Systems via Matrices - IB Mathematics Analysis and Approaches

A system of three linear equations in $x, y, z$ is inconsistent. Which of the following row-reduced echelon forms for its augmented matrix is NOT possible?
Answer: $\left[\begin{array}{ccc|c} 1 & -1 & 0 & 4 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right]$ 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 $[0 \quad 0 \quad 0 \mid c]$ where $c \neq 0$ . This represents the impossible equation $0 = c$ , which has no solution. Let's examine each option. Option A has the third row $[0 \quad 0 \quad 0 \mid 1]$ , representing $0 = 1$ - clearly inconsistent. Option C has $[0 \quad 0 \quad 0 \mid 1]$ in the third row, representing $0 = 1$ - also inconsistent. Option D has $[0 \quad 0 \quad 0 \mid 3]$ in the third row, representing $0 = 3$ - inconsistent as well. Option B, however, has the system: - $x - y = 4$ - $z = 3$ - $0 = 0$ The third row $[0 \quad 0 \quad 0 \mid 0]$ represents $0 = 0$ , which is always true. This system actually has infinitely many solutions (you can choose any value for $y$ , then $x = y + 4$ and $z = 3$ ). 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.
For which condition on $a$ and $b$ is the following system of equations consistent? $x - 2y + z = 3 \\ 2x + y - 2z = a \\ 4x - 3y = b$
Answer: $b = a + 6$ A system is consistent if it has at least one solution. We set up the augmented matrix and perform row reduction. $\left[\begin{array}{ccc|c} 1 & -2 & 1 & 3 \\ 2 & 1 & -2 & a \\ 4 & -3 & 0 & b \end{array}\right]$ Perform $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 4R_1$ . $\left[\begin{array}{ccc|c} 1 & -2 & 1 & 3 \\ 0 & 5 & -4 & a-6 \\ 0 & 5 & -4 & b-12 \end{array}\right]$ Next, perform $R_3 \to R_3 - R_2$ . $\left[\begin{array}{ccc|c} 1 & -2 & 1 & 3 \\ 0 & 5 & -4 & a-6 \\ 0 & 0 & 0 & (b-12) - (a-6) \end{array}\right]$ The last row simplifies to $[0 \ 0 \ 0 \ | \ b - a - 6]$ . For the system to be consistent, this last row must not be a contradiction. This means the last entry must be zero, as $0x+0y+0z$ must equal zero. So, $b - a - 6 = 0$ , which implies $b = a + 6$ . Distractor Rationale: - B: Sign error in $b-12 - (a-6) = b-12-a+6 = b-a-6$ . A common mistake is $b-12-a-6 = b-a-18$ leading to $b=a+18$ . Or sign error on the 6. - D: Incorrect rearrangement of the final equation. - A: Another possible arithmetic error.
Consider the system of linear equations: $x + y - z = 1 \\ 2x + 3y + kz = 3 \\ x + ky + 3z = 2$ For what value of $k$ does the system have infinitely many solutions?
Answer: $k = 2$ To determine the nature of the solution, we use Gaussian elimination. The augmented matrix is: $\left[\begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 2 & 3 & k & 3 \\ 1 & k & 3 & 2 \end{array}\right]$ After row reduction (as shown in the previous problem's explanation), the third row becomes: $[0 \quad 0 \quad -(k+3)(k-2) \quad | \quad 2-k]$ For the system to have infinitely many solutions, the last row must be of the form $[0 \ 0 \ 0 \ | \ 0]$ . This requires both the left side and the right side to be zero. $-(k+3)(k-2) = 0$ implies $k=-3$ or $k=2$ . $2-k = 0$ implies $k=2$ . The only value that satisfies both conditions is $k=2$ . When $k=2$ , the last row is $[0 \ 0 \ 0 \ | \ 0]$ , indicating infinitely many solutions. Distractor Rationale: - A: If $k=-3$ , the last row becomes $[0 \ 0 \ 0 \ | \ 5]$ , which corresponds to no solution. - B and D are based on potential algebraic errors.
