Linear Algebra Quiz: Cofactor Expansion And Minors
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Cofactor Expansion And MinorsQuestion 1 of 14

Let AA be a 3×33 \times 3 matrix given by

Let CijC_{ij} denote the cofactor of the entry aija_{ij}. What is the value of the expression a21C31+a22C32+a23C33a_{21}C_{31} + a_{22}C_{32} + a_{23}C_{33}?

-2
0
2
14
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Linear Algebra Quiz

Linear Algebra Quiz: Cofactor Expansion And Minors

Practice Cofactor Expansion And Minors in Linear Algebra 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 Cofactor Expansion And Minors, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

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Question 1

Let AA be a 3×33 \times 3 matrix given by

Let CijC_{ij} denote the cofactor of the entry aija_{ij}. What is the value of the expression a21C31+a22C32+a23C33a_{21}C_{31} + a_{22}C_{32} + a_{23}C_{33}?

  1. -2
  2. 0 (correct answer)
  3. 2
  4. 14
Explanation: This question tests a fundamental property of cofactor expansion. The sum of the products of the elements of one row with the cofactors of a different row is always zero. The expression a21C31+a22C32+a23C33a_{21}C_{31} + a_{22}C_{32} + a_{23}C_{33} uses the elements from row 2 (a21=3,a22=−1,a23=4a_{21}=3, a_{22}=-1, a_{23}=4) and the cofactors from row 3. Because the row indices do not match (2≠32 \neq 3), the result must be 0. For verification, the actual determinant of AA is found by expanding along a single row with its own cofactors, e.g., row 1: det⁡(A)=a11C11+a12C12+a13C13=2((−1)(2)−4(1))−1(3(2)−4(5))+0=2(−6)−1(−14)=−12+14=2\det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} = 2((-1)(2) - 4(1)) - 1(3(2) - 4(5)) + 0 = 2(-6) - 1(-14) = -12 + 14 = 2. Distractor (C) is the determinant of AA, which is a common mistake. Distractor (A) is the negative of the determinant. Distractor (D) might arise from a calculation error if one were to compute the cofactors and sum the products without recognizing the property.

Question 2

Consider the matrix AA:

In the calculation of det⁡(A)\det(A), the total value is a sum of contributions from the elements of any chosen row or column. What is the contribution to the determinant from the entry a23=6a_{23}=6?

  1. -84
  2. -14
  3. 14
  4. 84 (correct answer)
Explanation: The contribution of an entry aija_{ij} to the determinant is the term aijCija_{ij}C_{ij} in the cofactor expansion. Here we are interested in the entry a23=6a_{23}=6. First, we must calculate its cofactor, C23C_{23}. C23=(−1)2+3M23=−1⋅det⁡(143−2)C_{23} = (-1)^{2+3} M_{23} = -1 \cdot \det\begin{pmatrix} 1 & 4 \\ 3 & -2 \end{pmatrix}. The minor is M23=1(−2)−4(3)=−2−12=−14M_{23} = 1(-2) - 4(3) = -2 - 12 = -14. So, the cofactor is C23=−1(−14)=14C_{23} = -1(-14) = 14. The contribution to the determinant is a23C23=6×14=84a_{23}C_{23} = 6 \times 14 = 84. Distractor (C) is just the cofactor C23C_{23}. Distractor (B) is the minor M23M_{23}. Distractor (A) results from a sign error when multiplying the element by its cofactor.

Question 3

Let AA be the matrix

Let BB be the matrix formed by changing the entry a12a_{12} from 2 to 5. What is the value of det⁡(B)−det⁡(A)\det(B) - \det(A)?

  1. -15
  2. 5
  3. 15 (correct answer)
  4. 25
Explanation: The change in the determinant of a matrix when a single entry aija_{ij} is changed by an amount kk (i.e., aija_{ij} becomes aij+ka_{ij}+k) is given by k⋅Cijk \cdot C_{ij}. In this problem, the entry a12a_{12} is changed from 2 to 5. The change is k=5−2=3k = 5 - 2 = 3. The position is (i,j)=(1,2)(i,j)=(1,2). We need to calculate the cofactor C12C_{12} of the original matrix AA. C12=(−1)1+2M12=−1⋅det⁡(0516)C_{12} = (-1)^{1+2} M_{12} = -1 \cdot \det\begin{pmatrix} 0 & 5 \\ 1 & 6 \end{pmatrix} M12=0(6)−5(1)=−5M_{12} = 0(6) - 5(1) = -5. C12=−1(−5)=5C_{12} = -1(-5) = 5. The change in the determinant, det⁡(B)−det⁡(A)\det(B) - \det(A), is k⋅C12=3⋅5=15k \cdot C_{12} = 3 \cdot 5 = 15. Alternatively, one could compute both determinants directly: det⁡(A)=1(24−0)−2(0−5)+3(0−4)=24+10−12=22\det(A) = 1(24-0) - 2(0-5) + 3(0-4) = 24 + 10 - 12 = 22. B=(153045106)B = \begin{pmatrix} 1 & 5 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{pmatrix} det⁡(B)=1(24−0)−5(0−5)+3(0−4)=24+25−12=37\det(B) = 1(24-0) - 5(0-5) + 3(0-4) = 24 + 25 - 12 = 37. det⁡(B)−det⁡(A)=37−22=15\det(B) - \det(A) = 37 - 22 = 15. Distractor (B) is the value of the cofactor C12C_{12}. Distractor (A) comes from using the minor M12M_{12} instead of the cofactor C12C_{12} (3×−5=−153 \times -5 = -15). Distractor (D) might arise from multiplying the new entry value (5) by the cofactor (5×5=255 \times 5 = 25).

