Linear Algebra Quiz: Gaussian Elimination
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Gaussian EliminationQuestion 1 of 19

A system of linear equations is represented by the augmented matrix below. After performing the row operations R2→R2−2R1R_2 \to R_2 - 2R_1 and R3→R3+R1R_3 \to R_3 + R_1 to create zeros in the first column below the main diagonal, what is the resulting matrix?

(12−1∣3011∣20−21∣−7)\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 1 & 1 & | & 2 \\ 0 & -2 & 1 & | & -7 \end{pmatrix}
(12−1∣301−3∣20−21∣−7)\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 1 & -3 & | & 2 \\ 0 & -2 & 1 & | & -7 \end{pmatrix}
(12−1∣3011∣20−63∣−13)\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 1 & 1 & | & 2 \\ 0 & -6 & 3 & | & -13 \end{pmatrix}
(12−1∣3011∣80−21∣−10)\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 1 & 1 & | & 8 \\ 0 & -2 & 1 & | & -10 \end{pmatrix}
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Linear Algebra Quiz

Linear Algebra Quiz: Gaussian Elimination

Practice Gaussian Elimination 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 Gaussian Elimination, 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.

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Question 1

A system of linear equations is represented by the augmented matrix below. After performing the row operations R2→R2−2R1R_2 \to R_2 - 2R_1 and R3→R3+R1R_3 \to R_3 + R_1 to create zeros in the first column below the main diagonal, what is the resulting matrix?

  1. (12−1∣3011∣20−21∣−7)\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 1 & 1 & | & 2 \\ 0 & -2 & 1 & | & -7 \end{pmatrix} (correct answer)
  2. (12−1∣301−3∣20−21∣−7)\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 1 & -3 & | & 2 \\ 0 & -2 & 1 & | & -7 \end{pmatrix}
  3. (12−1∣3011∣20−63∣−13)\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 1 & 1 & | & 2 \\ 0 & -6 & 3 & | & -13 \end{pmatrix}
  4. (12−1∣3011∣80−21∣−10)\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 1 & 1 & | & 8 \\ 0 & -2 & 1 & | & -10 \end{pmatrix}
Explanation: The correct answer is obtained by applying the specified elementary row operations.
  1. R2→R2−2R1R_2 \to R_2 - 2R_1: The second row becomes [2,5,−1,8]−2[1,2,−1,3]=[2−2,5−4,−1−(−2),8−6]=[0,1,1,2][2, 5, -1, 8] - 2[1, 2, -1, 3] = [2-2, 5-4, -1-(-2), 8-6] = [0, 1, 1, 2].
  2. R3→R3+R1R_3 \to R_3 + R_1: The third row becomes [−1,−4,2,−10]+[1,2,−1,3]=[−1+1,−4+2,2+(−1),−10+3]=[0,−2,1,−7][-1, -4, 2, -10] + [1, 2, -1, 3] = [-1+1, -4+2, 2+(-1), -10+3] = [0, -2, 1, -7]. The resulting matrix is option A.
Distractor B is incorrect due to an arithmetic error in the second row: −1−2(−1)=−1+2=1-1 - 2(-1) = -1+2 = 1, not −3-3. Distractor C is incorrect due to an arithmetic error in the third row, specifically −4+2=−2-4+2=-2, not −6-6. Distractor D is incorrect because the row operations were not applied to the augmented part of the matrix (the last column).

Question 2

The row echelon form of an augmented matrix for a system of equations with variables x,y,zx, y, z is given by:

(12−3∣5012∣1000∣0)\begin{pmatrix} 1 & 2 & -3 & | & 5 \\ 0 & 1 & 2 & | & 1 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}

Which of the following correctly describes the general solution to the system?

