IB Mathematics: Analysis and Approaches Quiz: Systems Via Matrices
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Systems Via MatricesQuestion 1 of 20

Consider the system of equations:

x+y+2z=42x−y+z=2x + y + 2z = 4 \\ 2x - y + z = 2

Which of the following correctly describes the solution set?

The system has no solution.
The system has a unique solution.
The solutions form a plane.
The solutions form a line.
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Systems Via Matrices

Practice Systems Via Matrices in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Systems Via Matrices, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Question 1

Consider the system of equations:

x+y+2z=42x−y+z=2x + y + 2z = 4 \\ 2x - y + z = 2

Which of the following correctly describes the solution set?

  1. The system has no solution.
  2. The system has a unique solution.
  3. The solutions form a plane.
  4. The solutions form a line. (correct answer)
Explanation: When you encounter a system of linear equations with fewer equations than variables, you need to determine the geometric nature of the solution set by analyzing the system's structure. This system has 2 equations and 3 variables (xx, yy, zz), which immediately suggests infinitely many solutions rather than a unique solution. To find the solution set, solve for two variables in terms of the third. From the first equation: y=4−x−2zy = 4 - x - 2z. Substituting into the second equation: 2x−(4−x−2z)+z=22x - (4 - x - 2z) + z = 2, which simplifies to 3x+3z=63x + 3z = 6, or x=2−zx = 2 - z. Therefore: y=4−(2−z)−2z=2−zy = 4 - (2 - z) - 2z = 2 - z. The solution can be written parametrically as (x,y,z)=(2−t,2−t,t)(x, y, z) = (2 - t, 2 - t, t) where tt is any real number. This represents a line in 3D space. Option A is incorrect because the system is consistent—there's no contradiction when you solve it. Option B is wrong because systems with fewer equations than unknowns typically have infinitely many solutions, not a unique one. Option C misidentifies the geometric structure; while three variables suggest 3D space, the two constraints reduce the solution set from the full 3D space (which would be all of R3\mathbb{R}^3) down to a line, not a plane. Remember: For linear systems, count the relationships. With nn variables and rr independent equations where r<nr < n, expect an (n−r)(n-r)-dimensional solution set. Here, 3−2=13-2=1 dimension means a line.

Question 2

Consider the system of equations:

x+y+2z=42x−y+z=2x + y + 2z = 4 \\ 2x - y + z = 2

Which of the following correctly describes the solution set?

  1. The system has no solution.
  2. The system has a unique solution.
  3. The solutions form a plane.
  4. The solutions form a line. (correct answer)
Explanation: When you encounter a system of linear equations with fewer equations than variables, you need to determine the geometric nature of the solution set by analyzing the system's structure. This system has 2 equations and 3 variables (xx, yy, zz), which immediately suggests infinitely many solutions rather than a unique solution. To find the solution set, solve for two variables in terms of the third. From the first equation: y=4−x−2zy = 4 - x - 2z. Substituting into the second equation: 2x−(4−x−2z)+z=22x - (4 - x - 2z) + z = 2, which simplifies to 3x+3z=63x + 3z = 6, or x=2−zx = 2 - z. Therefore: y=4−(2−z)−2z=2−zy = 4 - (2 - z) - 2z = 2 - z. The solution can be written parametrically as (x,y,z)=(2−t,2−t,t)(x, y, z) = (2 - t, 2 - t, t) where tt is any real number. This represents a line in 3D space. Option A is incorrect because the system is consistent—there's no contradiction when you solve it. Option B is wrong because systems with fewer equations than unknowns typically have infinitely many solutions, not a unique one. Option C misidentifies the geometric structure; while three variables suggest 3D space, the two constraints reduce the solution set from the full 3D space (which would be all of R3\mathbb{R}^3) down to a line, not a plane. Remember: For linear systems, count the relationships. With nn variables and rr independent equations where r<nr < n, expect an (n−r)(n-r)-dimensional solution set. Here, 3−2=13-2=1 dimension means a line.

Question 3

The augmented matrix for a system of three linear equations in three variables is [A∣b][A|\mathbf{b}]. It is known that the rank of the coefficient matrix AA is 2, and the rank of the augmented matrix [A∣b][A|\mathbf{b}] is also 2. What can be concluded about the solution set of the system?

