Linear Algebra Quiz: Elementary Matrices
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Elementary MatricesQuestion 1 of 18

If PP and QQ are elementary matrices such that PA=QAPA = QA for some invertible matrix AA, which statement must be true?

PP and QQ perform equivalent row operations on any matrix
P=QP = Q is the only possibility
PP and QQ must both be row swap matrices
PQ−1=IPQ^{-1} = I and therefore P=QP = Q
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Linear Algebra Quiz

Linear Algebra Quiz: Elementary Matrices

Practice Elementary Matrices 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 Elementary Matrices, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

How to use this quiz

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.

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Question 1

If PP and QQ are elementary matrices such that PA=QAPA = QA for some invertible matrix AA, which statement must be true?

  1. PP and QQ perform equivalent row operations on any matrix
  2. P=QP = Q is the only possibility
  3. PP and QQ must both be row swap matrices
  4. PQ−1=IPQ^{-1} = I and therefore P=QP = Q (correct answer)
Explanation: From PA=QAPA = QA, we get (P−Q)A=0(P - Q)A = 0. Since AA is invertible, we can multiply on the right by A−1A^{-1} to get P−Q=0P - Q = 0, so P=QP = Q. Equivalently, PA=QAPA = QA implies PQ−1QA=QAPQ^{-1}QA = QA, so PQ−1=IPQ^{-1} = I, giving P=QP = Q. Choice A is wrong because we need P=QP = Q, not just equivalence. Choice B states the conclusion but not the reasoning. Choice C incorrectly restricts the type of elementary matrix.

Question 2

Let EE be the 3×33 \times 3 elementary matrix that performs the row operation R2−3R3→R2R_2 - 3R_3 \to R_2. If AA is any 3×n3 \times n matrix, which of the following represents the matrix product EAEA?

  1. Matrix AA with its second row replaced by the sum of its second row and 3 times its third row.
  2. Matrix AA with its third row replaced by the difference of its third row and 3 times its second row.
  3. Matrix AA with its second row replaced by the difference of its second row and 3 times its third row. (correct answer)
  4. Matrix AA with its second column replaced by the difference of its second column and 3 times its third column.
Explanation: The elementary matrix EE corresponding to the row operation Ri+kRj→RiR_i + kR_j \to R_i is the identity matrix with an additional entry kk at position (i,j)(i, j). For the operation R2−3R3→R2R_2 - 3R_3 \to R_2 on a 3×33 \times 3 matrix, i=2,j=3,k=−3i=2, j=3, k=-3. The matrix is E=(10001−3001)E = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{pmatrix}. Left-multiplying a matrix AA by EE (i.e., forming EAEA) performs this exact row operation on AA. Therefore, the product EAEA results in a new matrix where the second row of AA has been replaced by the original second row of AA minus 3 times the third row of AA.

Question 3

Let E1E_1 be the elementary matrix for the operation R1↔R2R_1 \leftrightarrow R_2, and E2E_2 be the matrix for R2−4R1→R2R_2 - 4R_1 \to R_2. If B=E2E1AB = E_2 E_1 A, which sequence of operations transforms matrix AA into matrix BB?

  1. First, subtract 4 times row 1 from row 2, then swap rows 1 and 2.
  2. First, swap rows 1 and 2, then subtract 4 times the new row 1 from the new row 2. (correct answer)
  3. First, swap rows 1 and 2, then subtract 4 times the original row 1 from the new row 2.
  4. First, subtract 4 times row 2 from row 1, then swap rows 1 and 2.
Explanation: Matrix multiplication acts from right to left. In the product E2E1AE_2 E_1 A, the matrix E1E_1 acts on AA first, and then the matrix E2E_2 acts on the result (E1A)(E_1 A). Therefore, the first operation is the one corresponding to E1E_1, which is swapping rows 1 and 2. The second operation is the one corresponding to E2E_2, which is subtracting 4 times row 1 from row 2. This second operation is applied to the matrix after the first operation has been completed. Thus, it's 4 times the new row 1 that is subtracted from the new row 2.