  1. The system has no solution.
  2. x=3+7tx = 3 + 7t, y=1−2ty = 1 - 2t, z=tz = t for any real number tt. (correct answer)
  3. x=3−7tx = 3 - 7t, y=1+2ty = 1 + 2t, z=tz = t for any real number tt.
  4. x=7x = 7, y=−1y = -1, and zz can be any real number.
Explanation: The last row, [0,0,0,∣,0][0, 0, 0, |, 0], corresponds to 0=00=0, which is always true. This indicates the system is dependent and has infinitely many solutions. Since there is no pivot in the third column, zz is a free variable. Let z=tz = t. We use back substitution. From row 2: y+2z=1  ⟹  y+2t=1  ⟹  y=1−2ty + 2z = 1 \implies y + 2t = 1 \implies y = 1 - 2t. From row 1: x+2y−3z=5  ⟹  x+2(1−2t)−3t=5  ⟹  x+2−4t−3t=5  ⟹  x−7t=3  ⟹  x=3+7tx + 2y - 3z = 5 \implies x + 2(1 - 2t) - 3t = 5 \implies x + 2 - 4t - 3t = 5 \implies x - 7t = 3 \implies x = 3 + 7t. The general solution is (x,y,z)=(3+7t,1−2t,t)(x, y, z) = (3+7t, 1-2t, t). Distractor A incorrectly interprets the row of zeros as an inconsistency. Distractor C contains sign errors that arise from incorrect back substitution (e.g., y=1+2ty = 1+2t instead of y=1−2ty = 1-2t). Distractor D incorrectly solves for xx and yy by ignoring the free variable zz in the first two equations, which is an incomplete analysis.

Question 3

A student performs Gaussian elimination on an augmented matrix. Their work is shown below. Identify the step in which the first error occurs.

Original Matrix:

(112∣924−3∣136−5∣0)\begin{pmatrix} 1 & 1 & 2 & | & 9 \\ 2 & 4 & -3 & | & 1 \\ 3 & 6 & -5 & | & 0 \end{pmatrix}

Step 1: Apply R2→R2−2R1R_2 \to R_2 - 2R_1. Result:

(112∣902−7∣−1736−5∣0)\begin{pmatrix} 1 & 1 & 2 & | & 9 \\ 0 & 2 & -7 & | & -17 \\ 3 & 6 & -5 & | & 0 \end{pmatrix}

Step 2: Apply R3→R3−3R1R_3 \to R_3 - 3R_1 to the result of Step 1. Result:

(112∣902−7∣−1703−11∣−25)\begin{pmatrix} 1 & 1 & 2 & | & 9 \\ 0 & 2 & -7 & | & -17 \\ 0 & 3 & -11 & | & -25 \end{pmatrix}

Step 3: Apply R3→R3−32R2R_3 \to R_3 - \frac{3}{2}R_2 to the result of Step 2. Result:

(112∣902−7∣−1700−0.5∣0.5)\begin{pmatrix} 1 & 1 & 2 & | & 9 \\ 0 & 2 & -7 & | & -17 \\ 0 & 0 & -0.5 & | & 0.5 \end{pmatrix}
  1. Step 1
  2. Step 2 (correct answer)
  3. Step 3
  4. There is no error in the student's work.
Explanation: Let's check each step. Step 1: R2→R2−2R1R_2 \to R_2 - 2R_1 applied to the original matrix gives [2,4,−3,1]−2[1,1,2,9]=[0,2,−7,1−18]=[0,2,−7,−17][2, 4, -3, 1] - 2[1, 1, 2, 9] = [0, 2, -7, 1-18] = [0, 2, -7, -17]. This step is correct. Step 2: R3→R3−3R1R_3 \to R_3 - 3R_1 applied to the original third row gives [3,6,−5,0]−3[1,1,2,9]=[0,3,−11,0−27]=[0,3,−11,−27][3, 6, -5, 0] - 3[1, 1, 2, 9] = [0, 3, -11, 0-27] = [0, 3, -11, -27]. The student's matrix shows a result of −25-25 in the augmented column instead of −27-27. This is the first error. Distractor A is incorrect because Step 1 was performed correctly. Distractor C is incorrect because, although Step 3 is arithmetically correct based on the erroneous matrix from Step 2, the first error occurred earlier. Distractor D is incorrect because an error exists in Step 2.