For which condition on $k$ does the following system of equations have a unique solution? $x + 2y + z = 1 \\ -x + 4y + z = 3 \\ 2x - 2y + kz = 0$
Answer: $k \neq 0$ A system of linear equations has a unique solution if and only if the determinant of the coefficient matrix is non-zero. The coefficient matrix $A$ is: $A = \begin{pmatrix} 1 & 2 & 1 \\ -1 & 4 & 1 \\ 2 & -2 & k \end{pmatrix}$ Calculating the determinant: $\det(A) = 1(4k - (-2)) - 2(-k - 2) + 1(2 - 8) \\ = (4k+2) - (-2k-4) + (-6) \\ = 4k+2+2k+4-6 \\ = 6k$ For a unique solution, $\det(A) \neq 0$ , which means $6k \neq 0$ , so $k \neq 0$ . Alternatively, using row reduction on the augmented matrix: $\left[\begin{array}{ccc|c} 1 & 2 & 1 & 1 \\ -1 & 4 & 1 & 3 \\ 2 & -2 & k & 0 \end{array}\right] \xrightarrow[R_3 \to R_3-2R_1]{R_2 \to R_2+R_1} \left[\begin{array}{ccc|c} 1 & 2 & 1 & 1 \\ 0 & 6 & 2 & 4 \\ 0 & -6 & k-2 & -2 \end{array}\right] \xrightarrow{R_3 \to R_3+R_2} \left[\begin{array}{ccc|c} 1 & 2 & 1 & 1 \\ 0 & 6 & 2 & 4 \\ 0 & 0 & k & 2 \end{array}\right]$ For a unique solution, the system must not have a row of zeros on the left side, which would lead to no solution or infinite solutions. Specifically, the pivot in the third row, $k$ , must be non-zero. Thus, $k \neq 0$ . Distractor Rationale: - A: $k=0$ is the condition for the system to not have a unique solution (in this case, it leads to no solution). - B: A specific incorrect value. - D: Incorrectly assumes the system always has a unique solution regardless of $k$ .
The augmented matrix for a system of three linear equations in three variables is $[A|\mathbf{b}]$ . It is known that the rank of the coefficient matrix $A$ is 2, and the rank of the augmented matrix $[A|\mathbf{b}]$ is also 2. What can be concluded about the solution set of the system?
Answer: The solutions form a line. 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 $\text{rank}(A) = \text{rank}([A|\mathbf{b}])$ , 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 $3 - 2 = 1$ 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 $\text{rank}(A) = 3$ (equal to the number of variables). With rank 2, you don't have enough independent constraints. Answer B is wrong because when $\text{rank}(A) = \text{rank}([A|\mathbf{b}])$ , the system is always consistent—inconsistency only occurs when $\text{rank}([A|\mathbf{b}]) > \text{rank}(A)$ . Answer C represents a common misconception: a plane would require rank 1, giving you $3 - 1 = 2$ free variables and thus a two-dimensional solution set. Remember this pattern: in an $n$ -variable system, if $\text{rank}(A) = \text{rank}([A|\mathbf{b}]) = r$ , then you have $n - r$ 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.
The solution set of the system $x + y + z = 6 \\ x + 2y + 3z = 14 \\ 2x + 3y + kz = 20$ forms a line. Find the value of $k$ .
Answer: $k=4$ 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 $k$ . From the first two equations: $x + y + z = 6$ $x + 2y + 3z = 14$ Subtracting the first from the second: $y + 2z = 8$ Now, let's eliminate $x$ from the third equation using the first: $2x + 3y + kz = 20$ $2(x + y + z) = 12$ , so $2x + 2y + 2z = 12$ Subtracting: $(2x + 3y + kz) - (2x + 2y + 2z) = 20 - 12$ This gives us: $y + (k-2)z = 8$ For the system to have infinitely many solutions, this equation must be identical to $y + 2z = 8$ . Therefore: $k - 2 = 2$ , so $k = 4$ . Looking at the wrong answers: A) $k = 2$ would give $y = 8$ , creating an inconsistent system. C) $k = 3$ would yield $y + z = 8$ , which contradicts $y + 2z = 8$ . D) $k = 5$ would produce $y + 3z = 8$ , 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.
The augmented matrix of a system of linear equations $(x, y, z)$ is reduced to the following form: $\left[\begin{array}{ccc|c} 1 & 0 & 2 & 5 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right]$ Which of the following represents the general solution to the system in terms of a parameter $t$ ?