Question 4

A 3×33 \times 3 matrix AA with det⁡(A)=4\det(A) = 4 is transformed into a matrix BB by the following sequence of row operations: R1↔R3R_1 \leftrightarrow R_3, then 2R2→R22R_2 \to R_2, then R3+5R2→R3R_3 + 5R_2 \to R_3. What is the determinant of BB?

  1. −8-8 (correct answer)
  2. 88
  3. −40-40
  4. −2-2
Explanation: The determinant of the resulting matrix is the product of the determinant of the original matrix and the determinants of the elementary matrices corresponding to the row operations.
  1. R1↔R3R_1 \leftrightarrow R_3: The determinant of this elementary matrix is −1-1.
  2. 2R2→R22R_2 \to R_2: The determinant of this elementary matrix is 22.
  3. R3+5R2→R3R_3 + 5R_2 \to R_3: The determinant of a row replacement elementary matrix is always 11. Therefore, det⁡(B)=(det⁡(E3))(det⁡(E2))(det⁡(E1))(det⁡(A))=(1)(2)(−1)(4)=−8\det(B) = (\det(E_3))(\det(E_2))(\det(E_1))(\det(A)) = (1)(2)(-1)(4) = -8.

Question 5

An invertible matrix AA can be row-reduced to the identity matrix II by a sequence of elementary row operations represented by elementary matrices E1,E2,…,EkE_1, E_2, \dots, E_k. If Ek…E2E1A=IE_k \dots E_2 E_1 A = I, which expression correctly represents AA?

  1. A=Ek…E2E1A = E_k \dots E_2 E_1
  2. A=Ek−1…E2−1E1−1A = E_k^{-1} \dots E_2^{-1} E_1^{-1}
  3. A=E1E2…EkA = E_1 E_2 \dots E_k
  4. A=E1−1E2−1…Ek−1A = E_1^{-1} E_2^{-1} \dots E_k^{-1} (correct answer)
Explanation: To solve for AA in the equation Ek…E2E1A=IE_k \dots E_2 E_1 A = I, we must multiply both sides by the inverses of the elementary matrices. To isolate AA, we must 'undo' the multiplications in the reverse order. We start by left-multiplying by Ek−1E_k^{-1}, then Ek−1−1E_{k-1}^{-1}, and so on, until we multiply by E1−1E_1^{-1}. This gives A=E1−1E2−1…Ek−1I=E1−1E2−1…Ek−1A = E_1^{-1} E_2^{-1} \dots E_k^{-1} I = E_1^{-1} E_2^{-1} \dots E_k^{-1}. This relies on the property that for invertible matrices P,QP, Q, (PQ)−1=Q−1P−1(PQ)^{-1} = Q^{-1}P^{-1}.

Question 6

Let EE be a 3×33 \times 3 elementary matrix that corresponds to subtracting 5 times row 1 from row 3 (R3−5R1→R3R_3 - 5R_1 \to R_3). What is the entry in the third row, first column of the matrix E2E^2?

  1. −10-10 (correct answer)
  2. −5-5
  3. 2525
  4. 00
Explanation: The elementary matrix for the operation R3−5R1→R3R_3 - 5R_1 \to R_3 is E=(100010−501)E = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -5 & 0 & 1 \end{pmatrix}. To find E2E^2, we multiply EE by itself: E2=(100010−501)(100010−501)E^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -5 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -5 & 0 & 1 \end{pmatrix}. The entry in the third row, first column of the product is found by taking the dot product of the third row of the first matrix and the first column of the second matrix: (−5)(1)+(0)(0)+(1)(−5)=−5−5=−10(-5)(1) + (0)(0) + (1)(-5) = -5 - 5 = -10. Conceptually, applying the operation twice means the total effect is R3−10R1→R3R_3 - 10R_1 \to R_3.