Question 4

In the context of solving a system of linear equations using an augmented matrix, which of the following describes a procedure that is NOT a valid elementary row operation?

  1. Interchanging the positions of any two rows.
  2. Multiplying all elements of a single row by the same non-zero constant.
  3. Replacing a row with the sum of that row and a non-zero multiple of another row.
  4. Replacing a row by adding a non-zero constant to each element of that row. (correct answer)
Explanation: The three valid elementary row operations are: 1. Swapping two rows (Ri↔RjR_i \leftrightarrow R_j). 2. Multiplying a row by a non-zero scalar (Ri→cRi,c≠0R_i \to cR_i, c \neq 0). 3. Adding a multiple of one row to another row (Ri→Ri+cRjR_i \to R_i + cR_j). Option D, adding a constant to each element (Ri→Ri+cR_i \to R_i + c), is not a valid operation because it does not correspond to an algebraic manipulation that preserves the solution set of the original system of equations. For example, it would change the equation x+y=2x+y=2 to x+1+y+1=2+1x+1+y+1=2+1, which is not equivalent. Distractors A, B, and C are the definitions of the three valid elementary row operations.

Question 5

A system of linear equations in variables x,y,zx, y, z has been reduced to the following row echelon form. What is the value of xx?

(2−31∣1501−2∣−7003∣9)\begin{pmatrix} 2 & -3 & 1 & | & 15 \\ 0 & 1 & -2 & | & -7 \\ 0 & 0 & 3 & | & 9 \end{pmatrix}
  1. x=4.5x = 4.5 (correct answer)
  2. x=3x = 3
  3. x=6x = 6
  4. x=7.5x = 7.5
Explanation: When you encounter a system in row echelon form, you're seeing the result of Gaussian elimination - a systematic way to solve linear systems by working backwards from the bottom row up through back-substitution. Start with the bottom row: 0x+0y+3z=90x + 0y + 3z = 9, which gives you z=3z = 3. Move up to the second row: 0x+1y−2z=−70x + 1y - 2z = -7. Substituting z=3z = 3: y−2(3)=−7y - 2(3) = -7, so y−6=−7y - 6 = -7, which means y=−1y = -1. Finally, use the top row: 2x−3y+z=152x - 3y + z = 15. Substituting your known values: 2x−3(−1)+3=152x - 3(-1) + 3 = 15, which simplifies to 2x+3+3=152x + 3 + 3 = 15, so 2x+6=152x + 6 = 15. Therefore 2x=92x = 9 and x=4.5x = 4.5. Looking at the wrong answers: B) x=3x = 3 likely comes from confusing the value of zz with xx, or from arithmetic errors in the back-substitution process. C) x=6x = 6 might result from incorrectly calculating 2x+6=152x + 6 = 15 as x+6=15x + 6 = 15 (forgetting the coefficient of 2), or from other algebraic mistakes. D) x=7.5x = 7.5 could arise from sign errors when substituting y=−1y = -1, perhaps treating it as positive. The correct answer is A) x=4.5x = 4.5. Remember: always work systematically from bottom to top in back-substitution, and double-check your arithmetic at each step. Many errors in these problems come from rushing through the algebraic manipulations rather than conceptual misunderstandings.

Question 6

Consider the system of equations: x+y+2z=3x + y + 2z = 3 x+2y+z=1x + 2y + z = 1 2x+y+z=02x + y + z = 0 After solving the system using Gaussian elimination, what is the value of zz?