  1. The system has a unique solution.
  2. The system has no solution.
  3. The solutions form a plane.
  4. The solutions form a line. (correct answer)
Explanation: When you encounter questions about the rank of matrices in linear systems, you're dealing with fundamental concepts that determine solution behavior. The rank tells you about linear independence and dimensionality. Here's the key insight: when rank(A)=rank([A∣b])\text{rank}(A) = \text{rank}([A|\mathbf{b}]), the system is consistent (has solutions). Since both ranks equal 2 in a system with 3 variables, you have 2 linearly independent equations constraining 3 unknowns. This means 3−2=13 - 2 = 1 free variable, creating a one-dimensional solution set—which is a line in 3D space. Why the other answers fail: Answer A is incorrect because a unique solution requires rank(A)=3\text{rank}(A) = 3 (equal to the number of variables). With rank 2, you don't have enough independent constraints. Answer B is wrong because when rank(A)=rank([A∣b])\text{rank}(A) = \text{rank}([A|\mathbf{b}]), the system is always consistent—inconsistency only occurs when rank([A∣b])>rank(A)\text{rank}([A|\mathbf{b}]) > \text{rank}(A). Answer C represents a common misconception: a plane would require rank 1, giving you 3−1=23 - 1 = 2 free variables and thus a two-dimensional solution set. Remember this pattern: in an nn-variable system, if rank(A)=rank([A∣b])=r\text{rank}(A) = \text{rank}([A|\mathbf{b}]) = r, then you have n−rn - r free variables. The solution set's dimension equals the number of free variables: 0 free variables = point (unique solution), 1 free variable = line, 2 free variables = plane, and so on.

Question 4

The system of equations

x+y+z=1x+2y+3z=4x+3y+kz=9x + y + z = 1 \\ x + 2y + 3z = 4 \\ x + 3y + kz = 9

has a unique solution for all real values of kk except for k=k0k = k_0. What is the nature of the solution when k=k0k = k_0?

  1. The system has no solution. (correct answer)
  2. The system has infinitely many solutions.
  3. The system has the unique solution (0,0,0).
  4. The nature of the solution cannot be determined.
Explanation: First, we find the value k0k_0 for which the system does not have a unique solution. This occurs when the determinant of the coefficient matrix is zero. A=(11112313k)A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & k \end{pmatrix} det⁡(A)=1(2k−9)−1(k−3)+1(3−2)=2k−9−k+3+1=k−5\det(A) = 1(2k - 9) - 1(k - 3) + 1(3 - 2) = 2k - 9 - k + 3 + 1 = k - 5 Setting det⁡(A)=0\det(A) = 0 gives k−5=0k - 5 = 0, so k0=5k_0 = 5. Now we must analyze the system for k=5k = 5. We use the augmented matrix: [111112341359]\left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 5 & 9 \end{array}\right] Perform row operations: R2→R2−R1R_2 \to R_2 - R_1 and R3→R3−R1R_3 \to R_3 - R_1. [111101230248]\left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 2 & 4 & 8 \end{array}\right] Next, perform R3→R3−2R2R_3 \to R_3 - 2R_2. [111101230008−2(3)]=[111101230002]\left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 8 - 2(3) \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 2 \end{array}\right] The last row represents the equation 0x+0y+0z=20x + 0y + 0z = 2, or 0=20 = 2, which is a contradiction. Therefore, when k=5k = 5, the system has no solution. Distractor Rationale:
  • B: This would be the case if the final row was [0 0 0 ∣ 0][0 \ 0 \ 0 \ | \ 0]. A student might assume that det⁡(A)=0\det(A)=0 always implies infinite solutions.
  • C: The trivial solution is irrelevant here.
  • D: The nature of the solution can be determined.

Question 5

The augmented matrix for a system of three linear equations in three variables is [A∣b][A|\mathbf{b}]. It is known that the rank of the coefficient matrix AA is 2, and the rank of the augmented matrix [A∣b][A|\mathbf{b}] is also 2. What can be concluded about the solution set of the system?

  1. The system has a unique solution.
  2. The system has no solution.
  3. The solutions form a plane.
  4. The solutions form a line. (correct answer)
Explanation: When you encounter questions about the rank of matrices in linear systems, you're dealing with fundamental concepts that determine solution behavior. The rank tells you about linear independence and dimensionality. Here's the key insight: when rank(A)=rank([A∣b])\text{rank}(A) = \text{rank}([A|\mathbf{b}]), the system is consistent (has solutions). Since both ranks equal 2 in a system with 3 variables, you have 2 linearly independent equations constraining 3 unknowns. This means 3−2=13 - 2 = 1 free variable, creating a one-dimensional solution set—which is a line in 3D space. Why the other answers fail: Answer A is incorrect because a unique solution requires rank(A)=3\text{rank}(A) = 3 (equal to the number of variables). With rank 2, you don't have enough independent constraints. Answer B is wrong because when rank(A)=rank([A∣b])\text{rank}(A) = \text{rank}([A|\mathbf{b}]), the system is always consistent—inconsistency only occurs when rank([A∣b])>rank(A)\text{rank}([A|\mathbf{b}]) > \text{rank}(A). Answer C represents a common misconception: a plane would require rank 1, giving you 3−1=23 - 1 = 2 free variables and thus a two-dimensional solution set. Remember this pattern: in an nn-variable system, if rank(A)=rank([A∣b])=r\text{rank}(A) = \text{rank}([A|\mathbf{b}]) = r, then you have n−rn - r free variables. The solution set's dimension equals the number of free variables: 0 free variables = point (unique solution), 1 free variable = line, 2 free variables = plane, and so on.