Question 7

A matrix AA is transformed into a matrix BB by the following sequence of operations on AA's rows:

  1. First, scale row 2 by a factor of 3.

  2. Second, add the new row 2 to row 1. Which single matrix PP satisfies the equation PA=BPA=B?

  1. (1103)\begin{pmatrix} 1 & 1 \\ 0 & 3 \end{pmatrix}
  2. (1303)\begin{pmatrix} 1 & 3 \\ 0 & 3 \end{pmatrix} (correct answer)
  3. (1301)\begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}
  4. (4001)\begin{pmatrix} 4 & 0 \\ 0 & 1 \end{pmatrix}
Explanation: Let E1E_1 be the matrix for the first operation (3R2→R23R_2 \to R_2) and E2E_2 be the matrix for the second operation (R1+R2→R1R_1 + R_2 \to R_1). For 2×22 \times 2 matrices, E1=(1003)E_1 = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix} and E2=(1101)E_2 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}. The operations are applied sequentially, so B=E2(E1A)=(E2E1)AB = E_2(E_1 A) = (E_2 E_1)A. The matrix PP is the product E2E1E_2 E_1. P=(1101)(1003)=(1(1)+1(0)1(0)+1(3)0(1)+1(0)0(0)+1(3))=(1303)P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 1(1)+1(0) & 1(0)+1(3) \\ 0(1)+1(0) & 0(0)+1(3) \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 0 & 3 \end{pmatrix}. The reversed product E1E2E_1 E_2 would give the incorrect matrix in choice A.

Question 8

The matrix A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} is transformed into B=(120−2)B = \begin{pmatrix} 1 & 2 \\ 0 & -2 \end{pmatrix} by left-multiplication with an elementary matrix EE. What is the matrix EE?

  1. (10−31)\begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix} (correct answer)
  2. (1031)\begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}
  3. (1−301)\begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix}
  4. (100−1/2)\begin{pmatrix} 1 & 0 \\ 0 & -1/2 \end{pmatrix}
Explanation: To transform matrix AA into matrix BB, we need to find the row operation that achieves this. The first row remains unchanged. The second row of BB can be obtained from the second row of AA by the operation R2−3R1→R2R_2 - 3R_1 \to R_2. Specifically, (3,4)−3×(1,2)=(3−3,4−6)=(0,−2)(3, 4) - 3 \times (1, 2) = (3-3, 4-6) = (0, -2). The elementary matrix EE for this operation is found by applying the same operation to the 2×22 \times 2 identity matrix I=(1001)I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. Applying R2−3R1→R2R_2 - 3R_1 \to R_2 to II yields E=(10−31)E = \begin{pmatrix} 1 & 0 \\ -3 & 1 \end{pmatrix}.

Question 9

Consider the matrix equation XA=BXA = B where A=(2103)A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} and B=(2163)B = \begin{pmatrix} 2 & 1 \\ 6 & 3 \end{pmatrix} . If XX must be an elementary matrix, which row operation does XX represent?

  1. Add 33 times row 11 to row 22 (correct answer)
  2. Add 22 times row 11 to row 22
  3. Multiply row 22 by 22
  4. Add 12\frac{1}{2} times row 22 to row 11
Explanation: We need X=BA−1X = BA^{-1}. First, A−1=1det⁡(A)(3−102)=16(3−102)A^{-1} = \frac{1}{\det(A)}\begin{pmatrix} 3 & -1 \\ 0 & 2 \end{pmatrix} = \frac{1}{6}\begin{pmatrix} 3 & -1 \\ 0 & 2 \end{pmatrix} . Then $$X = \begin{pmatrix} 2 & 1 \ 6 & 3 \end{pmatrix} \cdot \frac{1}{6}\begin{pmatrix} 3 & -1 \ 0 & 2 \end{pmatrix} = \frac{1}{6}\begin{pmatrix} 6 & 0 \ 18 & 6 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 3 & 1 \end{pmatrix}

Question 10

If matrices AA and BB are row equivalent, and E1,E2,…,EkE_1, E_2, \ldots, E_k are elementary matrices such that EkEk−1⋯E1A=BE_k E_{k-1} \cdots E_1 A = B, which statement about the matrix M=EkEk−1⋯E1M = E_k E_{k-1} \cdots E_1 is always true?