  1. z=1z = 1
  2. z=−2z = -2
  3. z=2z = 2 (correct answer)
  4. z=−1z = -1
Explanation: First, write the augmented matrix: $$ \begin{pmatrix} 1 & 1 & 2 & | & 3 \ 1 & 2 & 1 & | & 1 \ 2 & 1 & 1 & | & 0 \end{pmatrix} Step 1: $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - 2R_1$. $R_2$ becomes $[0, 1, -1, |-2]$. $R_3$ becomes $[0, -1, -3, |-6]$. The matrix is now: $$ \begin{pmatrix} 1 & 1 & 2 & | & 3 \\ 0 & 1 & -1 & | & -2 \\ 0 & -1 & -3 & | & -6 \end{pmatrix} Step 2: R3→R3+R2R_3 \to R_3 + R_2. R3R_3 becomes [0,0,−4,∣−8][0, 0, -4, |-8]. The matrix is now in row echelon form: $$ \begin{pmatrix} 1 & 1 & 2 & | & 3 \ 0 & 1 & -1 & | & -2 \ 0 & 0 & -4 & | & -8 \end{pmatrix} From the last row, we have the equation $-4z = -8$, which solves to $z=2$. Distractor A arises from an error in the last step, such as $-6 - (-2) = -4$, leading to $-4z=-4$. Distractor B arises from a sign error, such as $-4z = 8$. Distractor D arises from a similar arithmetic error in the final step of elimination.

Question 7

For a system of linear equations with the coefficient matrix AA shown below, what condition must be placed on the constant kk to ensure the system has a unique solution?

  1. The system has a unique solution for any real number kk.
  2. The system has a unique solution only if k=4k = 4.
  3. The system has a unique solution for all kk such that k≠4k \neq 4. (correct answer)
  4. The system has a unique solution for all kk such that k≠−4k \neq -4.
Explanation: A system has a unique solution if and only if its coefficient matrix can be reduced to a form with a pivot in every column. We perform Gaussian elimination on matrix AA. Step 1: R2→R2−2R1R_2 \to R_2 - 2R_1 and R3→R3−3R1R_3 \to R_3 - 3R_1. R2R_2 becomes [0,1,−5][0, 1, -5]. R3R_3 becomes [0,1,k−9][0, 1, k-9]. The matrix becomes: $$ \begin{pmatrix} 1 & -2 & 3 \ 0 & 1 & -5 \ 0 & 1 & k-9 \end{pmatrix} Step 2: $R_3 \to R_3 - R_2$. $R_3$ becomes $[0, 0, (k-9) - (-5)] = [0, 0, k-4]$. The matrix in row echelon form is: $$ \begin{pmatrix} 1 & -2 & 3 \\ 0 & 1 & -5 \\ 0 & 0 & k-4 \end{pmatrix} $$ For a unique solution, we need a pivot in the third column, which means the entry in position (3,3) must be non-zero. Thus, we require $k-4 \neq 0$, which implies $k \neq 4$. Distractor A is incorrect because if $k=4$, the last row becomes all zeros, leading to a dependent system (infinite or no solutions). Distractor B gives the condition for the system to NOT have a unique solution. Distractor D results from a sign error during the final elimination step: $(k-9) + (-5) = k-14$ or a similar mistake.

Question 8

Given the augmented matrix below, what is the correct first step in the standard Gaussian elimination algorithm?

(023∣41−15∣2231∣0)\begin{pmatrix} 0 & 2 & 3 & | & 4 \\ 1 & -1 & 5 & | & 2 \\ 2 & 3 & 1 & | & 0 \end{pmatrix}
  1. Perform the operation R1→R1+R2R_1 \to R_1 + R_2.
  2. Multiply the first row by any non-zero constant.
  3. The system cannot be solved as the leading coefficient is zero.
  4. Perform the operation R1↔R2R_1 \leftrightarrow R_2. (correct answer)
Explanation: When you encounter Gaussian elimination problems, your goal is to systematically transform the augmented matrix into row echelon form by creating zeros below the main diagonal. The standard algorithm works column by column from left to right, but there's a crucial requirement: you need a non-zero pivot element in each leading position. Looking at this matrix, the first column should have its pivot in the (1,1) position, but that entry is 0. Since you can't use zero as a pivot element (you can't divide by zero or use it to eliminate other entries), you must first get a non-zero element into that position. The most straightforward way is to swap the first row with another row that has a non-zero entry in the first column. Option D, performing R1↔R2R_1 \leftrightarrow R_2, does exactly this. After swapping, you'll have a 1 in the (1,1) position, giving you a perfect pivot to continue with elimination. Option A (R1→R1+R2R_1 \to R_1 + R_2) would put a 1 in the first position, but this isn't the standard first step when you have a zero pivot. Option B suggests multiplying the first row by a constant, but multiplying zero by any number still gives zero, so this doesn't solve the pivot problem. Option C is completely wrong—having a zero in the pivot position doesn't make the system unsolvable; you just need to rearrange the rows. Study tip: Always check your pivot positions first. If you encounter a zero pivot, immediately look for a row swap to fix it before proceeding with elimination operations.

