Linear Algebra Quiz: 2x2 And 3x3 Determinants
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2x2 And 3x3 DeterminantsQuestion 1 of 18

What is the determinant of the matrix B=(2−134−2615−7)B = \begin{pmatrix} 2 & -1 & 3 \\ 4 & -2 & 6 \\ 1 & 5 & -7 \end{pmatrix}?

28
-7
66
0
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Linear Algebra Quiz

Linear Algebra Quiz: 2x2 And 3x3 Determinants

Practice 2x2 And 3x3 Determinants 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 2x2 And 3x3 Determinants, 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.

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Question 1

What is the determinant of the matrix B=(2−134−2615−7)B = \begin{pmatrix} 2 & -1 & 3 \\ 4 & -2 & 6 \\ 1 & 5 & -7 \end{pmatrix}?

  1. 28
  2. -7
  3. 66
  4. 0 (correct answer)
Explanation: This problem can be solved by observing the relationship between the rows of the matrix. Notice that Row 2 is a scalar multiple of Row 1: (4,−2,6)=2×(2,−1,3)(4, -2, 6) = 2 \times (2, -1, 3). A fundamental property of determinants is that if one row (or column) of a matrix is a scalar multiple of another row (or column), the rows are linearly dependent, and the determinant of the matrix is 0. Without noticing this property, a full calculation would be: det⁡(B)=2((−2)(−7)−(6)(5))−(−1)((4)(−7)−(6)(1))+3((4)(5)−(−2)(1))\det(B) = 2((-2)(-7) - (6)(5)) - (-1)((4)(-7) - (6)(1)) + 3((4)(5) - (-2)(1)) =2(14−30)+1(−28−6)+3(20+2)= 2(14 - 30) + 1(-28 - 6) + 3(20 + 2) =2(−16)−34+3(22)= 2(-16) - 34 + 3(22) =−32−34+66=0= -32 - 34 + 66 = 0. Distractor A is the product of the diagonal elements. Distractor B is the sum of the diagonal elements (the trace). Distractor C is a plausible result of a sign error in the expansion.

Question 2

Let AA be a 3×33 \times 3 matrix with det⁡(A)=10\det(A) = 10. A new matrix BB is formed by performing the following sequence of row operations on AA:

  1. Swap Row 1 and Row 3. (R1↔R3R_1 \leftrightarrow R_3)
  2. Multiply Row 2 by a factor of 4. (R2→4R2R_2 \to 4R_2)
  3. Add -2 times Row 2 to Row 1. (R1→R1−2R2R_1 \to R_1 - 2R_2) What is the value of det⁡(B)\det(B)?
  1. 40
  2. -40 (correct answer)
  3. 80
  4. -10
Explanation: We need to track how each elementary row operation affects the determinant. Starting with det⁡(A)=10\det(A) = 10.
  1. Swap Row 1 and Row 3: Swapping any two rows of a matrix multiplies its determinant by -1. After this step, the new determinant is −1×det⁡(A)=−1×10=−10-1 \times \det(A) = -1 \times 10 = -10.
  2. Multiply Row 2 by a factor of 4: Multiplying a single row of a matrix by a scalar kk multiplies the determinant by kk. After this step, the new determinant is 4×(−10)=−404 \times (-10) = -40.
  3. Add -2 times Row 2 to Row 1: Adding a multiple of one row to another row does not change the determinant. After this step, the determinant remains −40-40.
Therefore, det⁡(B)=−40\det(B) = -40. Distractor A results from forgetting the sign change due to the row swap. Distractor C results from incorrectly believing the third operation multiplies the determinant by -2. Distractor D results from ignoring the effect of the second operation.

Question 3

Two students are asked to find det⁡(5832)\det\begin{pmatrix} 5 & 8 \\ 3 & 2 \end{pmatrix} . Student A computes (5)(2)−(8)(3)=10−24=−14(5)(2) - (8)(3) = 10 - 24 = -14. Student B computes (5)(2)−(3)(8)=10−24=−14(5)(2) - (3)(8) = 10 - 24 = -14. Which statement is correct?