Question 6

A company produces three products, P, Q, and R, using three machines, M1, M2, and M3. The hours required on each machine to produce one unit of each product are given below:

  • P requires 1 hour on M1, 2 hours on M2, and 1 hour on M3.
  • Q requires 2 hours on M1, 1 hour on M2, and 3 hours on M3.
  • R requires 3 hours on M1, 2 hours on M2, and 1 hour on M3. In a particular week, M1 is available for 11 hours, M2 for 9 hours, and M3 for 12 hours.

Let p,q,rp, q, r be the number of units of P, Q, and R produced, respectively. Find the value of pp.

  1. 1
  2. 2 (correct answer)
  3. 3
  4. 5
Explanation: Based on the information, we can set up a system of linear equations: p+2q+3r=11(Machine M1)2p+q+2r=9(Machine M2)p+3q+r=12(Machine M3)p + 2q + 3r = 11 \quad (\text{Machine M1}) \\ 2p + q + 2r = 9 \quad (\text{Machine M2}) \\ p + 3q + r = 12 \quad (\text{Machine M3}) We write this as an augmented matrix and solve: [12311212913112]\left[\begin{array}{ccc|c} 1 & 2 & 3 & 11 \\ 2 & 1 & 2 & 9 \\ 1 & 3 & 1 & 12 \end{array}\right] R2→R2−2R1R_2 \to R_2 - 2R_1 and R3→R3−R1R_3 \to R_3 - R_1: [123110−3−4−1301−21]\left[\begin{array}{ccc|c} 1 & 2 & 3 & 11 \\ 0 & -3 & -4 & -13 \\ 0 & 1 & -2 & 1 \end{array}\right] Swap R2R_2 and R3R_3: [1231101−210−3−4−13]\left[\begin{array}{ccc|c} 1 & 2 & 3 & 11 \\ 0 & 1 & -2 & 1 \\ 0 & -3 & -4 & -13 \end{array}\right] R3→R3+3R2R_3 \to R_3 + 3R_2: [1231101−2100−10−10]\left[\begin{array}{ccc|c} 1 & 2 & 3 & 11 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & -10 & -10 \end{array}\right] From the last row, −10r=−10  ⟹  r=1-10r = -10 \implies r=1. From the second row, q−2r=1  ⟹  q−2(1)=1  ⟹  q=3q - 2r = 1 \implies q - 2(1) = 1 \implies q = 3. From the first row, p+2q+3r=11  ⟹  p+2(3)+3(1)=11  ⟹  p+6+3=11  ⟹  p+9=11  ⟹  p=2p + 2q + 3r = 11 \implies p + 2(3) + 3(1) = 11 \implies p + 6 + 3 = 11 \implies p + 9 = 11 \implies p=2. The solution is p=2,q=3,r=1p=2, q=3, r=1. The question asks for the value of pp. Distractor Rationale:
  • A: This is the value of rr.
  • C: This is the value of qq.
  • D: This is the value of p+qp+q.

Question 7

The augmented matrix of a system of linear equations (x,y,z)(x, y, z) is reduced to the following form:

[102501−120000]\left[\begin{array}{ccc|c} 1 & 0 & 2 & 5 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right]

Which of the following represents the general solution to the system in terms of a parameter tt?

  1. x=5−2t,y=2+t,z=tx = 5 - 2t, y = 2 + t, z = t (correct answer)
  2. x=5+2t,y=2−t,z=tx = 5 + 2t, y = 2 - t, z = t
  3. x=5,y=2,z=0x = 5, y = 2, z = 0
  4. The system has no solution.
Explanation: The reduced row echelon form of the matrix corresponds to the following system of equations: x+2z=5y−z=20=0x + 2z = 5 \\ y - z = 2 \\ 0 = 0 The third equation 0=00=0 indicates that the system is dependent and has infinitely many solutions. We can introduce a parameter for one of the variables. Let z=tz = t, where tt is any real number. From the second equation, y−t=2y - t = 2, which gives y=2+ty = 2 + t. From the first equation, x+2t=5x + 2t = 5, which gives x=5−2tx = 5 - 2t. So, the general solution is (x,y,z)=(5−2t,2+t,t)(x, y, z) = (5 - 2t, 2 + t, t). Distractor Rationale:
  • B: Sign errors when solving for xx and yy from the equations.
  • C: This is a particular solution (when t=0t=0), but not the general solution.
  • D: This would be the conclusion if the last row were [0 0 0 ∣ c][0 \ 0 \ 0 \ | \ c] with c≠0c \neq 0. Misinterpreting the row of zeros.

Question 8

The augmented matrix for a system of linear equations is being row-reduced. After several steps, the matrix is:

[12−130−33−20−3k4]\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & -3 & 3 & -2 \\ 0 & -3 & k & 4 \end{array}\right]

For what value of kk will the system be inconsistent?

