Linear Algebra Quiz: Determinant Properties
18 questions · exam conditions
0:00
Determinant PropertiesQuestion 1 of 18

Let AA be a 4×44 \times 4 matrix. If swapping rows 1 and 3 of AA gives matrix BB, and then multiplying row 2 of BB by 12\frac{1}{2} gives matrix CC, which relationship correctly describes det⁡(C)\det(C) in terms of det⁡(A)\det(A)?

det⁡(C)=−12det⁡(A)\det(C) = -\frac{1}{2}\det(A)
det⁡(C)=12det⁡(A)\det(C) = \frac{1}{2}\det(A)
det⁡(C)=−2det⁡(A)\det(C) = -2\det(A)
det⁡(C)=2det⁡(A)\det(C) = 2\det(A)
← Back to quizzes

Linear Algebra Quiz

Linear Algebra Quiz: Determinant Properties

Practice Determinant Properties 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 Determinant Properties, 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.

All questions

Question 1

Let AA be a 4×44 \times 4 matrix. If swapping rows 1 and 3 of AA gives matrix BB, and then multiplying row 2 of BB by 12\frac{1}{2} gives matrix CC, which relationship correctly describes det⁡(C)\det(C) in terms of det⁡(A)\det(A)?

  1. det⁡(C)=−12det⁡(A)\det(C) = -\frac{1}{2}\det(A) (correct answer)
  2. det⁡(C)=12det⁡(A)\det(C) = \frac{1}{2}\det(A)
  3. det⁡(C)=−2det⁡(A)\det(C) = -2\det(A)
  4. det⁡(C)=2det⁡(A)\det(C) = 2\det(A)
Explanation: The correct answer is A. Row operations affect determinants systematically: swapping two rows multiplies the determinant by −1-1, and multiplying a row by a scalar kk multiplies the determinant by kk. Starting with det⁡(A)\det(A), swapping rows 1 and 3 gives det⁡(B)=−det⁡(A)\det(B) = -\det(A). Then multiplying row 2 by 12\frac{1}{2} gives det⁡(C)=12⋅det⁡(B)=12⋅(−det⁡(A))=−12det⁡(A)\det(C) = \frac{1}{2} \cdot \det(B) = \frac{1}{2} \cdot (-\det(A)) = -\frac{1}{2}\det(A). Choice B forgets the sign change from swapping. Choices C and D incorrectly use the reciprocal of the scaling factor.

Question 2

Let AA be a 3×33 \times 3 matrix with det⁡(A)=5\det(A) = 5. A new matrix BB is created by performing the following sequence of elementary row operations on AA:

  1. Swap Row 1 and Row 3 (R1↔R3R_1 \leftrightarrow R_3).
  2. Multiply Row 2 by 4 (R2→4R2R_2 \to 4R_2).
  3. Add -2 times Row 1 to Row 3 (R3→R3−2R1R_3 \to R_3 - 2R_1).

What is the value of det⁡(B)\det(B)?

  1. −20-20 (correct answer)
  2. 2020
  3. −5-5
  4. 4040
Explanation: Let's track the effect of each row operation on the determinant. Starting with det⁡(A)=5\det(A) = 5.
  1. The operation R1↔R3R_1 \leftrightarrow R_3 swaps two rows, which multiplies the determinant by −1-1. The new determinant is −1×5=−5-1 \times 5 = -5.
  2. The operation R2→4R2R_2 \to 4R_2 scales a single row by 4, which multiplies the current determinant by 4. The new determinant is 4×(−5)=−204 \times (-5) = -20.
  3. The operation R3→R3−2R1R_3 \to R_3 - 2R_1 adds a multiple of one row to another. This operation does not change the value of the determinant. The determinant remains −20-20. Therefore, det⁡(B)=−20\det(B) = -20.

Question 3

Let AA be a 3×33 \times 3 matrix with columns a⃗1,a⃗2,a⃗3\vec{a}_1, \vec{a}_2, \vec{a}_3. If det⁡(A)=3\det(A) = 3, and v⃗\vec{v} is a vector in R3\mathbb{R}^3 such that the determinant of the matrix with columns [v⃗,a⃗2,a⃗3][\vec{v}, \vec{a}_2, \vec{a}_3] is 4, what is the determinant of the matrix with columns [a⃗1+2v⃗,a⃗2,a⃗3][\vec{a}_1 + 2\vec{v}, \vec{a}_2, \vec{a}_3]?