Answer: $x = 5 - 2t, y = 2 + t, z = t$ The reduced row echelon form of the matrix corresponds to the following system of equations: $x + 2z = 5 \\ y - z = 2 \\ 0 = 0$ The third equation $0=0$ indicates that the system is dependent and has infinitely many solutions. We can introduce a parameter for one of the variables. Let $z = t$ , where $t$ is any real number. From the second equation, $y - t = 2$ , which gives $y = 2 + t$ . From the first equation, $x + 2t = 5$ , which gives $x = 5 - 2t$ . So, the general solution is $(x, y, z) = (5 - 2t, 2 + t, t)$ . Distractor Rationale: - B: Sign errors when solving for $x$ and $y$ from the equations. - C: This is a particular solution (when $t=0$ ), but not the general solution. - D: This would be the conclusion if the last row were $[0 \ 0 \ 0 \ | \ c]$ with $c \neq 0$ . Misinterpreting the row of zeros.
For what value of $c$ is the following system of equations consistent? $x + y = 3 \\ x - y = 1 \\ 2x + 3y = c$
Answer: $c=7$ For the system to be consistent, there must be a single point $(x,y)$ 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: $(x+y) + (x-y) = 3+1 \implies 2x = 4 \implies x=2$ . Substituting $x=2$ into the first equation: $2+y=3 \implies y=1$ . The point of intersection of the first two lines is $(2,1)$ . For the system to be consistent, this point must also lie on the third line, $2x+3y=c$ . Substituting $x=2$ and $y=1$ into the third equation: $2(2) + 3(1) = c$ $4 + 3 = c$ $c = 7$ . 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.
Let $p, q, r$ be the number of units of P, Q, and R produced, respectively. Find the value of $p$ .
Answer: 2 Based on the information, we can set up a system of linear equations: $p + 2q + 3r = 11 \quad (\text{Machine M1}) \\ 2p + q + 2r = 9 \quad (\text{Machine M2}) \\ p + 3q + r = 12 \quad (\text{Machine M3})$ We write this as an augmented matrix and solve: $\left[\begin{array}{ccc|c} 1 & 2 & 3 & 11 \\ 2 & 1 & 2 & 9 \\ 1 & 3 & 1 & 12 \end{array}\right]$ $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - R_1$ : $\left[\begin{array}{ccc|c} 1 & 2 & 3 & 11 \\ 0 & -3 & -4 & -13 \\ 0 & 1 & -2 & 1 \end{array}\right]$ Swap $R_2$ and $R_3$ : $\left[\begin{array}{ccc|c} 1 & 2 & 3 & 11 \\ 0 & 1 & -2 & 1 \\ 0 & -3 & -4 & -13 \end{array}\right]$ $R_3 \to R_3 + 3R_2$ : $\left[\begin{array}{ccc|c} 1 & 2 & 3 & 11 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & -10 & -10 \end{array}\right]$ From the last row, $-10r = -10 \implies r=1$ . From the second row, $q - 2r = 1 \implies q - 2(1) = 1 \implies q = 3$ . From the first row, $p + 2q + 3r = 11 \implies p + 2(3) + 3(1) = 11 \implies p + 6 + 3 = 11 \implies p + 9 = 11 \implies p=2$ . The solution is $p=2, q=3, r=1$ . The question asks for the value of $p$ . Distractor Rationale: - A: This is the value of $r$ . - C: This is the value of $q$ . - D: This is the value of $p+q$ .
A system of three linear equations in three variables has an infinite number of solutions. Which of the following geometric interpretations is NOT possible?
Answer: Two planes are parallel and distinct, and the third plane intersects them. An infinite number of solutions means there is at least a line of points common to all three planes. - A: If the three planes are coincident, every point on the plane is a solution. This is an infinite set of solutions (a plane of solutions). This is possible. - B: If the three planes intersect in a single line, every point on that line is a solution. This is an infinite set of solutions. This is possible. - C: If two planes are coincident, they are essentially the same plane. If a third plane intersects this plane, it does so in a line. Every point on this line is a solution. This is possible. - D: If two planes are parallel and distinct, they have no points in common. Therefore, there can be no point that lies on all three planes, regardless of the orientation of the third plane. This configuration results in zero solutions. Thus, it is not a possible interpretation for a system with infinitely many solutions. Distractor Rationale: - A, B, and C are all valid geometric interpretations for a system with infinitely many solutions. A student must recognize which configuration leads to no solution.