  1. k=−3k = -3
  2. k=3k = 3 (correct answer)
  3. k=4k = 4
  4. k=6k = 6
Explanation: To continue the row reduction from the given matrix, we can eliminate the first non-zero entry in the third row. We perform the row operation R3→R3−R2R_3 \to R_3 - R_2. [12−130−33−20−3−(−3)k−34−(−2)]=[12−130−33−200k−36]\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & -3 & 3 & -2 \\ 0 & -3-(-3) & k-3 & 4-(-2) \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & -3 & 3 & -2 \\ 0 & 0 & k-3 & 6 \end{array}\right] A system is inconsistent if its row-reduced form contains a row of the form [0 0 0 ∣ c][0 \ 0 \ 0 \ | \ c] where c≠0c \neq 0. In our case, the last row is [0 0 k−3 ∣ 6][0 \ 0 \ k-3 \ | \ 6]. For this to become a contradiction, the left side must be all zeros, and the right side non-zero. The right side is 6, which is non-zero. Thus, we need k−3=0k-3 = 0, which gives k=3k=3. Distractor Rationale:
  • A: A sign error when solving k−3=0k-3=0.
  • C: Mistaking the value on the right side of the third row for kk.
  • D: Mistaking the final value on the right side of the third row for kk.

Question 9

A student is solving the system x+2y=3,2x+3y=4x + 2y = 3, 2x + 3y = 4 using an augmented matrix. Their work is shown below. Step 0: [123234]\left[\begin{array}{cc|c} 1 & 2 & 3 \\ 2 & 3 & 4 \end{array}\right] Step 1: R2→R2−2R1R_2 \to R_2 - 2R_1 gives [12301−2]\left[\begin{array}{cc|c} 1 & 2 & 3 \\ 0 & 1 & -2 \end{array}\right] Step 2: R1→R1−2R2R_1 \to R_1 - 2R_2 gives [10701−2]\left[\begin{array}{cc|c} 1 & 0 & 7 \\ 0 & 1 & -2 \end{array}\right] From this, the student concludes the solution is (7,−2)(7, -2). In which step did the student make their first mistake?

  1. Step 1 (correct answer)
  2. Step 2
  3. The conclusion from the final matrix
  4. The student made no mistake.
Explanation: Let's check each step of the student's work. Step 0 is the correct setup of the augmented matrix. Step 1: The operation is R2→R2−2R1R_2 \to R_2 - 2R_1. The original R2R_2 is [2 3 ∣ 4][2 \ 3 \ | \ 4] and 2R12R_1 is 2×[1 2 ∣ 3]=[2 4 ∣ 6]2 \times [1 \ 2 \ | \ 3] = [2 \ 4 \ | \ 6]. So, R2−2R1=[2−23−4∣4−6]=[0−1∣−2]R_2 - 2R_1 = [2-2 \quad 3-4 \quad | \quad 4-6] = [0 \quad -1 \quad | \quad -2]. The student wrote [01∣−2][0 \quad 1 \quad | \quad -2]. This is a sign error on the '1'. This is the first mistake. Distractor Rationale:
  • B: Although the calculation in Step 2 is correct based on the erroneous matrix from Step 1, the first error occurred earlier.
  • C: The conclusion (7,−2)(7,-2) correctly interprets the final matrix, but the matrix itself is the result of an earlier error.
  • D: The student made a clear calculation error in Step 1.

Question 10

A company produces three products, P, Q, and R, using three machines, M1, M2, and M3. The hours required on each machine to produce one unit of each product are given below:

  • P requires 1 hour on M1, 2 hours on M2, and 1 hour on M3.
  • Q requires 2 hours on M1, 1 hour on M2, and 3 hours on M3.
  • R requires 3 hours on M1, 2 hours on M2, and 1 hour on M3. In a particular week, M1 is available for 11 hours, M2 for 9 hours, and M3 for 12 hours.

Let p,q,rp, q, r be the number of units of P, Q, and R produced, respectively. Find the value of pp.

  1. 1
  2. 2 (correct answer)
  3. 3
  4. 5
Explanation: Based on the information, we can set up a system of linear equations: p+2q+3r=11(Machine M1)2p+q+2r=9(Machine M2)p+3q+r=12(Machine M3)p + 2q + 3r = 11 \quad (\text{Machine M1}) \\ 2p + q + 2r = 9 \quad (\text{Machine M2}) \\ p + 3q + r = 12 \quad (\text{Machine M3}) We write this as an augmented matrix and solve: [12311212913112]\left[\begin{array}{ccc|c} 1 & 2 & 3 & 11 \\ 2 & 1 & 2 & 9 \\ 1 & 3 & 1 & 12 \end{array}\right] R2→R2−2R1R_2 \to R_2 - 2R_1 and R3→R3−R1R_3 \to R_3 - R_1: [123110−3−4−1301−21]\left[\begin{array}{ccc|c} 1 & 2 & 3 & 11 \\ 0 & -3 & -4 & -13 \\ 0 & 1 & -2 & 1 \end{array}\right] Swap R2R_2 and R3R_3: [1231101−210−3−4−13]\left[\begin{array}{ccc|c} 1 & 2 & 3 & 11 \\ 0 & 1 & -2 & 1 \\ 0 & -3 & -4 & -13 \end{array}\right] R3→R3+3R2R_3 \to R_3 + 3R_2: [1231101−2100−10−10]\left[\begin{array}{ccc|c} 1 & 2 & 3 & 11 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & -10 & -10 \end{array}\right] From the last row, −10r=−10  ⟹  r=1-10r = -10 \implies r=1. From the second row, q−2r=1  ⟹  q−2(1)=1  ⟹  q=3q - 2r = 1 \implies q - 2(1) = 1 \implies q = 3. From the first row, p+2q+3r=11  ⟹  p+2(3)+3(1)=11  ⟹  p+6+3=11  ⟹  p+9=11  ⟹  p=2p + 2q + 3r = 11 \implies p + 2(3) + 3(1) = 11 \implies p + 6 + 3 = 11 \implies p + 9 = 11 \implies p=2. The solution is p=2,q=3,r=1p=2, q=3, r=1. The question asks for the value of pp. Distractor Rationale:
  • A: This is the value of rr.
  • C: This is the value of qq.
  • D: This is the value of p+qp+q.