Question 9

After applying Gaussian elimination to the augmented matrix of a system of linear equations, the following matrix in row echelon form is obtained:

(1−23∣4015∣−2000∣6)\begin{pmatrix} 1 & -2 & 3 & | & 4 \\ 0 & 1 & 5 & | & -2 \\ 0 & 0 & 0 & | & 6 \end{pmatrix}

What can be concluded about the solution set of the system?

  1. The system has a unique solution where z=6z = 6.
  2. The system has infinitely many solutions with one free variable.
  3. The system is inconsistent and has no solution. (correct answer)
  4. The system has a unique solution, but it cannot be determined from this form.
Explanation: The third row of the matrix corresponds to the equation 0x+0y+0z=60x + 0y + 0z = 6, which simplifies to 0=60 = 6. This is a contradiction, indicating that the system of equations has no solution. Therefore, the system is inconsistent. Distractor A incorrectly interprets the last row as z=6z=6. The '6' is in the augmented part of the matrix, not the coefficient part. Distractor B is the conclusion one would draw if the last row were [0,0,0,∣,0][0, 0, 0, |, 0], which indicates a dependent system. Distractor D is incorrect because the row echelon form is sufficient to determine the nature of the solution; in this case, that no solution exists.

Question 10

Consider the following system of linear equations where kk is a real constant: x+y+z=1x + y + z = 1 2x+3y+2z=32x + 3y + 2z = 3 x+2y+kz=2x + 2y + kz = 2 For which value of kk does this system have infinitely many solutions?

  1. k=0k = 0
  2. No value of kk yields infinite solutions.
  3. k=−1k = -1
  4. k=1k = 1 (correct answer)
Explanation: When you encounter a system of linear equations with a parameter and need to find when it has infinitely many solutions, you're looking for the case where the system becomes consistent and dependent — meaning the equations don't contradict each other and at least one equation becomes redundant. Let's solve this systematically using elimination. From the first two equations: x+y+z=1x + y + z = 1 2x+3y+2z=32x + 3y + 2z = 3 Multiplying the first equation by 2 and subtracting from the second: (2x+3y+2z)−2(x+y+z)=3−2(1)(2x + 3y + 2z) - 2(x + y + z) = 3 - 2(1), which gives us y=1y = 1. Substituting y=1y = 1 back into the first equation: x+1+z=1x + 1 + z = 1, so x+z=0x + z = 0 or x=−zx = -z. Now check the third equation: x+2y+kz=2x + 2y + kz = 2. Substituting our results: (−z)+2(1)+kz=2(-z) + 2(1) + kz = 2, which simplifies to 2+z(k−1)=22 + z(k-1) = 2, or z(k−1)=0z(k-1) = 0. For infinitely many solutions, this equation must be satisfied for any value of zz, which only happens when k−1=0k - 1 = 0, so k=1k = 1. Looking at the wrong answers: (A) When k=0k = 0, we get −z=0-z = 0, forcing z=0z = 0 (unique solution, not infinite). (B) is incorrect since we just found k=1k = 1 works. (C) When k=−1k = -1, we get −2z=0-2z = 0, again forcing z=0z = 0 (unique solution). Study tip: For infinite solutions in parametric systems, look for the parameter value that makes a constraint equation become 0=00 = 0 rather than forcing a specific variable value.