For what value of $k$ does the following homogeneous system of equations have non-trivial solutions? $x + ky + z = 0 \\ 2x + y - z = 0 \\ 3x + 2y = 0$
Answer: $k = 1$ A homogeneous system $A\mathbf{x} = \mathbf{0}$ has non-trivial solutions (solutions other than $\mathbf{x} = \mathbf{0}$ ) if and only if the determinant of the coefficient matrix $A$ is zero. The coefficient matrix is: $A = \begin{pmatrix} 1 & k & 1 \\ 2 & 1 & -1 \\ 3 & 2 & 0 \end{pmatrix}$ We calculate the determinant of $A$ : $\det(A) = 1(1(0) - (-1)(2)) - k(2(0) - (-1)(3)) + 1(2(2) - 1(3)) \\ = 1(0 + 2) - k(0 + 3) + 1(4 - 3) \\ = 1(2) - k(3) + 1(1) \\ = 2 - 3k + 1 \\ = 3 - 3k$ For non-trivial solutions, we set $\det(A) = 0$ : $3 - 3k = 0 \implies 3 = 3k \implies k = 1$ Distractor Rationale: - A: A sign error in calculation. - B: A common test value that is incorrect here. - D: An error in solving $3-3k=0$ .
Consider a system of linear equations $A\mathbf{x} = \mathbf{b}$ , where $A$ is a $3 \times 3$ matrix. If $\det(A) = 0$ , which of the following statements must be true?
Answer: The system has either no solution or infinitely many solutions. When $\det(A) = 0$ , the matrix $A$ is singular. This means that its inverse $A^{-1}$ does not exist. Therefore, we cannot find a unique solution by calculating $\mathbf{x} = A^{-1}\mathbf{b}$ . The system does not have a unique solution. The two remaining possibilities are that the system is inconsistent (has no solution) or is dependent (has infinitely many solutions). Which of these cases occurs depends on the vector $\mathbf{b}$ . Geometrically, $\det(A)=0$ means the three planes are not independent; they could be parallel, intersect in a line, or be coincident, leading to no solution or infinitely many solutions, but never a single point of intersection. Distractor Rationale: - A: This is possible but not guaranteed. The system could have infinite solutions. - B: This is possible but not guaranteed. The system could have no solution. - D: The trivial solution $\mathbf{x} = \mathbf{0}$ is only guaranteed if the system is homogeneous (i.e., $\mathbf{b} = \mathbf{0}$ ). This is not stated in the problem.
The augmented matrix for a system of linear equations is being row-reduced. After several steps, the matrix is: $\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & -3 & 3 & -2 \\ 0 & -3 & k & 4 \end{array}\right]$ For what value of $k$ will the system be inconsistent?
Answer: $k = 3$ 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 $R_3 \to R_3 - R_2$ . $\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & -3 & 3 & -2 \\ 0 & -3-(-3) & k-3 & 4-(-2) \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & -3 & 3 & -2 \\ 0 & 0 & k-3 & 6 \end{array}\right]$ A system is inconsistent if its row-reduced form contains a row of the form $[0 \ 0 \ 0 \ | \ c]$ where $c \neq 0$ . In our case, the last row is $[0 \ 0 \ k-3 \ | \ 6]$ . 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 $k-3 = 0$ , which gives $k=3$ . Distractor Rationale: - A: A sign error when solving $k-3=0$ . - C: Mistaking the value on the right side of the third row for $k$ . - D: Mistaking the final value on the right side of the third row for $k$ .
Consider the system of linear equations: $x + y - z = 1 \\ 2x + 3y + kz = 3 \\ x + ky + 3z = 2$ For what value of $k$ does the system have no solution?
Answer: $k = -3$ To determine the nature of the solution, we can use Gaussian elimination on the augmented matrix. $\left[\begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 2 & 3 & k & 3 \\ 1 & k & 3 & 2 \end{array}\right]$ Perform row operations: $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - R_1$ . $\left[\begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 0 & 1 & k+2 & 1 \\ 0 & k-1 & 4 & 1 \end{array}\right]$ Next, perform $R_3 \to R_3 - (k-1)R_2$ . $\left[\begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 0 & 1 & k+2 & 1 \\ 0 & 0 & 4 - (k-1)(k+2) & 1 - (k-1) \end{array}\right]$ The third row simplifies to: $[0 \quad 0 \quad 4 - (k^2+k-2) \quad | \quad 2-k]$ $[0 \quad 0 \quad -k^2-k+6 \quad | \quad 2-k]$ $[0 \quad 0 \quad -(k+3)(k-2) \quad | \quad 2-k]$ For the system to have no solution, the last row must be of the form $[0 \ 0 \ 0 \ | \ c]$ where $c \neq 0$ . This requires the left side to be zero and the right side to be non-zero. $-(k+3)(k-2) = 0$ gives $k=-3$ or $k=2$ . $2-k \neq 0$ gives $k \neq 2$ . Both conditions are met only when $k=-3$ . When $k=-3$ , the last row is $[0 \ 0 \ 0 \ | \ 5]$ , indicating no solution. Distractor Rationale: - C: If $k=2$ , the last row becomes $[0 \ 0 \ 0 \ | \ 0]$ , which corresponds to infinitely many solutions. - B and D are based on potential algebraic errors during row reduction.
