Finite Mathematics Quiz: Systems Via Matrices And Row Reduction

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Systems Via Matrices And Row ReductionQuestion 1 of 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}

Unique solution at the point (3, 4, 0) where three planes intersect at exactly one point

Infinitely many solutions forming a line in 3D space where three planes intersect along a common line

No solution because the third equation represents an impossible constraint on the variables

Infinitely many solutions forming a plane in 3D space where two planes are coincident

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Finite Mathematics Quiz

Finite Mathematics Quiz: Systems Via Matrices And Row Reduction

Practice Systems Via Matrices And Row Reduction in Finite Mathematics 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 And Row Reduction, giving you a quick way to practice the rules, question types, and explanations that matter most for Finite Mathematics.

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Question 1

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}

Unique solution at the point (3, 4, 0) where three planes intersect at exactly one point
Infinitely many solutions forming a line in 3D space where three planes intersect along a common line (correct answer)
No solution because the third equation represents an impossible constraint on the variables
Infinitely many solutions forming a plane in 3D space where two planes are coincident

Explanation: 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.

Question 2

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}

$x_1 = 2 + \frac{1}{2}t, x_2 = t, x_3 = 1$ where $t$ is any real number (correct answer)
$x_1 = 1 + \frac{1}{2}s, x_2 = s, x_3 = 2$ where $s$ is any real number
$x_1 = 3 + \frac{1}{2}u, x_2 = u, x_3 = 1$ where $u$ is any real number
$x_1 = 2 - \frac{1}{2}v, x_2 = v, x_3 = 1$ where $v$ is any real number

Explanation: 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.

Question 3

A student is solving the system $$ \begin{cases} x + 2y - z = 4 \ 3x + 5y + 2z = 7 \ 2x + 4y - 2z = 8 \end{cases}

Only after performing back-substitution and finding that variables can take arbitrary values
After eliminating the first column below the pivot and observing that one equation becomes redundant
After completing the entire forward elimination phase and seeing the resulting triangular form
Immediately after forming the augmented matrix and noticing coefficient patterns in the original equations (correct answer)

Explanation: 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.

Question 4

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?

After the initial pivot operation and one row replacement involving the second row
After performing two row replacements and then scaling the second row appropriately
After completing all row replacements for the first column and beginning second column operations (correct answer)
After the final back-substitution phase when converting to reduced row echelon form

Explanation: Starting with the augmented matrix $$ \begin{bmatrix} 1 & 2 & 1 & | & 100 \ 2 & 1 & 3 & | & 150 \ 1 & -1 & 2 & | & 50 \end{bmatrix}

Question 5

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?

$k = 8$ because this makes the second row a multiple of the first row (correct answer)
$k = 12$ because this creates proportional coefficients across all three equations
$k = 6$ because this makes rows two and three identical after row operations
$k = 4$ because this eliminates contradictions between the first two equations

Explanation: 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.

Question 6

A student attempts to solve a system of equations by row-reducing the augmented matrix A. The student's work is shown below:

Initial Matrix: $A = \begin{pmatrix} 2 & 4 & -2 & | & 6 \\ 1 & 3 & 1 & | & 5 \\ 1 & 1 & -2 & | & -1 \end{pmatrix}$

Step 1: $R_1 \leftrightarrow R_2$ $\begin{pmatrix} 1 & 3 & 1 & | & 5 \\ 2 & 4 & -2 & | & 6 \\ 1 & 1 & -2 & | & -1 \end{pmatrix}$

Step 2: $R_2 - 2R_1 \to R_2$ $\begin{pmatrix} 1 & 3 & 1 & | & 5 \\ 0 & -2 & -4 & | & -4 \\ 1 & 1 & -2 & | & -1 \end{pmatrix}$

Step 3: $R_3 - R_1 \to R_3$ $\begin{pmatrix} 1 & 3 & 1 & | & 5 \\ 0 & -2 & -4 & | & -4 \\ 0 & -2 & -1 & | & -6 \end{pmatrix}$

Step 4: $R_3 - R_2 \to R_3$ $\begin{pmatrix} 1 & 3 & 1 & | & 5 \\ 0 & -2 & -4 & | & -4 \\ 0 & 0 & 3 & | & -2 \end{pmatrix}$

In which step did the student first make an error?

Step 1
Step 2
Step 3 (correct answer)
Step 4

Explanation: 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.

Question 7

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}

The system is inconsistent and has no solution.
The system has a unique solution where $z=0$ .
The original system must have contained the equation $0=0$ .
The system is dependent and has infinitely many solutions. (correct answer)

Explanation: 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.

Question 8

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}

$R_2 \leftrightarrow R_3$ (correct answer)
$\frac{1}{2}R_2 \to R_2$
$R_1 - 3R_3 \to R_1$
$R_2 + R_3 \to R_2$

Explanation: 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.

Question 9

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}

$c = 1$
$c = 3$ (correct answer)
$c = 9$
No value of $c$ exists.

Explanation: 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.

Question 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}

$x + 2y = -1$ and $3x - y = 11$ (correct answer)
$x + 2y = 1$ and $x - y = -5$
$x + 2y = -1$ and $2x + 4y = 5$
$x + 2y = -1$ and $-2x - 4y = 2$

Explanation: 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)

Question 11

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?

$\begin{pmatrix} 1 & 1 & 1 & | & 50 \\ 1 & -1 & 1 & | & 4 \\ 20 & 18 & 22 & | & 1008 \end{pmatrix}$
$\begin{pmatrix} 1 & 1 & 1 & | & 50 \\ -1 & 1 & -1 & | & -4 \\ 20 & 18 & 22 & | & 1008 \end{pmatrix}$ (correct answer)
$\begin{pmatrix} 1 & 1 & 1 & | & 50 \\ 1 & 1 & 1 & | & 4 \\ 20 & 18 & 22 & | & 1008 \end{pmatrix}$
$\begin{pmatrix} 1 & 1 & 1 & | & 50 \\ 1 & -1 & 1 & | & -4 \\ 18 & 20 & 22 & | & 1008 \end{pmatrix}$

Explanation: We need to translate each sentence into an equation.

'...sold a total of 50 cakes': $c + v + r = 50$ . This corresponds to the first row $[1, 1, 1, | , 50]$ .
'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 $[-1, 1, -1, | , -4]$ .
'...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.

Question 12

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}

The system has a unique solution at $x=5$ , $y=-1$ , $z=0$ .
The system has no solution because of the row of zeros.
For any real number $t$ , $x = 5 + 2t$ , $y = -1 - 3t$ , $z = t$ . (correct answer)
For any real number $t$ , $x = 5 + 2t$ , $y = -1 + 3t$ , $z = t$ .

Explanation: 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

Question 13

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}

A unique solution exists.
Infinitely many solutions exist. (correct answer)
There is no solution.
The only solution is the trivial solution $(0,0,0)$ .

Explanation: 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.

Question 14

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}

$k = -3$ (correct answer)
$k = 2$
$k = 1$
$k = 4$

Explanation: 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

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

Question 15

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}

$0$
$-2$
$-6$ (correct answer)
$-10$

Explanation: 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

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

-1 - 2 + (-3) = -6