  1. Both students made errors but coincidentally arrived at the same incorrect answer
  2. Student A used the correct method, Student B made an error but got lucky with the right answer
  3. Student B used the correct method, Student A made an error but got lucky with the right answer
  4. Both students used correct methods and arrived at the correct answer (correct answer)
Explanation: When you encounter a 2×2 determinant, you're applying a fundamental formula, but there's an important property about matrix operations that this question highlights: the determinant formula has some flexibility in how you can think about it. For a 2×2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, the determinant is ad−bcad - bc. Student A computed (5)(2)−(8)(3)(5)(2) - (8)(3), which follows this standard formula exactly: multiply the main diagonal elements and subtract the product of the off-diagonal elements. Student B computed (5)(2)−(3)(8)(5)(2) - (3)(8), which rearranges the subtraction term but gives the same result since multiplication is commutative: bc=cbbc = cb. Both approaches are mathematically valid because (8)(3)=(3)(8)=24(8)(3) = (3)(8) = 24. Option A is wrong because both students used valid mathematical reasoning—there were no errors. Option B incorrectly suggests Student B made an error, but rearranging factors in multiplication doesn't constitute an error. Option C makes the opposite mistake, claiming Student A was wrong when they used the textbook formula perfectly. Option D is correct because both students applied mathematically sound methods and reached the right answer of -14. Study tip: Remember that while the determinant formula ad−bcad - bc is typically written in a specific order, the commutative property of multiplication means you can rearrange factors within each term without changing the result. Focus on getting the sign pattern right (positive for main diagonal, negative for off-diagonal) rather than worrying about the exact order of multiplication.

Question 4

Matrix F=(k2−11k2−11k)F = \begin{pmatrix} k & 2 & -1 \\ 1 & k & 2 \\ -1 & 1 & k \end{pmatrix} has the property that det⁡(F)=0\det(F) = 0 for exactly three distinct values of kk. What is the sum of these three values?

  1. 66
  2. 33
  3. −3-3
  4. 00 (correct answer)
Explanation: When you encounter a matrix with a parameter and need to find when the determinant equals zero, you're looking for the characteristic polynomial's roots. The determinant being zero means the matrix is singular (non-invertible). To find det⁡(F)=0\det(F) = 0, calculate the determinant using cofactor expansion along the first row: =k(k2−2)−2(k+2)−(1+k)= k(k^2 - 2) - 2(k + 2) - (1 + k) =k3−2k−2k−4−1−k= k^3 - 2k - 2k - 4 - 1 - k =k3−5k−5= k^3 - 5k - 5 Setting this equal to zero: k3−5k−5=0k^3 - 5k - 5 = 0 This is a cubic equation, which can have at most 3 real roots. Since the problem states there are exactly 3 distinct values of kk, all three roots are real and distinct. By Vieta's formulas, for a cubic ak3+bk2+ck+d=0ak^3 + bk^2 + ck + d = 0, the sum of roots equals −ba-\frac{b}{a}. In our equation k3+0k2−5k−5=0k^3 + 0k^2 - 5k - 5 = 0, we have a=1a = 1 and b=0b = 0, so the sum of roots is −01=0-\frac{0}{1} = 0. Choice (A) 66 and (B) 33 are positive sums that might tempt you if you misapplied Vieta's formulas or made arithmetic errors. Choice (C) −3-3 could result from incorrectly thinking the sum relates to the constant term. Choice (D) 00 is correct because there's no k2k^2 term in our cubic. Study tip: Remember Vieta's formulas—the sum of roots of axn+bxn−1+...=0ax^n + bx^{n-1} + ... = 0 is −ba-\frac{b}{a}. When the second-highest degree term has coefficient zero, the sum of roots is zero.