Consider the system of equations: $x + y + 2z = 4 \\ 2x - y + z = 2$ Which of the following correctly describes the solution set?
Answer: The solutions form a line. 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 ( $x$ , $y$ , $z$ ), 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: $y = 4 - x - 2z$ . Substituting into the second equation: $2x - (4 - x - 2z) + z = 2$ , which simplifies to $3x + 3z = 6$ , or $x = 2 - z$ . Therefore: $y = 4 - (2 - z) - 2z = 2 - z$ . The solution can be written parametrically as $(x, y, z) = (2 - t, 2 - t, t)$ where $t$ 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 $\mathbb{R}^3$ ) down to a line, not a plane. Remember: For linear systems, count the relationships. With $n$ variables and $r$ independent equations where $r < n$ , expect an $(n-r)$ -dimensional solution set. Here, $3-2=1$ dimension means a line.
A student is solving the system $x + 2y = 3, 2x + 3y = 4$ using an augmented matrix. Their work is shown below. Step 0: $\left[\begin{array}{cc|c} 1 & 2 & 3 \\ 2 & 3 & 4 \end{array}\right]$ Step 1: $R_2 \to R_2 - 2R_1$ gives $\left[\begin{array}{cc|c} 1 & 2 & 3 \\ 0 & 1 & -2 \end{array}\right]$ Step 2: $R_1 \to R_1 - 2R_2$ gives $\left[\begin{array}{cc|c} 1 & 0 & 7 \\ 0 & 1 & -2 \end{array}\right]$ From this, the student concludes the solution is $(7, -2)$ . In which step did the student make their first mistake?
Answer: Step 1 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 $R_2 \to R_2 - 2R_1$ . The original $R_2$ is $[2 \ 3 \ | \ 4]$ and $2R_1$ is $2 \times [1 \ 2 \ | \ 3] = [2 \ 4 \ | \ 6]$ . So, $R_2 - 2R_1 = [2-2 \quad 3-4 \quad | \quad 4-6] = [0 \quad -1 \quad | \quad -2]$ . The student wrote $[0 \quad 1 \quad | \quad -2]$ . 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 $(7,-2)$ 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.
The system of equations $x + y + z = 1 \\ x + 2y + 3z = 4 \\ x + 3y + kz = 9$ has a unique solution for all real values of $k$ except for $k = k_0$ . What is the nature of the solution when $k = k_0$ ?
Answer: The system has no solution. First, we find the value $k_0$ for which the system does not have a unique solution. This occurs when the determinant of the coefficient matrix is zero. $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & k \end{pmatrix}$ $\det(A) = 1(2k - 9) - 1(k - 3) + 1(3 - 2) = 2k - 9 - k + 3 + 1 = k - 5$ Setting $\det(A) = 0$ gives $k - 5 = 0$ , so $k_0 = 5$ . Now we must analyze the system for $k = 5$ . We use the augmented matrix: $\left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 5 & 9 \end{array}\right]$ Perform row operations: $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ . $\left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 2 & 4 & 8 \end{array}\right]$ Next, perform $R_3 \to R_3 - 2R_2$ . $\left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 8 - 2(3) \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 2 \end{array}\right]$ The last row represents the equation $0x + 0y + 0z = 2$ , or $0 = 2$ , which is a contradiction. Therefore, when $k = 5$ , the system has no solution. Distractor Rationale: - B: This would be the case if the final row was $[0 \ 0 \ 0 \ | \ 0]$ . A student might assume that $\det(A)=0$ always implies infinite solutions. - C: The trivial solution is irrelevant here. - D: The nature of the solution can be determined.

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