  1. 77
  2. 88
  3. 1111 (correct answer)
  4. 1414
Explanation: The determinant is a linear function of each of its columns. Using the linearity property on the first column, we can write: det⁡([a⃗1+2v⃗,a⃗2,a⃗3])=det⁡([a⃗1,a⃗2,a⃗3])+det⁡([2v⃗,a⃗2,a⃗3])\det([\vec{a}_1 + 2\vec{v}, \vec{a}_2, \vec{a}_3]) = \det([\vec{a}_1, \vec{a}_2, \vec{a}_3]) + \det([2\vec{v}, \vec{a}_2, \vec{a}_3]). Further, we can factor out the scalar from the first column of the second term: det⁡([2v⃗,a⃗2,a⃗3])=2det⁡([v⃗,a⃗2,a⃗3])\det([2\vec{v}, \vec{a}_2, \vec{a}_3]) = 2 \det([\vec{v}, \vec{a}_2, \vec{a}_3]). We are given that det⁡([a⃗1,a⃗2,a⃗3])=det⁡(A)=3\det([\vec{a}_1, \vec{a}_2, \vec{a}_3]) = \det(A) = 3 and det⁡([v⃗,a⃗2,a⃗3])=4\det([\vec{v}, \vec{a}_2, \vec{a}_3]) = 4. Substituting these values, we get: det⁡([a⃗1+2v⃗,a⃗2,a⃗3])=3+2(4)=3+8=11\det([\vec{a}_1 + 2\vec{v}, \vec{a}_2, \vec{a}_3]) = 3 + 2(4) = 3 + 8 = 11.

Question 4

Let AA be a 4×44 \times 4 matrix with det⁡(A)=−2\det(A) = -2. What is the value of det⁡(3A)\det(3A)?

  1. −6-6
  2. −54-54
  3. −162-162 (correct answer)
  4. −128-128
Explanation: For an n×nn \times n matrix AA and a scalar cc, the property of determinants states that det⁡(cA)=cndet⁡(A)\det(cA) = c^n \det(A). In this case, AA is a 4×44 \times 4 matrix, so n=4n=4. The scalar is c=3c=3, and det⁡(A)=−2\det(A) = -2. Applying the formula: det⁡(3A)=34det⁡(A)=81×(−2)=−162\det(3A) = 3^4 \det(A) = 81 \times (-2) = -162. Distractor A, -6, results from the common mistake of calculating 3det⁡(A)3 \det(A). Distractor B, -54, results from incorrectly using n−1n-1 for the exponent: 34−1det⁡(A)3^{4-1} \det(A). Distractor D, -128, results from confusing the scalar and the dimension: 43det⁡(A)4^3 \det(A).

Question 5

Let A=(123456789)A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}. Let BB be the matrix obtained from AA by the sequence of row operations R2→R2−4R1R_2 \to R_2 - 4R_1 and then R3→R3−7R1R_3 \to R_3 - 7R_1. What is det⁡(B)\det(B)?

  1. 00 (correct answer)
  2. 11
  3. −9-9
  4. 66
Explanation: The row operations performed are of the type Ri→Ri−cRjR_i \to R_i - cR_j. These operations do not change the value of the determinant. Therefore, det⁡(B)=det⁡(A)\det(B) = \det(A). We can calculate det⁡(A)\det(A) directly: det⁡(A)=1(5⋅9−6⋅8)−2(4⋅9−6⋅7)+3(4⋅8−5⋅7)\det(A) = 1(5\cdot9 - 6\cdot8) - 2(4\cdot9 - 6\cdot7) + 3(4\cdot8 - 5\cdot7) det⁡(A)=1(45−48)−2(36−42)+3(32−35)\det(A) = 1(45 - 48) - 2(36 - 42) + 3(32 - 35) det⁡(A)=1(−3)−2(−6)+3(−3)=−3+12−9=0\det(A) = 1(-3) - 2(-6) + 3(-3) = -3 + 12 - 9 = 0. Alternatively, after performing the row operations on AA, we get: B=(1230−3−60−6−12)B = \begin{pmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -12 \end{pmatrix}. In matrix BB, the third row is a multiple of the second row (R3=2R2R_3 = 2R_2). If one row of a matrix is a multiple of another row, the rows are linearly dependent, and the determinant is 0.

