Linear Algebra Quiz: Determinants Via Row Reduction

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Determinants Via Row ReductionQuestion 1 of 16

Let $A$ be an $n \times n$ matrix with determinant $d$ . A matrix $B$ is formed by performing the operation $R_i \to R_i - kR_j$ on $A$ (where $i \neq j$ and $k \neq 0$ ), and then a matrix $C$ is formed by swapping rows $i$ and $j$ of $B$ . Which expression represents $\det(C)$ ?

d

-d

-kd

d - kd

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Linear Algebra Quiz

Linear Algebra Quiz: Determinants Via Row Reduction

Practice Determinants Via Row Reduction in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Determinants Via Row Reduction, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$d$
$-d$ (correct answer)
$-kd$
$d - kd$

Explanation: Let's analyze the operations step by step.

The transformation from $A$ to $B$ is $R_i \to R_i - kR_j$ . This is a row replacement operation, which does not change the determinant. Therefore, $\det(B) = \det(A) = d$ .
The transformation from $B$ to $C$ is swapping two distinct rows, $i$ and $j$ . A row swap multiplies the determinant by $-1$ . Therefore, $\det(C) = -1 \times \det(B) = -d$ .

The scalar

k

from the first operation does not affect the final determinant, which is a common point of confusion.

Question 2

Let $A$ and $B$ be $3 \times 3$ matrices. Suppose $B$ is obtained from $A$ by the sequence of operations: $R_1 \leftrightarrow R_2$ , followed by $R_3 \to 5R_3$ . If $\det(A) = 2$ , what is $\det(2A^T B^{-1})$ ?

$-\frac{8}{5}$ (correct answer)
$-\frac{2}{5}$
$\frac{8}{5}$
$16$

Explanation: This problem requires combining several determinant properties.

Find $\det(B)$ : The operation $R_1 \leftrightarrow R_2$ multiplies the determinant by $-1$ . The operation $R_3 \to 5R_3$ multiplies it by $5$ . So, $\det(B) = (-1)(5) \det(A) = -5(2) = -10$ .
Use determinant properties: We need $\det(2A^T B^{-1})$ . For $n \times n$ matrices, $\det(kM) = k^n \det(M)$ , $\det(M^T) = \det(M)$ , and $\det(M^{-1}) = 1/\det(M)$ .
Apply the properties: Here, $n=3$ . $\det(2A^T B^{-1}) = \det(2I \cdot A^T \cdot B^{-1})$ $= \det(2I) \cdot \det(A^T) \cdot \det(B^{-1})$ $= 2^3 \cdot \det(A) \cdot \frac{1}{\det(B)}$ $= 8 \cdot (2) \cdot \frac{1}{-10}$ $= 16 \cdot (-\frac{1}{10}) = -\frac{16}{10} = -\frac{8}{5}$ .

Question 3

Let $A$ be a $4 \times 4$ matrix with $\det(A) = 3$ . What is the determinant of the matrix $2A$ ?

$6$
$24$
$48$ (correct answer)
$243$

Explanation: This question tests the property of scaling an entire matrix. If

A

is an

n \times n

matrix and

k

is a scalar, then

\det(kA) = k^n \det(A)

. In this case,

A

is a

4 \times 4

matrix, so

n=4

. The scalar is

k=2

. We are given

\det(A) = 3

. Therefore,

\det(2A) = 2^4 \det(A) = 16 \times 3 = 48

. Distractor (A)

6

results from the common error of confusing this property with scaling a single row, which would give

2 \det(A)

. Distractor (B)

24

results from incorrectly using

n=3

instead of

n=4

, yielding

2^3 \det(A) = 8 \times 3 = 24

Question 4

Let $A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$ be a $3 \times 3$ matrix with $\det(A) = 4$ . Consider the matrix $B = \begin{pmatrix} g & h & i \\ d-2a & e-2b & f-2c \\ a & b & c \end{pmatrix}$ . What is the value of $\det(B)$ ?

$-8$
$-4$ (correct answer)
$4$
$8$

Explanation: To find

\det(B)

, we can express

B

in terms of

A

using elementary row operations and track the effect on the determinant. Let

R_1, R_2, R_3

be the rows of

A

. The rows of

B

are

R_3

R_2 - 2R_1

, and

R_1

. Let's start with

A

and apply operations to get to

B

Start with $A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$ , with $\det(A) = 4$ .
Swap $R_1$ and $R_3$ : $A' = \begin{pmatrix} g & h & i \\ d & e & f \\ a & b & c \end{pmatrix}$ . This operation multiplies the determinant by $-1$ , so $\det(A') = -\det(A) = -4$ .
Apply $R_2 \to R_2 - 2R_3$ to matrix $A'$ . The new second row is $(d,e,f) - 2(a,b,c) = (d-2a, e-2b, f-2c)$ . The resulting matrix is $B = \begin{pmatrix} g & h & i \\ d-2a & e-2b & f-2c \\ a & b & c \end{pmatrix}$ . This row replacement operation does not change the determinant.