Question 5

A student computes det⁡(4723)\det\begin{pmatrix} 4 & 7 \\ 2 & 3 \end{pmatrix} and gets -2. To check this answer, they compute $$\det\begin{pmatrix} 3 & 7 \ 2 & 4 \end{pmatrix}

  1. −2-2 (correct answer)
  2. 22
  3. −6-6
  4. 66
Explanation: The original determinant is 4⋅3−7⋅2=12−14=−24 \cdot 3 - 7 \cdot 2 = 12 - 14 = -2, so the student's answer is correct. For the swapped matrix: det⁡(3724)=3⋅4−7⋅2=12−14=−2\det\begin{pmatrix} 3 & 7 \\ 2 & 4 \end{pmatrix} = 3 \cdot 4 - 7 \cdot 2 = 12 - 14 = -2. The determinants are the same because swapping diagonal entries doesn't change the value ac−bdac - bd when the off-diagonal entries remain in the same positions. Choice B assumes the sign changes. Choice C assumes 3⋅4−7⋅2=12−18=−63 \cdot 4 - 7 \cdot 2 = 12 - 18 = -6. Choice D assumes 3⋅4−2⋅7=12−14=−23 \cdot 4 - 2 \cdot 7 = 12 - 14 = -2 but with wrong sign.

Question 6

For the matrix B=(cos⁡θ−sin⁡θ0sin⁡θcos⁡θ0002)B = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 2 \end{pmatrix} , what is det⁡(B)\det(B)?

  1. 2cos⁡θ2\cos\theta
  2. 22 (correct answer)
  3. cos⁡2θ+sin⁡2θ\cos^2\theta + \sin^2\theta
  4. 2(cos⁡2θ−sin⁡2θ)2(\cos^2\theta - \sin^2\theta)
Explanation: Using cofactor expansion along the third row: det⁡(B)=0⋅C31+0⋅C32+2⋅C33=2⋅det⁡(cos⁡θ−sin⁡θsin⁡θcos⁡θ)=2(cos⁡2θ−(−sin⁡θ)(sin⁡θ))=2(cos⁡2θ+sin⁡2θ)=2(1)=2\det(B) = 0 \cdot C_{31} + 0 \cdot C_{32} + 2 \cdot C_{33} = 2 \cdot \det\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} = 2(\cos^2\theta - (-\sin\theta)(\sin\theta)) = 2(\cos^2\theta + \sin^2\theta) = 2(1) = 2. Choice A incorrectly assumes only the cos⁡θ\cos\theta term matters. Choice C forgets the factor of 2 from the (3,3) entry. Choice D incorrectly computes the 2×2 determinant as cos⁡2θ−sin⁡2θ\cos^2\theta - \sin^2\theta.

Question 7

What is the determinant of the matrix R=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)R = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}?

  1. 1 (correct answer)
  2. -1
  3. 0
  4. cos⁡(2θ)\cos(2\theta)
Explanation: The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is ad−bcad-bc. For the given matrix RR, we have a=cos⁡θa = \cos\theta, b=−sin⁡θb = -\sin\theta, c=sin⁡θc = \sin\theta, and d=cos⁡θd = \cos\theta. Calculate the determinant: det⁡(R)=(cos⁡θ)(cos⁡θ)−(−sin⁡θ)(sin⁡θ)\det(R) = (\cos\theta)(\cos\theta) - (-\sin\theta)(\sin\theta) =cos⁡2θ+sin⁡2θ= \cos^2\theta + \sin^2\theta This is the fundamental Pythagorean identity in trigonometry, which states that cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1 for any angle θ\theta. Thus, the determinant is 1. Distractor D, cos⁡(2θ)\cos(2\theta), is the value of the similar-looking expression cos⁡2θ−sin⁡2θ\cos^2\theta - \sin^2\theta. Distractor B or C would result from sign errors or incorrect application of trigonometric identities.

Question 8

Let matrix M=(k−124k+1)M = \begin{pmatrix} k-1 & 2 \\ 4 & k+1 \end{pmatrix}. For which set of values of kk is the matrix MM singular?