Question 6

Let AA be a 3×33 \times 3 matrix with columns c⃗1,c⃗2,c⃗3\vec{c}_1, \vec{c}_2, \vec{c}_3, and let det⁡(A)=10\det(A)=10. What is the determinant of the matrix BB with columns [c⃗1+c⃗2,c⃗2+c⃗3,c⃗3+c⃗1][\vec{c}_1 + \vec{c}_2, \vec{c}_2 + \vec{c}_3, \vec{c}_3 + \vec{c}_1]?

  1. 00
  2. 1010
  3. 2020 (correct answer)
  4. 3030
Explanation: The columns of BB are linear combinations of the columns of AA. We can represent this transformation using matrix multiplication. Let B=[b⃗1,b⃗2,b⃗3]B = [\vec{b}_1, \vec{b}_2, \vec{b}_3]. Then: b⃗1=1c⃗1+1c⃗2+0c⃗3\vec{b}_1 = 1\vec{c}_1 + 1\vec{c}_2 + 0\vec{c}_3 b⃗2=0c⃗1+1c⃗2+1c⃗3\vec{b}_2 = 0\vec{c}_1 + 1\vec{c}_2 + 1\vec{c}_3 b⃗3=1c⃗1+0c⃗2+1c⃗3\vec{b}_3 = 1\vec{c}_1 + 0\vec{c}_2 + 1\vec{c}_3 This means B=AKB = A K, where K=(101110011)K = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}. Using the property det⁡(AK)=det⁡(A)det⁡(K)\det(AK) = \det(A)\det(K), we first find det⁡(K)\det(K): det⁡(K)=1(1⋅1−0⋅1)−0(1⋅1−0⋅0)+1(1⋅1−1⋅0)=1(1)−0+1(1)=2\det(K) = 1(1\cdot1 - 0\cdot1) - 0(1\cdot1 - 0\cdot0) + 1(1\cdot1 - 1\cdot0) = 1(1) - 0 + 1(1) = 2. Now, we can find det⁡(B)\det(B): det⁡(B)=det⁡(A)det⁡(K)=10×2=20\det(B) = \det(A) \det(K) = 10 \times 2 = 20.

Question 7

A 3×33 \times 3 matrix BB is formed from a matrix AA by first taking the transpose of AA, and then swapping the first and third columns of the resulting matrix. If det⁡(A)=k\det(A) = k, what is det⁡(B)\det(B)?

  1. kk
  2. −k-k (correct answer)
  3. 1/k1/k
  4. −1/k-1/k
Explanation: There are two operations applied to find BB from AA.
  1. Taking the transpose: Let C=ATC = A^T. A fundamental property of determinants is that det⁡(AT)=det⁡(A)\det(A^T) = \det(A). Therefore, det⁡(C)=k\det(C) = k.
  2. Swapping columns: Matrix BB is obtained by swapping columns 1 and 3 of CC. Swapping any two columns (or rows) of a matrix multiplies its determinant by -1. Therefore, det⁡(B)=−det⁡(C)\det(B) = -\det(C). Combining these steps, we get det⁡(B)=−det⁡(C)=−det⁡(AT)=−det⁡(A)=−k\det(B) = -\det(C) = -\det(A^T) = -\det(A) = -k.

Question 8

Let u⃗,v⃗,w⃗\vec{u}, \vec{v}, \vec{w} be vectors in R3\mathbb{R}^3. If the determinant of the matrix with columns [u⃗,v⃗,w⃗][\vec{u}, \vec{v}, \vec{w}] is 5, what is the determinant of the matrix with columns [u⃗+v⃗,v⃗+w⃗,w⃗−u⃗][\vec{u}+\vec{v}, \vec{v}+\vec{w}, \vec{w}-\vec{u}]?