  1. MM is an elementary matrix
  2. det⁡(M)=±1\det(M) = \pm 1
  3. MM is invertible and M−1=E1−1E2−1⋯Ek−1M^{-1} = E_1^{-1} E_2^{-1} \cdots E_k^{-1} (correct answer)
  4. MM represents a single elementary row operation
Explanation: When you encounter problems about row equivalent matrices and elementary operations, you're dealing with the fundamental relationship between matrix multiplication and row operations. The key insight is that performing a sequence of elementary row operations is equivalent to multiplying by a product of elementary matrices. Since EkEk−1⋯E1A=BE_k E_{k-1} \cdots E_1 A = B, the matrix M=EkEk−1⋯E1M = E_k E_{k-1} \cdots E_1 represents the composition of all these elementary operations. Each elementary matrix is invertible (you can always "undo" a row operation), and the product of invertible matrices is invertible. To find M−1M^{-1}, you reverse both the order and each individual operation: M−1=(EkEk−1⋯E1)−1=E1−1E2−1⋯Ek−1M^{-1} = (E_k E_{k-1} \cdots E_1)^{-1} = E_1^{-1} E_2^{-1} \cdots E_k^{-1}. This makes choice C correct. Choice A is wrong because MM is a product of multiple elementary matrices, not a single elementary matrix. Choice B is incorrect because while each elementary matrix has determinant ±1\pm 1 or a positive constant, their product could have any nonzero determinant. For example, if you scale a row by 2 twice, you get det⁡(M)=4\det(M) = 4. Choice D fails because MM represents multiple row operations combined, not a single operation. Remember: elementary matrices are the building blocks of all invertible matrices. When you see a product of elementary matrices, focus on the invertibility property and how inverses reverse both the order of operations and each individual operation.

Question 11

Let AA be a 3×33 \times 3 matrix with det⁡(A)=5\det(A) = 5. After applying the sequence of row operations: (i) add 33 times row 11 to row 22, (ii) multiply row 33 by −2-2, (iii) swap rows 22 and 33, the resulting matrix is BB. If EE is the single elementary matrix equivalent to this sequence, what is det⁡(E−1)\det(E^{-1})?

  1. 12\frac{1}{2} (correct answer)
  2. −12-\frac{1}{2}
  3. 22
  4. −2-2
Explanation: Let E1,E2,E3E_1, E_2, E_3 be the elementary matrices for operations (i), (ii), (iii) respectively. Then E=E3E2E1E = E_3E_2E_1 and B=EAB = EA. The determinants are: det⁡(E1)=1\det(E_1) = 1 (row addition), det⁡(E2)=−2\det(E_2) = -2 (row scaling), det⁡(E3)=−1\det(E_3) = -1 (row swap). So det⁡(E)=det⁡(E3)det⁡(E2)det⁡(E1)=(−1)(−2)(1)=2\det(E) = \det(E_3)\det(E_2)\det(E_1) = (-1)(-2)(1) = 2. Therefore det⁡(E−1)=1det⁡(E)=12\det(E^{-1}) = \frac{1}{\det(E)} = \frac{1}{2}. Choice B has the wrong sign. Choices C and D give det⁡(E)\det(E) instead of det⁡(E−1)\det(E^{-1}).

Question 12

Matrix AA can be reduced to matrix BB using exactly three elementary row operations: first multiply row 22 by 13\frac{1}{3}, then add −2-2 times row 11 to row 33, then swap rows 11 and 22. If det⁡(A)=12\det(A) = 12, what is det⁡(B)\det(B)?

