Systems via Matrices & Row Reduction - Finite Mathematics

1. A system of linear equations in variables $x, y,$ and $z$ is reduced to the augmented matrix shown below. Which of the following correctly describes the general solution to the system? $\begin{pmatrix} 1 & 0 & -2 & | & 5 \\ 0 & 1 & 3 & | & -1 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$
   Answer: For any real number $t$, $x = 5 + 2t$, $y = -1 - 3t$, $z = t$. The matrix is in reduced row-echelon form. The last row, $[0, 0, 0, | , 0]$, represents the equation $0x+0y+0z=0$, which is always true. This indicates that the system is dependent and has infinitely many solutions. The first row represents $x - 2z = 5$. The second row represents $y + 3z = -1$. Since the third column does not have a leading 1, $z$ is a free variable. We can set $z=t$ for any real number $t$. From the first equation, $x = 5 + 2z = 5 + 2t$. From the second equation, $y = -1 - 3z = -1 - 3t$. Thus, the general solution is $x = 5 + 2t$, $y = -1 - 3t$, $z=t$.
2. A manufacturing system can be modeled by the equations $x_1 + 2x_2 + x_3 = 100$, $2x_1 + x_2 + 3x_3 = 150$, and $x_1 - x_2 + 2x_3 = 50$. When solving this system using Gaussian elimination, at what step do you first obtain a leading coefficient of 1 in the second row?
   Answer: After completing all row replacements for the first column and beginning second column operations Starting with the augmented matrix $\begin{bmatrix} 1 & 2 & 1 & | & 100 \\ 2 & 1 & 3 & | & 150 \\ 1 & -1 & 2 & | & 50 \end{bmatrix}$, the first row already has a leading 1. For the first column elimination: R2 → R2 - 2R1 gives [0, -3, 1 | -50], and R3 → R3 - R1 gives [0, -3, 1 | -50]. Notice rows 2 and 3 are now identical. To get a leading 1 in the second row's second column position, we need R2 → R2/(-3), giving [0, 1, -1/3 | 50/3]. This occurs after we've completed all first column operations and begin working on the second column.
3. A system of equations has the augmented matrix $\begin{bmatrix} 1 & 3 & -2 & | & 4 \\ 2 & 6 & -4 & | & k \\ 1 & 3 & -2 & | & 6 \end{bmatrix}$. For which value of $k$ does this system represent a scenario where two of the original equations are essentially the same, but the system is still inconsistent?
   Answer: $k = 8$ because this makes the second row a multiple of the first row Looking at the matrix, we notice that row 2 has coefficients [2, 6, -4] which are exactly twice the coefficients [1, 3, -2] in row 1. For the system to have two equations that are essentially the same (proportional), the constant term in row 2 must also be twice the constant term in row 1. Since row 1 has constant term 4, we need k = 8. However, row 3 has the same coefficients as row 1 but a different constant term (6 ≠ 4), creating an inconsistency. When k = 8, rows 1 and 2 represent the same equation, but row 3 contradicts both, making the system inconsistent.
4. A system of three linear equations with variables $x, y,$ and $z$ is converted to an augmented matrix and row-reduced to the form shown below. What can be concluded about the solution set of the original system? $\begin{pmatrix} 1 & 0 & 3 & | & 4 \\ 0 & 1 & -1 & | & 2 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$
   Answer: The system is dependent and has infinitely many solutions. The last row of the matrix, $[0, 0, 0, | , 0]$, corresponds to the equation $0x + 0y + 0z = 0$, which simplifies to $0=0$. This is a true statement that provides no new information, indicating that one of the original equations was redundant (a linear combination of the others). The system is therefore dependent. The presence of a free variable ($z$, since its column lacks a leading 1) and the absence of any contradiction (like $0=c$ for $c\neq0$) means there are infinitely many solutions.
5. A bakery sells three types of cakes: chocolate, vanilla, and red velvet. On a particular day, they sold a total of 50 cakes. The number of vanilla cakes sold was 4 less than the sum of the chocolate and red velvet cakes sold. Revenue from chocolate cakes was $20 per cake, vanilla was $18 per cake, and red velvet was $22 per cake, for a total revenue of $1008. Let $c, v, r$ be the number of chocolate, vanilla, and red velvet cakes sold, respectively. Which augmented matrix correctly represents this system of equations?