  1. 00 (correct answer)
  2. 55
  3. 1010
  4. −5-5
Explanation: Let the columns of the new matrix be c⃗1′=u⃗+v⃗\vec{c}_1' = \vec{u}+\vec{v}, c⃗2′=v⃗+w⃗\vec{c}_2' = \vec{v}+\vec{w}, and c⃗3′=w⃗−u⃗\vec{c}_3' = \vec{w}-\vec{u}. The determinant of a matrix is zero if its columns are linearly dependent. Let's check for a linear combination of the columns that equals the zero vector. Consider the sum c⃗1′−c⃗2′+c⃗3′\vec{c}_1' - \vec{c}_2' + \vec{c}_3': (u⃗+v⃗)−(v⃗+w⃗)+(w⃗−u⃗)=u⃗+v⃗−v⃗−w⃗+w⃗−u⃗=0⃗(\vec{u}+\vec{v}) - (\vec{v}+\vec{w}) + (\vec{w}-\vec{u}) = \vec{u}+\vec{v}-\vec{v}-\vec{w}+\vec{w}-\vec{u} = \vec{0}. Since a non-trivial linear combination of the columns equals the zero vector (the coefficients are 1, -1, and 1), the columns are linearly dependent. Therefore, the determinant of the matrix is 0.

Question 9

Let D(r⃗1,r⃗2,r⃗3)D(\vec{r}_1, \vec{r}_2, \vec{r}_3) denote the determinant of a 3×33 \times 3 matrix with rows r⃗1,r⃗2,r⃗3\vec{r}_1, \vec{r}_2, \vec{r}_3. Which of the following expressions is a correct statement of the linearity property of determinants?

  1. D(cr⃗1,cr⃗2,cr⃗3)=cD(r⃗1,r⃗2,r⃗3)D(c\vec{r}_1, c\vec{r}_2, c\vec{r}_3) = c D(\vec{r}_1, \vec{r}_2, \vec{r}_3)
  2. D(r⃗1+s⃗,r⃗2+s⃗,r⃗3+s⃗)=D(r⃗1,r⃗2,r⃗3)+D(s⃗,s⃗,s⃗)D(\vec{r}_1 + \vec{s}, \vec{r}_2 + \vec{s}, \vec{r}_3 + \vec{s}) = D(\vec{r}_1, \vec{r}_2, \vec{r}_3) + D(\vec{s}, \vec{s}, \vec{s})
  3. D(r⃗1,r⃗2,r⃗3)=D(r⃗1,r⃗2,r⃗3)−D(r⃗1,r⃗1,r⃗3)D(\vec{r}_1, \vec{r}_2, \vec{r}_3) = D(\vec{r}_1, \vec{r}_2, \vec{r}_3) - D(\vec{r}_1, \vec{r}_1, \vec{r}_3)
  4. D(r⃗1,cr⃗2+s⃗2,r⃗3)=cD(r⃗1,r⃗2,r⃗3)+D(r⃗1,s⃗2,r⃗3)D(\vec{r}_1, c\vec{r}_2 + \vec{s}_2, \vec{r}_3) = c D(\vec{r}_1, \vec{r}_2, \vec{r}_3) + D(\vec{r}_1, \vec{s}_2, \vec{r}_3) (correct answer)
Explanation: When you encounter questions about determinant properties, focus on understanding that determinants are multilinear functions. This means they are linear in each row (or column) individually, but not necessarily linear across all rows simultaneously. The linearity property states that if you have a linear combination in any single row, the determinant distributes over that combination while keeping other rows fixed. Option D perfectly demonstrates this: D(r⃗1,cr⃗2+s⃗2,r⃗3)=cD(r⃗1,r⃗2,r⃗3)+D(r⃗1,s⃗2,r⃗3)D(\vec{r}_1, c\vec{r}_2 + \vec{s}_2, \vec{r}_3) = c D(\vec{r}_1, \vec{r}_2, \vec{r}_3) + D(\vec{r}_1, \vec{s}_2, \vec{r}_3). The determinant is linear in the second row while the first and third rows remain unchanged. Option A is incorrect because scaling all rows by factor cc scales the determinant by c3c^3, not cc. The correct relationship would be D(cr⃗1,cr⃗2,cr⃗3)=c3D(r⃗1,r⃗2,r⃗3)D(c\vec{r}_1, c\vec{r}_2, c\vec{r}_3) = c^3 D(\vec{r}_1, \vec{r}_2, \vec{r}_3). Option B fails because determinants aren't linear across multiple rows simultaneously. Adding the same vector to all rows doesn't follow any standard determinant property, and D(s⃗,s⃗,s⃗)=0D(\vec{s}, \vec{s}, \vec{s}) = 0 since identical rows make the determinant zero. Option C makes no mathematical sense—it's not even a proper equation since it claims something equals itself minus something else (unless that "something else" is zero, which isn't generally true). Study tip: Remember that determinants are linear in each individual row or column, but not across multiple rows/columns at once. Practice identifying which single row or column is being modified in linearity problems.