Question 11

During the process of solving a system of linear equations using Gaussian elimination, the following augmented matrix is obtained:

(13−2∣5014∣−10−29∣3)\begin{pmatrix} 1 & 3 & -2 & | & 5 \\ 0 & 1 & 4 & | & -1 \\ 0 & -2 & 9 & | & 3 \end{pmatrix}

Which of the following elementary row operations is the most appropriate next step to transform the matrix into row echelon form?

  1. R3→R3−2R2R_3 \to R_3 - 2R_2
  2. R3→R3+2R2R_3 \to R_3 + 2R_2 (correct answer)
  3. R3→R3+2R1R_3 \to R_3 + 2R_1
  4. R2↔R3R_2 \leftrightarrow R_3
Explanation: The goal of this step in Gaussian elimination is to create a zero in the third row, second column, below the pivot in the second row. The pivot is the '1' in position (2,2). To eliminate the '-2' in position (3,2), we must add a multiple of the second row to the third row. The correct operation is to add 2 times Row 2 to Row 3: R3→R3+2R2R_3 \to R_3 + 2R_2. This yields [0,−2,9,3]+2[0,1,4,−1]=[0,0,17,1][0, -2, 9, 3] + 2[0, 1, 4, -1] = [0, 0, 17, 1]. Distractor A uses the wrong sign and would result in −2−2(1)=−4-2 - 2(1) = -4, which does not create a zero. Distractor C incorrectly uses the first row as the pivot row, which would undo the zero already created in the first column of Row 3. Distractor D is an unnecessary row swap that moves the matrix away from, not towards, row echelon form.

Question 12

During Gaussian elimination on a 4×44 \times 4 system, a student obtains the following partially reduced matrix after two elimination steps:

[1203∣5014−1∣20025∣8000−3∣6]\begin{bmatrix} 1 & 2 & 0 & 3 & | & 5 \\ 0 & 1 & 4 & -1 & | & 2 \\ 0 & 0 & 2 & 5 & | & 8 \\ 0 & 0 & 0 & -3 & | & 6 \end{bmatrix}

What is the value of x4x_4 when the system is solved using back-substitution?

  1. x4=−2x_4 = -2 (correct answer)
  2. x4=2x_4 = 2
  3. x4=−12x_4 = -\frac{1}{2}
  4. x4=12x_4 = \frac{1}{2}
Explanation: In back-substitution, we start with the last equation: -3x₄ = 6, which gives x₄ = -2. The other choices represent common errors: x₄ = 2 ignores the negative coefficient, x₄ = -1/2 incorrectly inverts the fraction, and x₄ = 1/2 makes both errors (wrong sign and wrong fraction).

Question 13

A linear system has the augmented matrix:

[1−21∣301−3∣2000∣5]\begin{bmatrix} 1 & -2 & 1 & | & 3 \\ 0 & 1 & -3 & | & 2 \\ 0 & 0 & 0 & | & 5 \end{bmatrix}

This matrix is already in row echelon form. What can be concluded about the original system?

  1. The system has exactly one solution since it's in row echelon form
  2. The system has infinitely many solutions due to the zero row pattern
  3. The system has no solution because of the inconsistent final equation (correct answer)
  4. The system requires further elimination before conclusions can be drawn
Explanation: The third row represents the equation 0x₁ + 0x₂ + 0x₃ = 5, which simplifies to 0 = 5. This is a contradiction, making the system inconsistent and therefore having no solution. Choice A misses the contradiction in the last row. Choice B incorrectly associates zero rows with infinite solutions (true only when the augmented part is also zero). Choice D is wrong since the matrix is already in row echelon form.

Question 14

Consider the system represented by:

[111∣6231∣13120∣7]\begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 2 & 3 & 1 & | & 13 \\ 1 & 2 & 0 & | & 7 \end{bmatrix}

After complete Gaussian elimination to reduced row echelon form, what is the sum x1+x2+x3x_1 + x_2 + x_3?