   Answer: $\begin{pmatrix} 1 & 1 & 1 & | & 50 \\ -1 & 1 & -1 & | & -4 \\ 20 & 18 & 22 & | & 1008 \end{pmatrix}$ We need to translate each sentence into an equation. 1. '...sold a total of 50 cakes': $c + v + r = 50$. This corresponds to the first row $[1, 1, 1, | , 50]$. 2. 'The number of vanilla cakes sold was 4 less than the sum of the chocolate and red velvet cakes sold': $v = (c+r) - 4$. To put this in standard form ($Ac+Bv+Cr=D$), we rearrange it: $-c + v - r = -4$. This corresponds to the second row $[-1, 1, -1, | , -4]$. 3. '...total revenue of $1008': 20c + 18v + 22r = 1008$. This corresponds to the third row $[20, 18, 22, | , 1008]$. Combining these rows gives the matrix in option B.
6. Consider the following system of linear equations where $k$ is a constant. For what value of $k$ does this system have no solution? $\begin{cases} x + y - z = 1 \\ 2x + 3y + kz = 3 \\ x + ky + 3z = 2 \end{cases}$
   Answer: $k = -3$ First, write the augmented matrix for the system: $\begin{pmatrix} 1 & 1 & -1 & | & 1 \\ 2 & 3 & k & | & 3 \\ 1 & k & 3 & | & 2 \end{pmatrix}$ Perform row operations: $R_2 - 2R_1 \to R_2$ and $R_3 - R_1 \to R_3$. This yields: $\begin{pmatrix} 1 & 1 & -1 & | & 1 \\ 0 & 1 & k+2 & | & 1 \\ 0 & k-1 & 4 & | & 1 \end{pmatrix}$ Next, perform $R_3 - (k-1)R_2 \to R_3$. The new third row will be: $[0, 0, 4 - (k-1)(k+2), | , 1 - (k-1)]$ $[0, 0, 4 - (k^2+k-2), | , 2-k]$ $[0, 0, -k^2-k+6, | , 2-k]$ For the system to have no solution, the last row must be of the form $[0, 0, 0, | , c]$ where $c \neq 0$. We need $-k^2-k+6=0$ and $2-k \neq 0$. $-(k^2+k-6)=0 \implies -(k+3)(k-2)=0$, so $k=-3$ or $k=2$. The condition $2-k \neq 0$ means $k \neq 2$. Therefore, the only value for which there is no solution is $k=-3$.
7. In which step did the student first make an error?
   Answer: Step 3 Let's check each step: Step 1: $R_1 \leftrightarrow R_2$. Swapping row 1 and row 2 is a valid operation and is performed correctly. Step 2: $R_2 - 2R_1 \to R_2$. Using the matrix from Step 1, the new Row 2 should be $[2, 4, -2, | , 6] - 2 \times [1, 3, 1, | , 5] = [2-2, 4-6, -2-2, | , 6-10] = [0, -2, -4, | , -4]$. This is correct. Step 3: $R_3 - R_1 \to R_3$. Using the matrix from Step 2, the new Row 3 should be $[1, 1, -2, | , -1] - [1, 3, 1, | , 5] = [1-1, 1-3, -2-1, | , -1-5] = [0, -2, -3, | , -6]$. The student's matrix shows $[0, -2, -1, | , -6]$. The calculation for the third column, $-2-1=-3$, is incorrect. The student calculated $-2-1=-1$. This is the first error. Step 4: The calculation in this step is correct based on the student's incorrect matrix from Step 3, but the error had already occurred.
8. For what value of the constant $c$ will the following system of linear equations have infinitely many solutions? $\begin{cases} x - 2y + 3z = 2 \\ 2x + y - z = 1 \\ 3x - y + 2z = c \end{cases}$
   Answer: $c = 3$ We set up the augmented matrix and perform row reduction. $\begin{pmatrix} 1 & -2 & 3 & | & 2 \\ 2 & 1 & -1 & | & 1 \\ 3 & -1 & 2 & | & c \end{pmatrix}$ Perform $R_2 - 2R_1 \to R_2$ and $R_3 - 3R_1 \to R_3$. New $R_2$: $[2,1,-1,1] - 2[1,-2,3,2] = [0, 5, -7, -3]$. New $R_3$: $[3,-1,2,c] - 3[1,-2,3,2] = [0, 5, -7, c-6]$. The matrix becomes: $\begin{pmatrix} 1 & -2 & 3 & | & 2 \\ 0 & 5 & -7 & | & -3 \\ 0 & 5 & -7 & | & c-6 \end{pmatrix}$ For infinitely many solutions, we need to obtain a row of all zeros. Perform the operation $R_3 - R_2 \to R_3$. New $R_3$: $[0, 5, -7, c-6] - [0, 5, -7, -3] = [0, 0, 0, (c-6) - (-3)] = [0, 0, 0, c-3]$. For the row to be all zeros, we must have $c-3 = 0$, which means $c=3$. If $c \neq 3$, the system would have no solution.