  1. {3,−3}\{3, -3\} (correct answer)
  2. {7,−7}\{\sqrt{7}, -\sqrt{7}\}
  3. {1,−1}\{1, -1\}
  4. {}\{\}
Explanation: A matrix is singular if its determinant is equal to zero. The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is ad−bcad-bc. For matrix MM, the determinant is det⁡(M)=(k−1)(k+1)−(2)(4)\det(M) = (k-1)(k+1) - (2)(4). Simplify the expression: det⁡(M)=(k2−1)−8=k2−9\det(M) = (k^2 - 1) - 8 = k^2 - 9. Set the determinant to zero to find the values of kk for which the matrix is singular: k2−9=0k^2 - 9 = 0 (k−3)(k+3)=0(k-3)(k+3) = 0 This gives two solutions: k=3k=3 and k=−3k=-3. Therefore, the set of values is {3,−3}\{3, -3\}. Distractor B is incorrect; it results from the calculation error (k−1)(k+1)+(2)(4)=k2+7=0(k-1)(k+1) + (2)(4) = k^2+7=0. Distractor C is incorrect; it results from incorrectly solving by setting the diagonal terms to zero individually. Distractor D is incorrect; it would be the answer if the quadratic equation had no real solutions.

Question 9

What is the area of a triangle with vertices at coordinates (1,2)(1, 2), (4,3)(4, 3), and (2,5)(2, 5)?

  1. 4 (correct answer)
  2. 5
  3. 8
  4. 10
Explanation: The area of a triangle with vertices (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), and (x3,y3)(x_3, y_3) can be found using the formula: Area =12∣det⁡(M)∣= \frac{1}{2} |\det(M)|, where M=(x1y11x2y21x3y31)M = \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix}. Let's set up the matrix with the given vertices: M=(121431251)M = \begin{pmatrix} 1 & 2 & 1 \\ 4 & 3 & 1 \\ 2 & 5 & 1 \end{pmatrix}. Now, calculate the determinant of MM: det⁡(M)=1(3⋅1−5⋅1)−2(4⋅1−2⋅1)+1(4⋅5−2⋅3)\det(M) = 1(3 \cdot 1 - 5 \cdot 1) - 2(4 \cdot 1 - 2 \cdot 1) + 1(4 \cdot 5 - 2 \cdot 3) =1(3−5)−2(4−2)+1(20−6)= 1(3-5) - 2(4-2) + 1(20-6) =1(−2)−2(2)+1(14)= 1(-2) - 2(2) + 1(14) =−2−4+14=8= -2 - 4 + 14 = 8. The area is 12∣8∣=4\frac{1}{2} |8| = 4. An alternative method is to translate one vertex to the origin, e.g., (1,2)(1,2) to (0,0)(0,0). The other vertices become (4−1,3−2)=(3,1)(4-1, 3-2) = (3,1) and (2−1,5−2)=(1,3)(2-1, 5-2) = (1,3). The area is half the absolute value of the determinant of the matrix formed by these new vectors: Area =12∣det⁡(3113)∣=12∣(3)(3)−(1)(1)∣=12∣9−1∣=12(8)=4= \frac{1}{2} |\det\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}| = \frac{1}{2} |(3)(3)-(1)(1)| = \frac{1}{2} |9-1| = \frac{1}{2}(8) = 4. Distractor C is the value of the determinant before multiplying by 12\frac{1}{2}. Distractor B results from an error in the 2x2 determinant calculation, computing ad+bcad+bc instead of ad−bcad-bc, leading to 12∣9+1∣=5\frac{1}{2}|9+1|=5.

Question 10

Let AA be a 2×22 \times 2 matrix such that det⁡(A)=5\det(A) = 5. What is the value of det⁡(3A)\det(3A)?

  1. 15
  2. 45 (correct answer)
  3. 5
  4. 135
Explanation: For any n×nn \times n matrix AA and any scalar kk, the determinant of the matrix kAkA is given by the property det⁡(kA)=kndet⁡(A)\det(kA) = k^n \det(A). In this problem, AA is a 2×22 \times 2 matrix, so n=2n=2. The scalar is k=3k=3, and we are given det⁡(A)=5\det(A)=5. Applying the formula: det⁡(3A)=32det⁡(A)\det(3A) = 3^2 \det(A) =9×5= 9 \times 5 =45= 45. Distractor A is the result of the common mistake of thinking det⁡(kA)=kdet⁡(A)\det(kA) = k \det(A). Distractor C incorrectly assumes the scalar has no effect. Distractor D is the result of incorrectly assuming the matrix is 3×33 \times 3, which would yield 33det⁡(A)=27×5=1353^3 \det(A) = 27 \times 5 = 135.

