[0, 0, 0, 2]

. The third row represents

0x_1 + 0x_2 + 0x_3 = 2

, which is impossible. This makes the system inconsistent (no solution). The student incorrectly focused on the zero row from the second equation and missed that the third row creates a contradiction.

Question 17

A student solving $\begin{bmatrix} 1 & 3 & 2 & | & 1 \\ 0 & 0 & 1 & | & 2 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}$ writes the solution as $x_1 = 1 - 3x_2 - 2x_3 = 1 - 3t - 2(2) = -3 - 3t$ , $x_2 = t$ , $x_3 = 2$ . Which error did the student make in this back-substitution process?

Incorrectly identified $x_2$ as the free variable when $x_3$ should be the parameter
Made a computational error when substituting $x_3 = 2$ into the expression for $x_1$
Failed to recognize that $x_3$ is determined by the second equation, not a free parameter (correct answer)
Used the wrong sign when isolating $x_1$ from the first equation of the reduced system

Explanation: From the reduced matrix, the second row gives

x_3 = 2

(determined, not free), and the first row gives

x_1 + 3x_2 + 2x_3 = 1

. Since

x_3 = 2

is fixed, only

x_2

is free. The correct solution is

x_1 = 1 - 3x_2 - 2(2) = -3 - 3t

x_2 = t

x_3 = 2

. The student's arithmetic is actually correct, but they conceptually treated

x_3

as free when it's determined by the pivot in the second row.

Question 18

A student performs row reduction on the augmented matrix $\begin{bmatrix} 2 & -4 & 6 & | & 8 \\ 0 & 3 & -9 & | & 12 \\ 1 & -2 & 3 & | & 4 \end{bmatrix}$ and obtains $\begin{bmatrix} 1 & -2 & 3 & | & 4 \\ 0 & 1 & -3 & | & 4 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}$ . The student concludes that the system has a unique solution $x_1 = 10$ , $x_2 = 4$ , $x_3 = 0$ . What is the primary error in this analysis?

The row reduction contains arithmetic errors that led to an incorrect reduced form of the matrix
The student failed to recognize that the system has infinitely many solutions due to the free variable $x_3$ (correct answer)
The student incorrectly identified the system as inconsistent when it actually has a unique solution
The student performed back-substitution incorrectly, using wrong values for the leading variables in each row

Explanation: The reduced matrix shows that

x_1 - 2x_2 + 3x_3 = 4

and

x_2 - 3x_3 = 4

. Since there are 3 variables but only 2 non-zero equations,

x_3

is a free variable, meaning the system has infinitely many solutions of the form

x_1 = 10 + 3t

x_2 = 4 + 3t

x_3 = t

where

t

is any real number. The student's error was assuming

x_3 = 0

and concluding there was a unique solution.