  1. 44
  2. 1212
  3. −4-4 (correct answer)
  4. −12-12
Explanation: When you encounter questions about determinants and elementary row operations, remember that each type of row operation affects the determinant in a predictable way. This is a fundamental relationship in linear algebra that connects matrix transformations to determinant values. Let's trace through each operation applied to matrix AA to find det⁡(B)\det(B). Starting with det⁡(A)=12\det(A) = 12: Operation 1: Multiply row 2 by 13\frac{1}{3}. When you multiply a row by a scalar kk, the determinant gets multiplied by kk. So det⁡=12×13=4\det = 12 \times \frac{1}{3} = 4. Operation 2: Add −2-2 times row 1 to row 3. Adding a multiple of one row to another row leaves the determinant unchanged. So det⁡=4\det = 4. Operation 3: Swap rows 1 and 2. Swapping any two rows multiplies the determinant by −1-1. So det⁡(B)=4×(−1)=−4\det(B) = 4 \times (-1) = -4. Looking at the wrong answers: (A) 4 forgets the final row swap that changes the sign. (B) 12 incorrectly assumes row operations don't change the determinant at all. (D) -12 likely applies the sign change from the row swap but forgets that multiplying row 2 by 13\frac{1}{3} reduces the determinant. Key strategy: Memorize the three determinant rules for row operations: scaling a row multiplies the determinant by that scalar, adding multiples of rows doesn't change it, and swapping rows multiplies by −1-1. Apply them sequentially in the order given.

Question 13

Let EE be the elementary matrix corresponding to the row operation R1+7R3→R1R_1 + 7R_3 \to R_1 for 3×33 \times 3 matrices. Which matrix represents E−1E^{-1}?

  1. The elementary matrix for the operation R1+17R3→R1R_1 + \frac{1}{7}R_3 \to R_1.
  2. The elementary matrix for the operation R3−7R1→R3R_3 - 7R_1 \to R_3.
  3. The elementary matrix for the operation R1−7R3→R1R_1 - 7R_3 \to R_1. (correct answer)
  4. The elementary matrix for the operation R1+7R3→R1R_1 + 7R_3 \to R_1.
Explanation: The inverse of an elementary matrix corresponds to the inverse elementary row operation. The operation is adding 7 times row 3 to row 1. The inverse operation that would undo this is subtracting 7 times row 3 from row 1. This corresponds to the row operation R1−7R3→R1R_1 - 7R_3 \to R_1. Distractor A incorrectly uses the rule for the inverse of a scaling operation. Distractor B confuses the roles of the rows. Distractor D incorrectly suggests the matrix is its own inverse, which is only true for row-swapping matrices.

Question 14

A square matrix AA is transformed into a matrix BB by a single elementary row operation. If matrix BB is singular, which of the following statements must be true?

  1. The elementary row operation must have been scaling a row by zero.
  2. The matrix AA must be the identity matrix.
  3. The original matrix AA must also be singular. (correct answer)
  4. The elementary matrix corresponding to the operation must be singular.
Explanation: Let the elementary row operation be represented by the elementary matrix EE. Then B=EAB = EA. The determinant of BB is det⁡(B)=det⁡(E)det⁡(A)\det(B) = \det(E)\det(A). Since BB is singular, det⁡(B)=0\det(B) = 0. This gives det⁡(E)det⁡(A)=0\det(E)\det(A) = 0. All elementary matrices are invertible, so their determinants are never zero (det⁡(E)≠0\det(E) \neq 0). For the product to be zero, it must be that det⁡(A)=0\det(A) = 0. A matrix with a determinant of zero is singular. Therefore, the original matrix AA must also be singular.

Question 15

A 2×22 \times 2 matrix AA is transformed by the row operation R1−2R2→R1R_1 - 2R_2 \to R_1 to produce the matrix B=(1031)B = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}. What is the original matrix AA?