9. Without solving the system completely, determine the nature of the solution set for the system represented by the following augmented matrix. $\begin{pmatrix} 1 & -1 & 2 & | & 3 \\ -2 & 3 & -1 & | & -4 \\ -1 & 2 & 1 & | & -1 \end{pmatrix}$
   Answer: Infinitely many solutions exist. To determine the nature of the solution, we can begin the row reduction process. Start with the given matrix: $\begin{pmatrix} 1 & -1 & 2 & | & 3 \\ -2 & 3 & -1 & | & -4 \\ -1 & 2 & 1 & | & -1 \end{pmatrix}$ Perform $R_2 + 2R_1 \to R_2$ and $R_3 + R_1 \to R_3$. New $R_2$: $[-2, 3, -1, -4] + 2[1, -1, 2, 3] = [-2+2, 3-2, -1+4, -4+6] = [0, 1, 3, 2]$. New $R_3$: $[-1, 2, 1, -1] + [1, -1, 2, 3] = [-1+1, 2-1, 1+2, -1+3] = [0, 1, 3, 2]$. The matrix becomes: $\begin{pmatrix} 1 & -1 & 2 & | & 3 \\ 0 & 1 & 3 & | & 2 \\ 0 & 1 & 3 & | & 2 \end{pmatrix}$ Notice that Row 2 and Row 3 are identical. The next step, $R_3 - R_2 \to R_3$, will result in a row of all zeros: $[0, 0, 0, | , 0]$. This indicates that the system is dependent and has infinitely many solutions.
10. Row reduction of the augmented matrix for a system of two linear equations in two variables, $x$ and $y$, yields the matrix shown below. Which of the following could have been the original system of equations? $\begin{pmatrix} 1 & 0 & | & 3 \\ 0 & 1 & | & -2 \end{pmatrix}$
    Answer: $x + 2y = -1$ and $3x - y = 11$ The given reduced row-echelon form matrix corresponds to the solution $x=3$ and $y=-2$. Any original system of equations that is row-equivalent to this must have this unique solution. We can test each of the given systems by substituting $x=3$ and $y=-2$. A) For $x+2y=-1$: $3 + 2(-2) = 3-4 = -1$. (Correct) For $3x-y=11$: $3(3) - (-2) = 9+2 = 11$. (Correct) Since both equations are satisfied, this system has the solution $(3, -2)$ and could be the original system. B) For $x+2y=1$: $3+2(-2) = -1 \neq 1$. (Incorrect) C) This system is inconsistent ($2(x+2y) = 2(-1) = -2$, not 5), so it has no solution. (Incorrect) D) This system is dependent (the second equation is -2 times the first), so it has infinitely many solutions. (Incorrect)
11. Consider the system represented by $\begin{bmatrix} 2 & -1 & 1 & | & 5 \\ 4 & -2 & 3 & | & 11 \\ 6 & -3 & 2 & | & 14 \end{bmatrix}$. After row reduction, this matrix is equivalent to $\begin{bmatrix} 1 & -\frac{1}{2} & \frac{1}{2} & | & \frac{5}{2} \\ 0 & 0 & 1 & | & 1 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}$. What is the complete solution set expressed in parametric form?
    Answer: $x_1 = 2 + \frac{1}{2}t, x_2 = t, x_3 = 1$ where $t$ is any real number From the reduced matrix, we have the equations x₁ - (1/2)x₂ + (1/2)x₃ = 5/2 and x₃ = 1. Since x₃ = 1, substituting into the first equation: x₁ - (1/2)x₂ + (1/2)(1) = 5/2, which gives x₁ - (1/2)x₂ = 2, so x₁ = 2 + (1/2)x₂. Since x₂ is a free variable (let x₂ = t), the parametric solution is x₁ = 2 + (1/2)t, x₂ = t, x₃ = 1.
12. The augmented matrix for a linear system is shown below. Which of the following row operations is a valid and strategically sound next step in the process of Gaussian elimination to achieve row-echelon form? $\begin{pmatrix} 1 & 3 & -2 & | & 1 \\ 0 & 0 & 2 & | & 4 \\ 0 & 1 & 5 & | & 3 \end{pmatrix}$
    Answer: $R_2 \leftrightarrow R_3$ The goal of Gaussian elimination is to transform the matrix into row-echelon form. This involves creating a triangular structure with leading 1s (pivots) for each row, moving from top to bottom. The first column is already in the desired form, with a pivot in the first row. For the second column, we need a non-zero entry in the second row to serve as a pivot. Currently, the entry is 0. The standard algorithm dictates that we should swap with a lower row to bring a non-zero entry into the pivot position. Row 3 has a 1 in the second column. Therefore, swapping Row 2 and Row 3 ($R_2 \leftrightarrow R_3$) is the most direct and strategically sound next step. Other options are either less efficient or steps that should be performed later in the process.