Question 11

Find the value of xx for which the determinant of matrix C=(10−23x12−14)C = \begin{pmatrix} 1 & 0 & -2 \\ 3 & x & 1 \\ 2 & -1 & 4 \end{pmatrix} is equal to 7.

  1. 1/4
  2. 7/4
  3. 0 (correct answer)
  4. -7/4
Explanation: To find the value of xx, we first need to compute the determinant of matrix CC in terms of xx. It is efficient to use cofactor expansion along the first row, as it contains a zero. det⁡(C)=1⋅det⁡(x1−14)−0⋅det⁡(3124)+(−2)⋅det⁡(3x2−1)\det(C) = 1 \cdot \det\begin{pmatrix} x & 1 \\ -1 & 4 \end{pmatrix} - 0 \cdot \det\begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix} + (-2) \cdot \det\begin{pmatrix} 3 & x \\ 2 & -1 \end{pmatrix} Now compute the 2×22 \times 2 determinants: det⁡(C)=1((x)(4)−(1)(−1))−0+(−2)((3)(−1)−(x)(2))\det(C) = 1((x)(4) - (1)(-1)) - 0 + (-2)((3)(-1) - (x)(2)) =1(4x+1)−2(−3−2x)= 1(4x + 1) - 2(-3 - 2x) =4x+1+6+4x= 4x + 1 + 6 + 4x =8x+7= 8x + 7. We are given that the determinant is equal to 7. So, we set up the equation: 8x+7=78x + 7 = 7 8x=08x = 0 x=0x = 0. Distractor A results from the calculation error 8x+5=78x+5=7, which leads to 8x=28x=2 and x=1/4x=1/4. Distractors B and D result from other common algebraic errors when solving for xx.

Question 12

Given matrices A=(3122)A = \begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix} and B=(x41−1)B = \begin{pmatrix} x & 4 \\ 1 & -1 \end{pmatrix}, if det⁡(AB)=−24\det(AB) = -24, what is the value of xx?

  1. 10
  2. -1
  3. -4
  4. 2 (correct answer)
Explanation: This problem is most efficiently solved using the property that for any two square matrices AA and BB of the same size, det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B). First, calculate the determinant of matrix AA: det⁡(A)=(3)(2)−(1)(2)=6−2=4\det(A) = (3)(2) - (1)(2) = 6 - 2 = 4. Next, calculate the determinant of matrix BB in terms of xx: det⁡(B)=(x)(−1)−(4)(1)=−x−4\det(B) = (x)(-1) - (4)(1) = -x - 4. Now, use the determinant property: det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A) \det(B) −24=(4)(−x−4)-24 = (4)(-x - 4) Divide both sides by 4: −6=−x−4-6 = -x - 4 Add 4 to both sides: −2=−x-2 = -x Multiply by -1: x=2x = 2. Distractor A, x=10x=10, results from a sign error in det⁡(B)\det(B), yielding 4(−x+4)=−244(-x+4)=-24, so −x+4=−6-x+4=-6, and −x=−10-x=-10. Distractor B, x=−1x=-1, results from a sign error in det⁡(A)\det(A), yielding 8(−x−4)=−248(-x-4)=-24, so −x−4=−3-x-4=-3, and −x=1-x=1.

Question 13

For which value of cc does the system of linear equations below lack a unique solution?