  1. x1+x2+x3=4x_1 + x_2 + x_3 = 4
  2. x1+x2+x3=5x_1 + x_2 + x_3 = 5
  3. x1+x2+x3=7x_1 + x_2 + x_3 = 7
  4. x1+x2+x3=6x_1 + x_2 + x_3 = 6 (correct answer)
Explanation: When you encounter a system of linear equations in matrix form, you're being asked to find the values of the variables that satisfy all equations simultaneously. The key insight here is that you don't actually need to solve for each individual variable - you can find their sum directly from the first equation. Looking at the augmented matrix, the first row represents the equation x1+x2+x3=6x_1 + x_2 + x_3 = 6. This immediately tells you that the sum equals 6, regardless of the individual values of each variable. While you could perform complete Gaussian elimination to find x1=1x_1 = 1, x2=2x_2 = 2, and x3=3x_3 = 3, the question specifically asks only for their sum, making the calculation unnecessary. Let's examine why the other answers are wrong. Choice A (x1+x2+x3=4x_1 + x_2 + x_3 = 4) and choice B (x1+x2+x3=5x_1 + x_2 + x_3 = 5) likely come from arithmetic errors during elimination or misreading the augmented matrix. Choice C (x1+x2+x3=7x_1 + x_2 + x_3 = 7) might result from accidentally using the constant from the third equation instead of the first, or from calculation mistakes during row operations. The correct answer is D: x1+x2+x3=6x_1 + x_2 + x_3 = 6. Study tip: When a question asks for the sum of variables in a system where one equation has all coefficients equal to 1, look for that equation first - it gives you the sum directly. This saves time and reduces the chance of computational errors during lengthy elimination procedures.

Question 15

A student applies Gaussian elimination to solve: 3x+2y−z=13x + 2y - z = 1 6x+4y−2z=36x + 4y - 2z = 3 9x+6y−3z=39x + 6y - 3z = 3

After the first elimination step (eliminating xx from equations 2 and 3), what does the student discover about the system?

  1. The system reduces to two independent equations with infinite solutions
  2. The second equation becomes inconsistent, indicating no solution exists (correct answer)
  3. All equations become dependent, requiring parametric solution methods
  4. The elimination creates new pivot positions for continued solving
Explanation: After eliminating x: R2 - 2R1 gives 0x + 0y + 0z = 1 (since 6x + 4y - 2z - 2(3x + 2y - z) = 3 - 2(1) = 1), and R3 - 3R1 gives 0x + 0y + 0z = 0. The second equation reduces to 0 = 1, which is impossible, making the system inconsistent. Choice A misses the contradiction. Choice C incorrectly assumes dependency. Choice D ignores the inconsistency revealed by elimination.

Question 16

Consider the homogeneous system Ax=0Ax = 0 where AA is a 5×75 \times 7 matrix. After applying Gaussian elimination, the reduced matrix has exactly 3 pivot positions. How many free variables does the solution set contain?

  1. 2 free variables, since rank determines degrees of freedom directly
  2. 3 free variables, since the number equals the number of pivots
  3. 4 free variables, since variables minus pivots gives degrees of freedom (correct answer)
  4. 5 free variables, since the system dimension determines the count
Explanation: For a homogeneous system with n variables and rank r, the number of free variables is n - r. Here, we have 7 variables and rank 3, so 7 - 3 = 4 free variables. Choice A incorrectly suggests rank directly determines free variables. Choice B confuses free variables with pivot variables. Choice D incorrectly uses the number of equations instead of variables minus rank.

Question 17

In solving a system using Gaussian elimination, a student reaches this reduced form:

[102−1∣301−12∣10000∣0]\begin{bmatrix} 1 & 0 & 2 & -1 & | & 3 \\ 0 & 1 & -1 & 2 & | & 1 \\ 0 & 0 & 0 & 0 & | & 0 \end{bmatrix}

If x3=tx_3 = t and x4=sx_4 = s are chosen as free variables, what is the general solution for x1x_1?