Question 10

Let AA be an n×nn \times n matrix with rows r⃗1,r⃗2,…,r⃗n\vec{r}_1, \vec{r}_2, \ldots, \vec{r}_n. Let BB be a matrix where, for some i≠ji \neq j, the ii-th row of AA has been replaced by the jj-th row of AA, while all other rows remain the same. Which of the following must be true?

  1. det⁡(B)=det⁡(A)\det(B) = \det(A)
  2. det⁡(B)=−det⁡(A)\det(B) = -\det(A)
  3. det⁡(B)=0\det(B) = 0 (correct answer)
  4. det⁡(B)\det(B) cannot be determined from the given information.
Explanation: In the new matrix BB, the ii-th row is now a copy of the jj-th row, i.e., b⃗i=r⃗j\vec{b}_i = \vec{r}_j. The jj-th row of BB is also the jj-th row of AA, so b⃗j=r⃗j\vec{b}_j = \vec{r}_j. This means matrix BB has two identical rows (the ii-th and jj-th rows). A matrix with two identical rows has linearly dependent rows, and its determinant is always 0. This can be shown by performing the row operation Ri→Ri−RjR_i \to R_i - R_j on BB, which results in a row of zeros at row ii. This operation does not change the determinant, and the determinant of a matrix with a zero row is 0.

Question 11

Let AA be a 3×33 \times 3 matrix with det⁡(A)=5\det(A) = 5. If matrix BB is obtained from AA by first multiplying the second row by −2-2, then adding 3 times the first row to the third row, what is det⁡(B)\det(B)?

  1. −10-10 (correct answer)
  2. 1010
  3. −30-30
  4. 3030
Explanation: The correct answer is A. Row operations affect determinants as follows: (1) Multiplying a row by a scalar kk multiplies the determinant by kk, and (2) Adding a multiple of one row to another row does not change the determinant. First, multiplying the second row by −2-2 changes the determinant from 55 to 5×(−2)=−105 \times (-2) = -10. Then, adding 3 times the first row to the third row leaves the determinant unchanged at −10-10. Choice B incorrectly uses the absolute value. Choice C incorrectly multiplies by both −2-2 and 33. Choice D combines both errors from B and C.

Question 12

A student claims that for any 3×33 \times 3 matrix AA, if BB is obtained by adding the first row to the second row and then subtracting the third row from the first row, then det⁡(B)=det⁡(A)\det(B) = \det(A). Which statement best evaluates this claim?

  1. The claim is correct only when the matrix AA is invertible
  2. The claim is incorrect because subtracting a row changes the sign of the determinant
  3. The claim is incorrect because the order of operations affects the final determinant value
  4. The claim is correct because both operations preserve the determinant value (correct answer)
Explanation: When you encounter questions about row operations and determinants, focus on which specific operations preserve the determinant value versus those that change it. Let's analyze the two operations step by step. First, adding the first row to the second row is an elementary row operation of type "add a multiple of one row to another row." This operation always preserves the determinant value. Second, subtracting the third row from the first row is also an "add a multiple of one row to another row" operation (adding −1-1 times the third row to the first row). This also preserves the determinant value. Since both operations preserve the determinant independently, performing them in sequence means det⁡(B)=det⁡(A)\det(B) = \det(A), making the claim correct. Looking at the wrong answers: Choice A incorrectly suggests that invertibility matters here. Elementary row operations that preserve determinants work the same way regardless of whether the matrix is invertible or singular. Choice B contains a fundamental misconception—subtracting a row (which is adding a multiple of one row to another) does not change the determinant's sign. You might be thinking of row swapping, which does flip the sign. Choice C incorrectly suggests that the order of these determinant-preserving operations matters. Since both operations preserve the determinant value, their order is irrelevant to the final result. Remember this key pattern: Adding any multiple of one row to a different row always preserves the determinant, while swapping rows changes the sign, and scaling a row by a constant multiplies the determinant by that constant.