Question 11

The system of equations

x+y+z=1x+2y+3z=4x+3y+kz=9x + y + z = 1 \\ x + 2y + 3z = 4 \\ x + 3y + kz = 9

has a unique solution for all real values of kk except for k=k0k = k_0. What is the nature of the solution when k=k0k = k_0?

  1. The system has no solution. (correct answer)
  2. The system has infinitely many solutions.
  3. The system has the unique solution (0,0,0).
  4. The nature of the solution cannot be determined.
Explanation: First, we find the value k0k_0 for which the system does not have a unique solution. This occurs when the determinant of the coefficient matrix is zero. A=(11112313k)A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & k \end{pmatrix} det⁡(A)=1(2k−9)−1(k−3)+1(3−2)=2k−9−k+3+1=k−5\det(A) = 1(2k - 9) - 1(k - 3) + 1(3 - 2) = 2k - 9 - k + 3 + 1 = k - 5 Setting det⁡(A)=0\det(A) = 0 gives k−5=0k - 5 = 0, so k0=5k_0 = 5. Now we must analyze the system for k=5k = 5. We use the augmented matrix: [111112341359]\left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 5 & 9 \end{array}\right] Perform row operations: R2→R2−R1R_2 \to R_2 - R_1 and R3→R3−R1R_3 \to R_3 - R_1. [111101230248]\left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 2 & 4 & 8 \end{array}\right] Next, perform R3→R3−2R2R_3 \to R_3 - 2R_2. [111101230008−2(3)]=[111101230002]\left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 8 - 2(3) \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 2 \end{array}\right] The last row represents the equation 0x+0y+0z=20x + 0y + 0z = 2, or 0=20 = 2, which is a contradiction. Therefore, when k=5k = 5, the system has no solution. Distractor Rationale:
  • B: This would be the case if the final row was [0 0 0 ∣ 0][0 \ 0 \ 0 \ | \ 0]. A student might assume that det⁡(A)=0\det(A)=0 always implies infinite solutions.
  • C: The trivial solution is irrelevant here.
  • D: The nature of the solution can be determined.

Question 12

The augmented matrix of a system of linear equations (x,y,z)(x, y, z) is reduced to the following form:

[102501−120000]\left[\begin{array}{ccc|c} 1 & 0 & 2 & 5 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right]

Which of the following represents the general solution to the system in terms of a parameter tt?

  1. x=5−2t,y=2+t,z=tx = 5 - 2t, y = 2 + t, z = t (correct answer)
  2. x=5+2t,y=2−t,z=tx = 5 + 2t, y = 2 - t, z = t
  3. x=5,y=2,z=0x = 5, y = 2, z = 0
  4. The system has no solution.
Explanation: The reduced row echelon form of the matrix corresponds to the following system of equations: x+2z=5y−z=20=0x + 2z = 5 \\ y - z = 2 \\ 0 = 0 The third equation 0=00=0 indicates that the system is dependent and has infinitely many solutions. We can introduce a parameter for one of the variables. Let z=tz = t, where tt is any real number. From the second equation, y−t=2y - t = 2, which gives y=2+ty = 2 + t. From the first equation, x+2t=5x + 2t = 5, which gives x=5−2tx = 5 - 2t. So, the general solution is (x,y,z)=(5−2t,2+t,t)(x, y, z) = (5 - 2t, 2 + t, t). Distractor Rationale:
  • B: Sign errors when solving for xx and yy from the equations.
  • C: This is a particular solution (when t=0t=0), but not the general solution.
  • D: This would be the conclusion if the last row were [0 0 0 ∣ c][0 \ 0 \ 0 \ | \ c] with c≠0c \neq 0. Misinterpreting the row of zeros.