{(c−2)x+5y=103x+cy=15\begin{cases} (c-2)x + 5y = 10 \\ 3x + cy = 15 \end{cases}
  1. 3
  2. -5
  3. 2
  4. 5 (correct answer)
Explanation: A system of linear equations of the form Ax=bA\mathbf{x} = \mathbf{b} has a unique solution if and only if the determinant of the coefficient matrix AA is non-zero. Therefore, the system will lack a unique solution (i.e., have no solution or infinitely many solutions) if the determinant of the coefficient matrix is zero. The coefficient matrix for this system is A=(c−253c)A = \begin{pmatrix} c-2 & 5 \\ 3 & c \end{pmatrix}. We need to find the value of cc for which det⁡(A)=0\det(A) = 0. det⁡(A)=(c−2)(c)−(5)(3)\det(A) = (c-2)(c) - (5)(3) =c2−2c−15= c^2 - 2c - 15 Set the determinant to zero and solve for cc: c2−2c−15=0c^2 - 2c - 15 = 0 Factor the quadratic equation: (c−5)(c+3)=0(c-5)(c+3) = 0 The values of cc for which the system does not have a unique solution are c=5c=5 and c=−3c=-3. Of the choices provided, c=5c=5 is listed. Distractor C, c=2c=2, is a value that makes an entry in the matrix zero but does not make the determinant zero. Distractors A and B result from sign errors when factoring the quadratic equation.

Question 14

What is the determinant of the matrix A=(58−40−29003)A = \begin{pmatrix} 5 & 8 & -4 \\ 0 & -2 & 9 \\ 0 & 0 & 3 \end{pmatrix}?

  1. 6
  2. 30
  3. -30 (correct answer)
  4. 0
Explanation: The given matrix AA is an upper triangular matrix because all the entries below the main diagonal are zero. A property of triangular matrices (both upper and lower) is that their determinant is the product of the elements on the main diagonal. So, det⁡(A)=(5)×(−2)×(3)\det(A) = (5) \times (-2) \times (3). det⁡(A)=−10×3=−30\det(A) = -10 \times 3 = -30. Alternatively, one could use cofactor expansion along the first column: det⁡(A)=5⋅det⁡(−2903)−0⋅(...)+0⋅(...)=5((−2)(3)−(9)(0))=5(−6)=−30\det(A) = 5 \cdot \det\begin{pmatrix} -2 & 9 \\ 0 & 3 \end{pmatrix} - 0 \cdot (...) + 0 \cdot (...) = 5((-2)(3) - (9)(0)) = 5(-6) = -30. Distractor A is the sum of the diagonal elements (the trace). Distractor B is the correct magnitude but with an incorrect sign. Distractor D could result from a calculation error or incorrectly assuming the presence of zeros makes the determinant zero.

Question 15

The determinant of a 3×33 \times 3 matrix AA is found to be 0. Which of the following statements must be true about the matrix AA?

  1. The matrix contains a row of zeros.
  2. Two columns of the matrix are identical.
  3. The column vectors of the matrix are coplanar. (correct answer)
  4. The trace of the matrix is zero.
Explanation: A determinant of zero for a 3×33 \times 3 matrix indicates several equivalent conditions. The matrix is singular (not invertible), its rows are linearly dependent, and its columns are linearly dependent. Geometrically, the absolute value of the determinant of a 3×33 \times 3 matrix represents the volume of the parallelepiped formed by its three column (or row) vectors. If the determinant is zero, this volume is zero. This happens if and only if the three vectors lie in the same plane (i.e., they are coplanar). Let's analyze the distractors:
  • A: A matrix with a row of zeros has a determinant of 0, but a matrix can have a determinant of 0 without a row of zeros (e.g., if one row is a multiple of another). So, this is a sufficient but not a necessary condition.
  • B: A matrix with two identical columns has a determinant of 0, but this is also a sufficient, not a necessary condition.
  • C: The column vectors being coplanar is the geometric definition of linear dependence in 3D, which is equivalent to the determinant being zero. This statement must be true.
  • D: The trace (the sum of the main diagonal elements) has no required relationship with the determinant being zero. For example, the identity matrix has a trace of 3 and a determinant of 1, while (111222345)\begin{pmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 4 & 5 \end{pmatrix} has a determinant of 0 but a trace of 8.