  1. x1=3−2t+sx_1 = 3 - 2t + s (correct answer)
  2. x1=3+2t−sx_1 = 3 + 2t - s
  3. x1=3−2t−sx_1 = 3 - 2t - s
  4. x1=3+2t+sx_1 = 3 + 2t + s
Explanation: From the first row: x₁ + 0x₂ + 2x₃ - 1x₄ = 3, so x₁ = 3 - 2x₃ + x₄ = 3 - 2t + s. Choice B has the wrong sign on the x₃ term. Choice C has the wrong sign on the x₄ term. Choice D has wrong signs on both terms. These represent common sign errors when moving terms to the other side of the equation.

Question 18

During Gaussian elimination, a student encounters this situation:

[13−2∣4005∣1502−1∣7]\begin{bmatrix} 1 & 3 & -2 & | & 4 \\ 0 & 0 & 5 & | & 15 \\ 0 & 2 & -1 & | & 7 \end{bmatrix}

To continue the elimination process properly, what is the most appropriate next step?

  1. Proceed with back-substitution since sufficient pivots are established
  2. Exchange rows 2 and 3 to position the next pivot correctly (correct answer)
  3. Multiply row 2 by 1/5 to normalize the pivot element
  4. Use row 2 to eliminate the third column entry in row 3
Explanation: In Gaussian elimination, we need pivots to appear in descending diagonal positions. Currently, row 2 has a zero in the second column where we need a pivot, while row 3 has a non-zero entry (2) in that position. We must swap rows 2 and 3 to bring the pivot into the correct position before continuing. Choice A is premature since we haven't established proper row echelon form. Choice C addresses normalization but doesn't fix the structural issue. Choice D attempts elimination before establishing the proper pivot structure.

Question 19

What is the reduced row echelon form (RREF) of the following matrix?

  1. (101012)\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \end{pmatrix} (correct answer)
  2. (137012)\begin{pmatrix} 1 & 3 & 7 \\ 0 & 1 & 2 \end{pmatrix}
  3. (1013012)\begin{pmatrix} 1 & 0 & 13 \\ 0 & 1 & 2 \end{pmatrix}
  4. (100010)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}
Explanation: When you encounter a matrix and need to find its reduced row echelon form (RREF), you're applying a systematic process to create the simplest possible form where each leading entry (pivot) is 1, and all entries above and below each pivot are 0. Let's work through matrix A step by step. Starting with: The matrix already has pivots in the correct positions: the leading 1 in row 1, column 1, and the leading 1 in row 2, column 2. However, we need to eliminate the entry above the second pivot. We need to make the 3 in position (1,2) equal to 0. Using the row operation R₁ ← R₁ - 3R₂:
  • New row 1: (1, 3, 7) - 3(0, 1, 2) = (1, 0, 1)
  • Row 2 remains: (0, 1, 2)
This gives us the RREF: $$ \begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 2 \end{pmatrix} AnswerBistheoriginalmatrix,notyetinRREFsincethe3abovethesecondpivothasn′tbeeneliminated.AnswerCincorrectlycalculatestheeliminationstep—thiswouldresultfromadding3R2insteadofsubtracting.AnswerDrepresentsanimpossibleRREFforthismatrixsinceitwouldrequireadditionalcolumnsofzerosthatdon′texistintheoriginalmatrix.Remember:RREFrequiresleading1′swithzerosaboveandbeloweachpivot.Alwaysworksystematicallyfromlefttoright,eliminatingentriesaboveexistingpivots. Answer B is the original matrix, not yet in RREF since the 3 above the second pivot hasn't been eliminated. Answer C incorrectly calculates the elimination step—this would result from adding 3R₂ instead of subtracting. Answer D represents an impossible RREF for this matrix since it would require additional columns of zeros that don't exist in the original matrix. Remember: RREF requires leading 1's with zeros above and below each pivot. Always work systematically from left to right, eliminating entries above existing pivots.