Question 13

A student is solving the system x+2y=3,2x+3y=4x + 2y = 3, 2x + 3y = 4 using an augmented matrix. Their work is shown below. Step 0: [123234]\left[\begin{array}{cc|c} 1 & 2 & 3 \\ 2 & 3 & 4 \end{array}\right] Step 1: R2→R2−2R1R_2 \to R_2 - 2R_1 gives [12301−2]\left[\begin{array}{cc|c} 1 & 2 & 3 \\ 0 & 1 & -2 \end{array}\right] Step 2: R1→R1−2R2R_1 \to R_1 - 2R_2 gives [10701−2]\left[\begin{array}{cc|c} 1 & 0 & 7 \\ 0 & 1 & -2 \end{array}\right] From this, the student concludes the solution is (7,−2)(7, -2). In which step did the student make their first mistake?

  1. Step 1 (correct answer)
  2. Step 2
  3. The conclusion from the final matrix
  4. The student made no mistake.
Explanation: Let's check each step of the student's work. Step 0 is the correct setup of the augmented matrix. Step 1: The operation is R2→R2−2R1R_2 \to R_2 - 2R_1. The original R2R_2 is [2 3 ∣ 4][2 \ 3 \ | \ 4] and 2R12R_1 is 2×[1 2 ∣ 3]=[2 4 ∣ 6]2 \times [1 \ 2 \ | \ 3] = [2 \ 4 \ | \ 6]. So, R2−2R1=[2−23−4∣4−6]=[0−1∣−2]R_2 - 2R_1 = [2-2 \quad 3-4 \quad | \quad 4-6] = [0 \quad -1 \quad | \quad -2]. The student wrote [01∣−2][0 \quad 1 \quad | \quad -2]. This is a sign error on the '1'. This is the first mistake. Distractor Rationale:
  • B: Although the calculation in Step 2 is correct based on the erroneous matrix from Step 1, the first error occurred earlier.
  • C: The conclusion (7,−2)(7,-2) correctly interprets the final matrix, but the matrix itself is the result of an earlier error.
  • D: The student made a clear calculation error in Step 1.

Question 14

The augmented matrix for a system of linear equations is being row-reduced. After several steps, the matrix is:

[12−130−33−20−3k4]\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & -3 & 3 & -2 \\ 0 & -3 & k & 4 \end{array}\right]

For what value of kk will the system be inconsistent?

  1. k=−3k = -3
  2. k=3k = 3 (correct answer)
  3. k=4k = 4
  4. k=6k = 6
Explanation: To continue the row reduction from the given matrix, we can eliminate the first non-zero entry in the third row. We perform the row operation R3→R3−R2R_3 \to R_3 - R_2. [12−130−33−20−3−(−3)k−34−(−2)]=[12−130−33−200k−36]\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & -3 & 3 & -2 \\ 0 & -3-(-3) & k-3 & 4-(-2) \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & -3 & 3 & -2 \\ 0 & 0 & k-3 & 6 \end{array}\right] A system is inconsistent if its row-reduced form contains a row of the form [0 0 0 ∣ c][0 \ 0 \ 0 \ | \ c] where c≠0c \neq 0. In our case, the last row is [0 0 k−3 ∣ 6][0 \ 0 \ k-3 \ | \ 6]. For this to become a contradiction, the left side must be all zeros, and the right side non-zero. The right side is 6, which is non-zero. Thus, we need k−3=0k-3 = 0, which gives k=3k=3. Distractor Rationale:
  • A: A sign error when solving k−3=0k-3=0.
  • C: Mistaking the value on the right side of the third row for kk.
  • D: Mistaking the final value on the right side of the third row for kk.

Question 15

For what value of cc is the following system of equations consistent?

x+y=3x−y=12x+3y=cx + y = 3 \\ x - y = 1 \\ 2x + 3y = c
  1. c=1c=1
  2. c=3c=3
  3. c=5c=5
  4. c=7c=7 (correct answer)
Explanation: For the system to be consistent, there must be a single point (x,y)(x,y) that satisfies all three equations. We can solve the first two equations to find the point of intersection of the first two lines. Adding the first two equations: (x+y)+(x−y)=3+1  ⟹  2x=4  ⟹  x=2(x+y) + (x-y) = 3+1 \implies 2x = 4 \implies x=2. Substituting x=2x=2 into the first equation: 2+y=3  ⟹  y=12+y=3 \implies y=1. The point of intersection of the first two lines is (2,1)(2,1). For the system to be consistent, this point must also lie on the third line, 2x+3y=c2x+3y=c. Substituting x=2x=2 and y=1y=1 into the third equation: 2(2)+3(1)=c2(2) + 3(1) = c 4+3=c4 + 3 = c c=7c = 7. Distractor Rationale:
  • A, B, C: Incorrect values that could arise from arithmetic errors or from misinterpreting the problem, for example, by using the constants from the first two equations.