Therefore,

\det(B) = \det(A') = -4

Question 5

A $3 \times 3$ matrix $A$ is transformed into a matrix $B$ by the following sequence of elementary row operations: first $R_1 \to \frac{1}{2}R_1$ , then $R_2 \leftrightarrow R_3$ , and finally $R_3 \to R_3 - 5R_1$ . If $\det(B) = 10$ , what is $\det(A)$ ?

$-20$ (correct answer)
$-5$
$5$
$20$

Explanation: We can find the relationship between

\det(A)

and

\det(B)

by tracking the effect of each operation:

$R_1 \to \frac{1}{2}R_1$ : This scales the determinant by $\frac{1}{2}$ .
$R_2 \leftrightarrow R_3$ : This scales the determinant by $-1$ .
$R_3 \to R_3 - 5R_1$ : This does not change the determinant.

Combining these effects, we have

\det(B) = (\frac{1}{2})(-1)(1) \times \det(A) = -\frac{1}{2}\det(A)

. We are given

\det(B) = 10

. So,

10 = -\frac{1}{2}\det(A)

. Solving for

\det(A)

, we multiply both sides by

-2

\det(A) = 10 \times (-2) = -20

Question 6

A $4 \times 4$ matrix $A$ with $\det(A)=10$ is converted to an upper triangular matrix $U$ using the following sequence of row operations: $R_2 \to R_2 - 3R_1$ , then $R_3 \leftrightarrow R_4$ , and finally $R_4 \to 5R_4$ . What is the product of the diagonal entries of $U$ ?

$-50$ (correct answer)
$-10$
$-2$
$50$

Explanation: The product of the diagonal entries of an upper triangular matrix

U

is equal to its determinant,

\det(U)

. We need to find the relationship between

\det(A)

and

\det(U)

by tracking the effects of the row operations.

$R_2 \to R_2 - 3R_1$ : This is a row replacement operation, which does not change the determinant (multiplies by 1).
$R_3 \leftrightarrow R_4$ : This is a row swap, which multiplies the determinant by $-1$ .
$R_4 \to 5R_4$ : This is row scaling, which multiplies the determinant by $5$ .

Combining these effects, we have

\det(U) = (1) \times (-1) \times (5) \times \det(A)

. Given

\det(A) = 10

, we calculate:

\det(U) = -5 \times 10 = -50

Question 7

Let $A$ be a $5 \times 5$ matrix with $\det(A) = -2$ . Matrix $B$ is obtained from $A$ by performing the operations $R_1 \leftrightarrow R_3$ , then $R_4 \to 3R_4$ , and finally $R_2 \to R_2 + R_5$ . What is the value of $\det(B)$ ?

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

Explanation: We determine the effect of each row operation on the determinant:

$R_1 \leftrightarrow R_3$ : Swapping two rows multiplies the determinant by $-1$ .
$R_4 \to 3R_4$ : Multiplying a row by a scalar $k=3$ multiplies the determinant by $3$ .
$R_2 \to R_2 + R_5$ : Adding a multiple of one row to another does not change the determinant (this is equivalent to multiplying by $1$ ).

The determinant of

B

is the product of these factors and the determinant of

A

\det(B) = (-1) \times (3) \times (1) \times \det(A) = -3 \times (-2) = 6

Question 8

A $4 \times 4$ matrix $M$ has determinant $-8$ . Through row reduction to row echelon form, exactly 3 row interchanges were performed, 2 rows were scaled by factors of $\frac{1}{2}$ and $\frac{1}{3}$ respectively, and row additions were used to create zeros. What is the determinant of the resulting row echelon form matrix?

$\frac{4}{3}$
$-\frac{4}{3}$ (correct answer)
$\frac{1}{3}$
$-\frac{1}{3}$

Explanation: Starting with det(M) = -8, we track how each operation affects the determinant: 3 row interchanges multiply by (-1)³ = -1; scaling by 1/2 multiplies by 1/2; scaling by 1/3 multiplies by 1/3; row additions don't change the determinant. So the final determinant is (-8)(-1)(1/2)(1/3) = -4/3. Choice A misses the sign change from odd number of row swaps. Choice C incorrectly calculates the scaling effect. Choice D has the wrong magnitude.