Question 13

Consider matrices AA and BB where BB is obtained from AA by the sequence of operations: (1) multiply row 1 by kk, (2) swap rows 2 and 3, (3) add 5 times row 1 to row 2. If det⁡(A)=6\det(A) = 6 and det⁡(B)=−18\det(B) = -18, what is the value of kk?

  1. 12\frac{1}{2}
  2. −12-\frac{1}{2}
  3. 33 (correct answer)
  4. −3-3
Explanation: The correct answer is C. Track how each operation affects the determinant: (1) Multiplying row 1 by kk changes the determinant from 66 to 6k6k. (2) Swapping rows 2 and 3 changes the determinant from 6k6k to −6k-6k. (3) Adding 5 times row 1 to row 2 leaves the determinant unchanged at −6k-6k. Since det⁡(B)=−18\det(B) = -18, we have −6k=−18-6k = -18, which gives k=3k = 3. Choice A gives the wrong magnitude. Choice B has the wrong sign and magnitude. Choice D has the correct magnitude but wrong sign, resulting from forgetting that the row swap introduces a negative sign.

Question 14

If det⁡(abcdefghi)=7\det\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = 7, what is the value of $$\det\begin{pmatrix} a+2d & b+2e & c+2f \ d & e & f \ 3g & 3h & 3i \end{pmatrix}

  1. 4242
  2. 2121 (correct answer)
  3. 1414
  4. 77
Explanation: The correct answer is B. Using linearity of determinants in the first row: det⁡(a+2db+2ec+2fdef3g3h3i)=det⁡(abcdef3g3h3i)+2det⁡(defdef3g3h3i)\det\begin{pmatrix} a+2d & b+2e & c+2f \\ d & e & f \\ 3g & 3h & 3i \end{pmatrix} = \det\begin{pmatrix} a & b & c \\ d & e & f \\ 3g & 3h & 3i \end{pmatrix} + 2\det\begin{pmatrix} d & e & f \\ d & e & f \\ 3g & 3h & 3i \end{pmatrix} . The second determinant equals 0 because rows 1 and 2 are identical. For the first determinant, factoring out 3 from the third row gives 3det⁡(abcdefghi)=3×7=213\det\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = 3 \times 7 = 21. Choice A incorrectly multiplies by 6 instead of 3. Choice C incorrectly multiplies by 2. Choice D ignores the row scaling entirely.

Question 15

A matrix BB is obtained from a 3×33 \times 3 matrix AA by the single row operation R2→R2−3R1R_2 \to R_2 - 3R_1. If det⁡(B)=12\det(B) = 12, what is det⁡(2A)\det(2A)?

  1. 2424
  2. 1212
  3. 3232
  4. 9696 (correct answer)
Explanation: This is a two-step problem. First, we find det⁡(A)\det(A), then we find det⁡(2A)\det(2A). Step 1: The row operation R2→R2−3R1R_2 \to R_2 - 3R_1 adds a multiple of one row to another. This type of operation does not change the value of the determinant. Therefore, det⁡(A)=det⁡(B)=12\det(A) = \det(B) = 12. Step 2: We need to find det⁡(2A)\det(2A). For an n×nn \times n matrix AA and a scalar cc, the property is det⁡(cA)=cndet⁡(A)\det(cA) = c^n \det(A). Here, AA is a 3×33 \times 3 matrix, so n=3n=3, and the scalar is c=2c=2. det⁡(2A)=23det⁡(A)=8×12=96\det(2A) = 2^3 \det(A) = 8 \times 12 = 96. Distractor A (24) is the result of the common error 2det⁡(A)2\det(A). Distractor C (32) results from first incorrectly concluding det⁡(A)=12/(−3)\det(A) = 12/(-3) or 12/312/3 and then correctly applying the scaling property.

Question 16

What is the determinant of the matrix M=(1234056700892469)M = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 5 & 6 & 7 \\ 0 & 0 & 8 & 9 \\ 2 & 4 & 6 & 9 \end{pmatrix}?