Question 16

Evaluate the determinant of the matrix M=(xyzx+1y+1z+1x+2y+2z+2)M = \begin{pmatrix} x & y & z \\ x+1 & y+1 & z+1 \\ x+2 & y+2 & z+2 \end{pmatrix}.

  1. 1
  2. -2
  3. 0 (correct answer)
  4. x+y+zx+y+z
Explanation: The determinant can be evaluated using row operations, which simplify the matrix without changing the determinant's value (for operations of the type Ri→Ri+kRjR_i \to R_i + kR_j). Let's perform the following row operations:
  1. Replace Row 2 with Row 2 - Row 1 (R2→R2−R1R_2 \to R_2 - R_1).
  2. Replace Row 3 with Row 3 - Row 1 (R3→R3−R1R_3 \to R_3 - R_1).
This gives a new matrix M′M' with the same determinant: R2−R1=((x+1)−x,(y+1)−y,(z+1)−z)=(1,1,1)R_2 - R_1 = ((x+1)-x, (y+1)-y, (z+1)-z) = (1, 1, 1) R3−R1=((x+2)−x,(y+2)−y,(z+2)−z)=(2,2,2)R_3 - R_1 = ((x+2)-x, (y+2)-y, (z+2)-z) = (2, 2, 2) So, M′=(xyz111222)M' = \begin{pmatrix} x & y & z \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{pmatrix}. In matrix M′M', Row 3 is a scalar multiple of Row 2 (specifically, R3=2R2R_3 = 2R_2). A matrix with one row being a multiple of another has linearly dependent rows, and its determinant is 0. Therefore, det⁡(M)=det⁡(M′)=0\det(M) = \det(M') = 0. The other choices represent plausible but incorrect guesses if one does not see the structural property of the matrix.

Question 17

The determinant of matrix $$E = \begin{pmatrix} 2 & 1 & 3 \ 1 & 2 & 1 \ 3 & 1 & 2 \end{pmatrix}

  1. Cofactor expansion along any row, since all entries are small positive integers
  2. Cofactor expansion along the main diagonal, since it contains the largest entries
  3. Row operations to create zeros before expansion, since no row or column has zeros initially (correct answer)
  4. Direct application of the Rule of Sarrus, since this is a 3×3 matrix with a symmetric pattern
Explanation: Since matrix EE has no zeros, cofactor expansion along any row or column requires computing three 2×2 determinants. However, using row operations to create zeros first would reduce this to computing only one 2×2 determinant. Choice A ignores the computational advantage of creating zeros. Choice B incorrectly suggests expanding along the diagonal (not a standard method). Choice D mentions Sarrus's rule, which works but isn't more efficient than expansion with zeros created first.

Question 18

If det⁡(abcd)=7\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = 7 and det⁡(2a2b3c3d)=k⋅7\det\begin{pmatrix} 2a & 2b \\ 3c & 3d \end{pmatrix} = k \cdot 7, what is the value of kk?

  1. k=4k = 4
  2. k=5k = 5
  3. k=6k = 6 (correct answer)
  4. k=12k = 12
Explanation: We can factor out scalars from rows: det⁡(2a2b3c3d)=det⁡(2003)⋅det⁡(abcd)\det\begin{pmatrix} 2a & 2b \\ 3c & 3d \end{pmatrix} = \det\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} \cdot \det\begin{pmatrix} a & b \\ c & d \end{pmatrix} . Actually, this is incorrect. The proper approach uses the fact that multiplying a row by a scalar multiplies the determinant by that scalar. So det⁡(2a2b3c3d)=2⋅3⋅det⁡(abcd)=6⋅7=42\det\begin{pmatrix} 2a & 2b \\ 3c & 3d \end{pmatrix} = 2 \cdot 3 \cdot \det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = 6 \cdot 7 = 42. Therefore k=6k = 6. Choice A assumes k=2⋅2=4k = 2 \cdot 2 = 4. Choice B assumes k=2+3=5k = 2 + 3 = 5. Choice D assumes k=2⋅2⋅3=12k = 2 \cdot 2 \cdot 3 = 12.