Question 16

For which condition on aa and bb is the following system of equations consistent?

x−2y+z=32x+y−2z=a4x−3y=bx - 2y + z = 3 \\ 2x + y - 2z = a \\ 4x - 3y = b
  1. a+b=18a + b = 18
  2. b=a−6b = a - 6
  3. b=a+6b = a + 6 (correct answer)
  4. a=b+6a = b + 6
Explanation: A system is consistent if it has at least one solution. We set up the augmented matrix and perform row reduction. [1−21321−2a4−30b]\left[\begin{array}{ccc|c} 1 & -2 & 1 & 3 \\ 2 & 1 & -2 & a \\ 4 & -3 & 0 & b \end{array}\right] Perform R2→R2−2R1R_2 \to R_2 - 2R_1 and R3→R3−4R1R_3 \to R_3 - 4R_1. [1−21305−4a−605−4b−12]\left[\begin{array}{ccc|c} 1 & -2 & 1 & 3 \\ 0 & 5 & -4 & a-6 \\ 0 & 5 & -4 & b-12 \end{array}\right] Next, perform R3→R3−R2R_3 \to R_3 - R_2. [1−21305−4a−6000(b−12)−(a−6)]\left[\begin{array}{ccc|c} 1 & -2 & 1 & 3 \\ 0 & 5 & -4 & a-6 \\ 0 & 0 & 0 & (b-12) - (a-6) \end{array}\right] The last row simplifies to [0 0 0 ∣ b−a−6][0 \ 0 \ 0 \ | \ b - a - 6]. For the system to be consistent, this last row must not be a contradiction. This means the last entry must be zero, as 0x+0y+0z0x+0y+0z must equal zero. So, b−a−6=0b - a - 6 = 0, which implies b=a+6b = a + 6. Distractor Rationale:
  • B: Sign error in b−12−(a−6)=b−12−a+6=b−a−6b-12 - (a-6) = b-12-a+6 = b-a-6. A common mistake is b−12−a−6=b−a−18b-12-a-6 = b-a-18 leading to b=a+18b=a+18. Or sign error on the 6.
  • D: Incorrect rearrangement of the final equation.
  • A: Another possible arithmetic error.

Question 17

Consider the system of linear equations:

x+y−z=12x+3y+kz=3x+ky+3z=2x + y - z = 1 \\ 2x + 3y + kz = 3 \\ x + ky + 3z = 2

For what value of kk does the system have no solution?

  1. k=−3k = -3 (correct answer)
  2. k=−2k = -2
  3. k=2k = 2
  4. k=3k = 3
Explanation: To determine the nature of the solution, we can use Gaussian elimination on the augmented matrix. [11−1123k31k32]\left[\begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 2 & 3 & k & 3 \\ 1 & k & 3 & 2 \end{array}\right] Perform row operations: R2→R2−2R1R_2 \to R_2 - 2R_1 and R3→R3−R1R_3 \to R_3 - R_1. [11−1101k+210k−141]\left[\begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 0 & 1 & k+2 & 1 \\ 0 & k-1 & 4 & 1 \end{array}\right] Next, perform R3→R3−(k−1)R2R_3 \to R_3 - (k-1)R_2. [11−1101k+21004−(k−1)(k+2)1−(k−1)]\left[\begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 0 & 1 & k+2 & 1 \\ 0 & 0 & 4 - (k-1)(k+2) & 1 - (k-1) \end{array}\right] The third row simplifies to: [004−(k2+k−2)∣2−k][0 \quad 0 \quad 4 - (k^2+k-2) \quad | \quad 2-k] [00−k2−k+6∣2−k][0 \quad 0 \quad -k^2-k+6 \quad | \quad 2-k] [00−(k+3)(k−2)∣2−k][0 \quad 0 \quad -(k+3)(k-2) \quad | \quad 2-k] For the system to have no solution, the last row must be of the form [0 0 0 ∣ c][0 \ 0 \ 0 \ | \ c] where c≠0c \neq 0. This requires the left side to be zero and the right side to be non-zero. −(k+3)(k−2)=0-(k+3)(k-2) = 0 gives k=−3k=-3 or k=2k=2. 2−k≠02-k \neq 0 gives k≠2k \neq 2. Both conditions are met only when k=−3k=-3. When k=−3k=-3, the last row is [0 0 0 ∣ 5][0 \ 0 \ 0 \ | \ 5], indicating no solution. Distractor Rationale:
  • C: If k=2k=2, the last row becomes [0 0 0 ∣ 0][0 \ 0 \ 0 \ | \ 0], which corresponds to infinitely many solutions.
  • B and D are based on potential algebraic errors during row reduction.

Question 18

The solution set of the system

x+y+z=6x+2y+3z=142x+3y+kz=20x + y + z = 6 \\ x + 2y + 3z = 14 \\ 2x + 3y + kz = 20

forms a line. Find the value of kk.