  1. 00
  2. 4040 (correct answer)
  3. 360360
  4. −40-40
Explanation: We can simplify the matrix using a row operation that does not change the determinant. Notice that the fourth row is very similar to twice the first row. Let's perform the row operation R4→R4−2R1R_4 \to R_4 - 2R_1. The new fourth row is: (2,4,6,9)−2(1,2,3,4)=(2−2,4−4,6−6,9−8)=(0,0,0,1)(2, 4, 6, 9) - 2(1, 2, 3, 4) = (2-2, 4-4, 6-6, 9-8) = (0, 0, 0, 1). Since this operation does not change the determinant, we have: det⁡(M)=det⁡(1234056700890001)\det(M) = \det \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 5 & 6 & 7 \\ 0 & 0 & 8 & 9 \\ 0 & 0 & 0 & 1 \end{pmatrix}. This new matrix is upper triangular. The determinant of a triangular matrix is the product of its diagonal entries. det⁡(M)=1×5×8×1=40\det(M) = 1 \times 5 \times 8 \times 1 = 40. Distractor A (0) is a trap for those who incorrectly assume R4R_4 is exactly 2R12R_1. Distractor C (360) is the product of the diagonal entries of the original matrix, which is not triangular.

Question 17

If AA is a 4×44 \times 4 matrix and det⁡(A)=0\det(A) = 0, which of the following statements must be true?

  1. The columns of AA are linearly independent.
  2. The equation Ax⃗=0⃗A\vec{x} = \vec{0} has only the trivial solution x⃗=0⃗\vec{x}=\vec{0}.
  3. Every element in the matrix AA is zero.
  4. One row of AA can be expressed as a linear combination of the other rows. (correct answer)
Explanation: When you encounter a determinant problem, you're dealing with one of the most fundamental concepts in linear algebra that reveals crucial information about a matrix's properties and the linear independence of its rows and columns. If det⁡(A)=0\det(A) = 0, the matrix is called singular or non-invertible. This immediately tells you that the rows (and columns) of the matrix are linearly dependent, meaning at least one row can be written as a linear combination of the others. This is exactly what option D states, making it correct. Let's examine why the other options fail. Option A claims the columns are linearly independent, but this is false – when det⁡(A)=0\det(A) = 0, the columns are actually linearly dependent. Option B suggests the homogeneous equation Ax⃗=0⃗A\vec{x} = \vec{0} has only the trivial solution, but this is wrong too. A zero determinant guarantees that this equation has infinitely many non-trivial solutions, forming the null space of A. Option C makes an extreme claim that every element is zero. While the zero matrix does have determinant zero, many non-zero matrices also have zero determinants – for example, any matrix with two identical rows. The key insight is that det⁡(A)=0\det(A) = 0 is equivalent to linear dependence among rows (or columns). Think of the determinant as a "linear independence detector" – it's zero precisely when the vectors that form the matrix's rows or columns don't span the full space. Remember: Zero determinant always means linear dependence, which opens the door to non-trivial solutions and eliminates invertibility.

Question 18

If det⁡(2u2v2wxyz3u+x3v+y3w+z)=48\det\begin{pmatrix} 2u & 2v & 2w \\ x & y & z \\ 3u+x & 3v+y & 3w+z \end{pmatrix} = 48, what is $$\det\begin{pmatrix} u & v & w \ x & y & z \ u & v & w \end{pmatrix}

  1. 2424
  2. 88
  3. 1212
  4. 00 (correct answer)
Explanation: When you encounter determinant problems involving matrix transformations, focus on how row operations and linear combinations affect determinant values. Let's work with the given matrix systematically. First, notice that the third row is 3u+x,3v+y,3w+z3u+x, 3v+y, 3w+z, which can be written as 3(u,v,w)+(x,y,z)3(u,v,w) + (x,y,z). Using the linearity property of determinants, we can split this: The second determinant equals zero because it has two identical rows (the second and third rows are the same). So we have: Now factor out constants: factor 2 from the first row and 3 from the third row: But the target matrix has identical first and third rows, making its determinant zero. Choice A (24) assumes you divided 48 by 2 only. Choice B (8) assumes you divided by 6 but ignored the repeated rows. Choice C (12) assumes you divided by 4 somehow. Key insight: Any matrix with two identical rows has determinant zero, regardless of other calculations. Always check for this pattern first.