Question 9

During the row reduction of matrix $G$ to row echelon form, the pivot in the second row is created by first adding $3$ times row 1 to row 2, then scaling the new row 2 by $\frac{1}{7}$ . If this scaling step were skipped, the determinant of the resulting matrix would be $21$ . What is the determinant after the scaling step is performed?

$3$ (correct answer)
$21$
$7$
$147$

Explanation: The row addition (adding 3 times row 1 to row 2) does not change the determinant, so after this step the determinant is still the same as the original. If skipping the scaling gives determinant 21, this means the determinant after just the row addition is 21. When we then scale row 2 by 1/7, the determinant becomes 21 × (1/7) = 3. Choice B is the determinant before scaling. Choice C would result from scaling by 1/3 instead of 1/7. Choice D would result from multiplying by 7 instead of dividing by 7.

Question 10

Matrix $C$ can be reduced to row echelon form using only row additions (no scaling or row interchanges needed). If the row echelon form has pivots on the main diagonal with values $2, -1, 3, 4$ , what is $\det(C)$ ?

$8$
$10$
$24$
$-24$ (correct answer)

Explanation: When you encounter a question about determinants and row operations, remember that different row operations affect the determinant in specific ways. Row additions leave the determinant unchanged, while row scaling multiplies it by the scaling factor, and row interchanges multiply it by -1. Since matrix

C

can be reduced to row echelon form using only row additions, the determinant remains unchanged throughout the process. This means

\det(C)

equals the determinant of its row echelon form. For any matrix in row echelon form, the determinant equals the product of the diagonal entries (the pivots). With pivots

2, -1, 3, 4

, we calculate:

\det(C) = 2 \times (-1) \times 3 \times 4 = -24

. Looking at the wrong answers: Choice A (

8

) incorrectly omits the negative sign, perhaps by miscalculating

2 \times (-1) = -2

as positive. Choice B (

10

) might result from adding the pivots instead of multiplying them:

2 + (-1) + 3 + 4 = 8

, though this doesn't match exactly. Choice C (

24

) makes the most common error—correctly multiplying the absolute values but forgetting that one pivot is negative, so the result should be negative. The key insight is that row additions preserve determinants, making this calculation straightforward. Always remember that for triangular or row echelon matrices, the determinant is simply the product of diagonal entries. Watch for sign errors when negative pivots are involved—they're the most frequent mistake on determinant problems.

Question 11

Matrix $A$ is obtained from matrix $B$ by performing the following sequence of row operations: (1) multiply row 2 by $-3$ , (2) add 4 times row 1 to row 3, (3) interchange rows 1 and 4. If $\det(B) = 5$ , what is $\det(A)$ ?

$15$ (correct answer)
$-15$
$60$
$-60$

Explanation: Each row operation affects the determinant as follows: (1) Multiplying row 2 by -3 multiplies the determinant by -3, giving det = 5(-3) = -15. (2) Adding 4 times row 1 to row 3 does not change the determinant, so det = -15. (3) Interchanging rows 1 and 4 multiplies the determinant by -1, giving det = (-15)(-1) = 15. Choice B forgets the final row interchange. Choice C incorrectly multiplies by 4 from the row addition. Choice D combines the errors from B and C.

Question 12

Matrix $E$ undergoes row reduction where exactly $n$ row swaps are needed to avoid zero pivots, and the final row echelon form has determinant $6$ . If $\det(E) = -6$ , what can be concluded about $n$ ?

$n$ must be even and no row scaling was performed during reduction
$n$ must be even and exactly one row was scaled by $-1$ during reduction
$n$ must be odd and no row scaling was performed during reduction (correct answer)
$n$ must be odd and exactly one row was scaled by $-1$ during reduction

Explanation: When you encounter determinant problems involving row operations, remember that each elementary row operation affects the determinant in a predictable way: row swaps multiply the determinant by

-1

, and scaling a row by factor

k

multiplies the determinant by

k

. Let's trace what happens to the determinant during reduction. Starting with

\det(E) = -6

, after

n

row swaps, the determinant becomes

(-1)^n \cdot (-6)

. If row scaling occurs, we multiply by the scaling factors. The final determinant equals

6

. So we have:

(-1)^n \cdot (-6) \cdot (\text{scaling factors}) = 6

This simplifies to:

(-1)^{n+1} \cdot 6 \cdot (\text{scaling factors}) = 6

Therefore:

(-1)^{n+1} \cdot (\text{scaling factors}) = 1

For this equation to hold, we need

(-1)^{n+1} = 1

and scaling factors =

1

. Since

(-1)^{n+1} = 1

when

n+1

is even,

n

must be odd. The scaling factors equal

1

means no net scaling occurred. Answer C correctly identifies that

n

is odd with no row scaling. Answer A incorrectly claims

n

is even. Answer B wrongly suggests

n

is even and includes unnecessary scaling by

-1

. Answer D incorrectly includes scaling by

-1

, which would give us

(-1)^{n+1} \cdot (-1) = -(-1)^{n+1} = 1

, requiring

n

to be even instead. Study tip: Always track how each row operation transforms the determinant step-by-step. Row swaps contribute factors of

-1

, while scaling contributes the scaling factor itself.

Question 13

Let $A$ be a $3 \times 3$ matrix such that $\det(A) = 8$ . A new matrix $B$ is formed by performing the row operation $R_2 \to 4R_2$ on $A$ . What is $\det(B)$ ?

$2$
$8$
$32$ (correct answer)
$512$

Explanation: When a single row of a matrix is multiplied by a scalar

k

, the determinant of the new matrix is

k

times the determinant of the original matrix. In this case, matrix

B

is obtained from

A

by multiplying the second row by

k=4

. Therefore,

\det(B) = 4 \times \det(A) = 4 \times 8 = 32

. Distractor (D)

512

is incorrect; it results from the common mistake of calculating

\det(4A) = 4^3 \det(A) = 64 \times 8 = 512

, which applies when the entire matrix is scaled, not just a single row.

Question 14

Let $A$ be a $4 \times 4$ matrix with $\det(A) = 5$ . If matrix $B$ is obtained from $A$ by swapping rows 2 and 4, and then adding 3 times row 1 to row 3, what is the determinant of $B$ ?

$-15$
$-5$ (correct answer)
$5$
$15$

Explanation: The determinant of a matrix is affected by elementary row operations in specific ways. Swapping two rows multiplies the determinant by

-1

. Adding a multiple of one row to another row does not change the determinant.

The first operation is swapping rows 2 and 4 ( $R_2 \leftrightarrow R_4$ ). This changes the determinant by a factor of $-1$ . After this step, the determinant is $5 \times (-1) = -5$ .
The second operation is adding 3 times row 1 to row 3 ( $R_3 \to R_3 + 3R_1$ ). This operation has no effect on the value of the determinant.

Therefore,

\det(B) = -1 \times \det(A) = -5

Question 15

Let $A = \begin{pmatrix} 1 & 3 & -1 \\ 0 & 1 & 4 \\ 0 & 0 & -2 \end{pmatrix}$ . A matrix $B$ is obtained from $A$ by applying the row operation $R_1 \to R_1 - 3R_2$ . What is $\det(B)$ ?

$-6$
$-2$ (correct answer)
$4$
$6$

Explanation: The matrix

A

is an upper triangular matrix. The determinant of a triangular matrix is the product of its diagonal entries. So,

\det(A) = (1)(1)(-2) = -2

. The matrix

B

is obtained by the row operation

R_1 \to R_1 - 3R_2

. This operation adds a multiple of one row to another row. Such operations do not change the value of the determinant. Therefore,

\det(B) = \det(A) = -2

. A student might waste time by computing the new matrix

B = \begin{pmatrix} 1 & 0 & -13 \\ 0 & 1 & 4 \\ 0 & 0 & -2 \end{pmatrix}

and then finding its determinant, which is still

(1)(1)(-2) = -2

Question 16

Consider the matrix $A = \begin{pmatrix} x & y & z \\ u & v & w \\ 2x-u & 2y-v & 2z-w \end{pmatrix}$ . What is the value of $\det(A)$ ?

$0$ (correct answer)
$1$
$2$
Cannot be determined without the values of the variables.

Explanation: The determinant of a matrix is zero if its rows (or columns) are linearly dependent. Let the rows of

A

R_1 = (x, y, z)

R_2 = (u, v, w)

, and

R_3 = (2x-u, 2y-v, 2z-w)

. We can see that

R_3 = 2R_1 - R_2

. We can prove the determinant is zero using row operations that do not change the determinant's value. Let's perform the operation

R_3 \to R_3 - 2R_1 + R_2

. The new third row will be:

(2x-u) - 2(x) + (u) = 0

(2y-v) - 2(y) + (v) = 0

(2z-w) - 2(z) + (w) = 0

The resulting matrix has a row of zeros. The determinant of any matrix with a zero row is 0. Since the row operations did not change the determinant, the original determinant must also be 0.