  1. k=2k=2
  2. k=4k=4 (correct answer)
  3. k=3k=3
  4. k=5k=5
Explanation: When you encounter a system of linear equations where the solution set forms a line, you're dealing with a system that has infinitely many solutions. This happens when the system is consistent (the equations don't contradict each other) but has dependent equations - meaning one equation can be derived from the others. To find when this occurs, we need the coefficient matrix to have rank 2 (since we want a line, not a unique point). Let's use elimination to find the value of kk. From the first two equations: x+y+z=6x + y + z = 6 x+2y+3z=14x + 2y + 3z = 14 Subtracting the first from the second: y+2z=8y + 2z = 8 Now, let's eliminate xx from the third equation using the first: 2x+3y+kz=202x + 3y + kz = 20 2(x+y+z)=122(x + y + z) = 12, so 2x+2y+2z=122x + 2y + 2z = 12 Subtracting: (2x+3y+kz)−(2x+2y+2z)=20−12(2x + 3y + kz) - (2x + 2y + 2z) = 20 - 12 This gives us: y+(k−2)z=8y + (k-2)z = 8 For the system to have infinitely many solutions, this equation must be identical to y+2z=8y + 2z = 8. Therefore: k−2=2k - 2 = 2, so k=4k = 4. Looking at the wrong answers: A) k=2k = 2 would give y=8y = 8, creating an inconsistent system. C) k=3k = 3 would yield y+z=8y + z = 8, which contradicts y+2z=8y + 2z = 8. D) k=5k = 5 would produce y+3z=8y + 3z = 8, also inconsistent. The answer is B. Study tip: When a linear system's solution set forms a line, look for the value that makes the reduced equations identical - this ensures infinite solutions along that line.

Question 19

Consider the system of linear equations:

x+y−z=12x+3y+kz=3x+ky+3z=2x + y - z = 1 \\ 2x + 3y + kz = 3 \\ x + ky + 3z = 2

For what value of kk does the system have infinitely many solutions?

  1. k=−3k = -3
  2. k=−2k = -2
  3. k=2k = 2 (correct answer)
  4. k=3k = 3
Explanation: To determine the nature of the solution, we use Gaussian elimination. The augmented matrix is: [11−1123k31k32]\left[\begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 2 & 3 & k & 3 \\ 1 & k & 3 & 2 \end{array}\right] After row reduction (as shown in the previous problem's explanation), the third row becomes: [00−(k+3)(k−2)∣2−k][0 \quad 0 \quad -(k+3)(k-2) \quad | \quad 2-k] For the system to have infinitely many solutions, the last row must be of the form [0 0 0 ∣ 0][0 \ 0 \ 0 \ | \ 0]. This requires both the left side and the right side to be zero. −(k+3)(k−2)=0-(k+3)(k-2) = 0 implies k=−3k=-3 or k=2k=2. 2−k=02-k = 0 implies k=2k=2. The only value that satisfies both conditions is k=2k=2. When k=2k=2, the last row is [0 0 0 ∣ 0][0 \ 0 \ 0 \ | \ 0], indicating infinitely many solutions. Distractor Rationale:
  • A: If k=−3k=-3, the last row becomes [0 0 0 ∣ 5][0 \ 0 \ 0 \ | \ 5], which corresponds to no solution.
  • B and D are based on potential algebraic errors.

Question 20

A system of three linear equations in x,y,zx, y, z is inconsistent. Which of the following row-reduced echelon forms for its augmented matrix is NOT possible?

  1. [102501320001]\left[\begin{array}{ccc|c} 1 & 0 & 2 & 5 \\ 0 & 1 & 3 & 2 \\ 0 & 0 & 0 & 1 \end{array}\right]
  2. [1−10400130000]\left[\begin{array}{ccc|c} 1 & -1 & 0 & 4 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right] (correct answer)
  3. [120000100001]\left[\begin{array}{ccc|c} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]
  4. [1002012−10003]\left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\ 0 & 1 & 2 & -1 \\ 0 & 0 & 0 & 3 \end{array}\right]
Explanation: When you encounter questions about inconsistent systems of linear equations, focus on what makes a system have no solutions: contradictory equations that cannot all be satisfied simultaneously. An inconsistent system in row-reduced echelon form will have at least one row of the form [000∣c][0 \quad 0 \quad 0 \mid c] where c≠0c \neq 0. This represents the impossible equation 0=c0 = c, which has no solution. Let's examine each option. Option A has the third row [000∣1][0 \quad 0 \quad 0 \mid 1], representing 0=10 = 1 - clearly inconsistent. Option C has [000∣1][0 \quad 0 \quad 0 \mid 1] in the third row, representing 0=10 = 1 - also inconsistent. Option D has [000∣3][0 \quad 0 \quad 0 \mid 3] in the third row, representing 0=30 = 3 - inconsistent as well. Option B, however, has the system:
  • x−y=4x - y = 4
  • z=3z = 3
  • 0=00 = 0
The third row [000∣0][0 \quad 0 \quad 0 \mid 0] represents 0=00 = 0, which is always true. This system actually has infinitely many solutions (you can choose any value for yy, then x=y+4x = y + 4 and z=3z = 3). This makes the system consistent, not inconsistent. Since the question asks which form is NOT possible for an inconsistent system, option B is correct because it represents a consistent system with infinitely many solutions. Study tip: For inconsistency, look for rows where all coefficients are zero but the constant term is non-zero. If the constant is also zero, the system remains consistent.