13. A student is solving the system $\begin{cases} x + 2y - z = 4 \\ 3x + 5y + 2z = 7 \\ 2x + 4y - 2z = 8 \end{cases}$ using matrix row operations. After the first elimination step, they claim the system has infinitely many solutions. At what point in their row reduction process would this conclusion become apparent?
    Answer: Immediately after forming the augmented matrix and noticing coefficient patterns in the original equations When solving systems of linear equations, you can often determine the nature of the solution (unique, none, or infinitely many) by examining the coefficient patterns in the original equations before doing any row operations. Looking at this system, notice that the third equation $2x + 4y - 2z = 8$ is exactly twice the first equation $x + 2y - z = 4$. When you multiply the first equation by 2, you get $2x + 4y - 2z = 8$, which is identical to the third equation. This means the third equation is redundant—it provides no new information beyond what the first equation already tells us. Since we effectively have only two independent equations but three unknowns, the system must have infinitely many solutions. This becomes apparent immediately upon inspection of the augmented matrix, making D correct. A is wrong because back-substitution comes much later in the process, and you don't need to reach that stage to recognize infinite solutions when equations are obviously dependent. B is incorrect because the redundancy is visible before any elimination steps—you don't need to eliminate the first column to see it. C is also wrong because you don't need to complete forward elimination; the relationship between equations 1 and 3 is immediately obvious. Study tip: Before diving into row operations, always scan the original equations for obvious relationships like one equation being a multiple of another. This quick check can save you time and immediately reveal whether the system has infinitely many solutions or no solution.
14. For the system of equations given below, what is the value of the expression $x - y + z$? $\begin{cases} x + 2y - z = 6 \\ 2x - y + 3z = -13 \\ 3x - 2y + 3z = -16 \end{cases}$
    Answer: $-6$ First, solve the system. The augmented matrix is: $\begin{pmatrix} 1 & 2 & -1 & | & 6 \\ 2 & -1 & 3 & | & -13 \\ 3 & -2 & 3 & | & -16 \end{pmatrix}$ Perform row reduction: $R_2 - 2R_1 \to R_2$ gives $[0, -5, 5, | , -25]$. $R_3 - 3R_1 \to R_3$ gives $[0, -8, 6, | , -34]$. The matrix becomes: $\begin{pmatrix} 1 & 2 & -1 & | & 6 \\ 0 & -5 & 5 & | & -25 \\ 0 & -8 & 6 & | & -34 \end{pmatrix}$ Simplify $R_2$ by $R_2 / -5 \to R_2$: $[0, 1, -1, | , 5]$. The matrix is now: $\begin{pmatrix} 1 & 2 & -1 & | & 6 \\ 0 & 1 & -1 & | & 5 \\ 0 & -8 & 6 & | & -34 \end{pmatrix}$ Perform $R_3 + 8R_2 \to R_3$: $[0, 0, -2, | , 6]$. From the new $R_3$, we have $-2z = 6$, so $z=-3$. Substitute $z=-3$ into the simplified $R_2$ equation: $y - z = 5 \implies y - (-3) = 5 \implies y=2$. Substitute $y=2$ and $z=-3$ into the $R_1$ equation: $x+2(2)-(-3) = 6 \implies x+4+3=6 \implies x=-1$. The solution is $x=-1, y=2, z=-3$. The value of $x - y + z$ is $-1 - 2 + (-3) = -6$.
15. In solving a system using row reduction, a student obtains the matrix $\begin{bmatrix} 1 & 0 & 2 & | & 3 \\ 0 & 1 & -1 & | & 4 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}$. If the original system had three equations in three unknowns, which statement correctly describes both the solution set and the geometric interpretation?
    Answer: Infinitely many solutions forming a line in 3D space where three planes intersect along a common line The reduced matrix shows that we have two equations in three unknowns: x + 2z = 3 and y - z = 4, with z as a free parameter. We can write the solution as x = 3 - 2t, y = 4 + t, z = t where t is any real number. This represents a line in 3D space. Geometrically, the original three planes intersect along this line. The third row [0 0 0 | 0] indicates that one of the original equations was redundant (a linear combination of the other two), so effectively we only have two independent planes, and their intersection is a line.

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