  1. (−5−231)\begin{pmatrix} -5 & -2 \\ 3 & 1 \end{pmatrix}
  2. (1237)\begin{pmatrix} 1 & 2 \\ 3 & 7 \end{pmatrix}
  3. (7231)\begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix} (correct answer)
  4. (1−23−5)\begin{pmatrix} 1 & -2 \\ 3 & -5 \end{pmatrix}
Explanation: The row operation corresponds to the equation EA=BEA=B, where EE is the elementary matrix for R1−2R2→R1R_1 - 2R_2 \to R_1, which is E=(1−201)E = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}. To find AA, we must compute A=E−1BA = E^{-1}B. The inverse operation is R1+2R2→R1R_1 + 2R_2 \to R_1, so the inverse matrix is E−1=(1201)E^{-1} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}. Thus, A=(1201)(1031)=(1(1)+2(3)1(0)+2(1)0(1)+1(3)0(0)+1(1))=(7231)A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix} = \begin{pmatrix} 1(1)+2(3) & 1(0)+2(1) \\ 0(1)+1(3) & 0(0)+1(1) \end{pmatrix} = \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix}.

Question 16

Which of the following 3×33 \times 3 matrices is NOT an elementary matrix?

  1. (100001010)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}
  2. (1000100−21)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{pmatrix}
  3. (100050001)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{pmatrix}
  4. (100011000)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix} (correct answer)
Explanation: An elementary matrix is obtained by performing a single elementary row operation on an identity matrix. All elementary matrices are invertible. Choice A is obtained from the identity matrix by swapping rows 2 and 3 (R2↔R3R_2 \leftrightarrow R_3). Choice B is obtained by the operation R3−2R2→R3R_3 - 2R_2 \to R_3. Choice C is obtained by scaling row 2 by 5 (5R2→R25R_2 \to R_2). Choice D has a row of zeros. Its determinant is 0, so it is not invertible. No single elementary row operation performed on an identity matrix can create a row of zeros. Therefore, it is not an elementary matrix.

Question 17

Let A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} and E=(1501)E = \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}. The matrix product AEAE results in which transformation on matrix AA?

  1. Replacing row 1 with the sum of row 1 and 5 times row 2.
  2. Replacing column 2 with the sum of column 2 and 5 times column 1. (correct answer)
  3. Replacing row 2 with the sum of row 2 and 5 times row 1.
  4. Replacing column 1 with the sum of column 1 and 5 times column 2.
Explanation: When an elementary matrix EE multiplies a matrix AA on the right (forming AEAE), it performs column operations on AA. The product is AE=(abcd)(1501)=(a5a+bc5c+d)AE = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} a & 5a+b \\ c & 5c+d \end{pmatrix}. The first column is unchanged. The new second column is the original second column plus 5 times the original first column. This corresponds to the column operation C2+5C1→C2C_2 + 5C_1 \to C_2. Left-multiplication (EAEA) results in row operations.

Question 18

Let EE be an elementary matrix of size 4×44 \times 4. If det⁡(E)=−1\det(E) = -1 and E2=IE^2 = I, what can be concluded about the row operation represented by EE?

  1. EE multiplies some row by −1-1
  2. EE swaps two distinct rows (correct answer)
  3. EE adds a multiple of one row to another row
  4. EE could represent any of the three types of elementary operations
Explanation: Elementary matrices have determinant: (1) det⁡=1\det = 1 for row addition, (2) det⁡=c\det = c for multiplying a row by c≠0c \neq 0, (3) det⁡=−1\det = -1 for row swaps. Since det⁡(E)=−1\det(E) = -1, EE is a row swap matrix. Row swap matrices satisfy E2=IE^2 = I, confirming this. Choice A would require det⁡(E)=−1\det(E) = -1 and E2=IE^2 = I, but multiplying by −1-1 gives det⁡=−1\det = -1 yet E2≠IE^2 \neq I. Choice C gives det⁡=1\det = 1. Choice D ignores the constraints.