Question 4

Let AA be a 3×33 \times 3 matrix with det⁡(A)=4\det(A) = 4. Let CC be the matrix whose entry in the ii-th row and jj-th column is the cofactor CijC_{ij} of aija_{ij}. What is the determinant of the matrix CC?

  1. 4
  2. 8
  3. 16 (correct answer)
  4. 64
Explanation: This question relates the determinant of a matrix to the determinant of its cofactor matrix. The adjugate of AA, denoted adj(A)\mathrm{adj}(A), is the transpose of the cofactor matrix CC. Thus, adj(A)=CT\mathrm{adj}(A) = C^T. A key property is that A⋅adj(A)=det⁡(A)⋅IA \cdot \mathrm{adj}(A) = \det(A) \cdot I. Taking the determinant of both sides gives: det⁡(A⋅adj(A))=det⁡(det⁡(A)⋅I)\det(A \cdot \mathrm{adj}(A)) = \det(\det(A) \cdot I) det⁡(A)⋅det⁡(adj(A))=(det⁡(A))n\det(A) \cdot \det(\mathrm{adj}(A)) = (\det(A))^n, where nn is the dimension of the matrix. Dividing by det⁡(A)\det(A) (since it's non-zero), we get: det⁡(adj(A))=(det⁡(A))n−1\det(\mathrm{adj}(A)) = (\det(A))^{n-1}. Since the determinant of a matrix is equal to the determinant of its transpose, det⁡(C)=det⁡(CT)=det⁡(adj(A))\det(C) = \det(C^T) = \det(\mathrm{adj}(A)). Therefore, det⁡(C)=(det⁡(A))n−1\det(C) = (\det(A))^{n-1}. For this problem, n=3n=3 and det⁡(A)=4\det(A)=4. So, det⁡(C)=(4)3−1=42=16\det(C) = (4)^{3-1} = 4^2 = 16. Distractor (A) incorrectly assumes det⁡(C)=det⁡(A)\det(C)=\det(A). Distractor (D) incorrectly uses the exponent nn instead of n−1n-1, calculating 43=644^3=64. Distractor (B) might arise from other incorrect formulas like 2det⁡(A)2\det(A) or (n−1)det⁡(A)(n-1)\det(A).

Question 5

Consider the block matrix S=(2010030240−100105)S = \begin{pmatrix} 2 & 0 & 1 & 0 \\ 0 & 3 & 0 & 2 \\ 4 & 0 & -1 & 0 \\ 0 & 1 & 0 & 5 \end{pmatrix} . When computing the cofactor C13C_{13} using the standard definition, which of the following 3×33 \times 3 determinants must be evaluated?

  1. (correct answer)
Explanation: The cofactor C13=(−1)1+3⋅M13C_{13} = (-1)^{1+3} \cdot M_{13} where M13M_{13} is the minor obtained by deleting row 1 and column 3 from matrix SS. Deleting row 1 removes the first row (2010)\begin{pmatrix} 2 & 0 & 1 & 0 \end{pmatrix}, and deleting column 3 removes the third column (10−10)\begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix}. This leaves us with the submatrix formed by rows 2, 3, 4 and columns 1, 2, 4, which is $$ \begin{pmatrix} 0 & 3 & 2 \ 4 & 0 & 0 \ 0 & 1 & 5 \end{pmatrix}

Question 6

What is the determinant of the matrix AA?

  1. -120
  2. -40
  3. 40
  4. 120 (correct answer)
Explanation: To calculate the determinant of a 4x4 matrix, it is strategic to expand along a row or column with the most zeros. Column 3 has three zeros, so we will expand along it. det⁡(A)=a13C13+a23C23+a33C33+a43C43\det(A) = a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33} + a_{43}C_{43} Since a13,a23,a43a_{13}, a_{23}, a_{43} are all zero, this simplifies to: det⁡(A)=a33C33=3⋅(−1)3+3M33=3M33\det(A) = a_{33}C_{33} = 3 \cdot (-1)^{3+3} M_{33} = 3 M_{33}. Now we must calculate the minor M33M_{33}: M33=det⁡(214−152116)M_{33} = \det \begin{pmatrix} 2 & 1 & 4 \\ -1 & 5 & 2 \\ 1 & 1 & 6 \end{pmatrix} Expanding along the first row: M33=2(5(6)−2(1))−1((−1)(6)−2(1))+4((−1)(1)−5(1))M_{33} = 2(5(6) - 2(1)) - 1((-1)(6) - 2(1)) + 4((-1)(1) - 5(1)) M33=2(30−2)−1(−6−2)+4(−1−5)M_{33} = 2(30 - 2) - 1(-6 - 2) + 4(-1 - 5) M33=2(28)−1(−8)+4(−6)=56+8−24=40M_{33} = 2(28) - 1(-8) + 4(-6) = 56 + 8 - 24 = 40. Finally, det⁡(A)=3⋅M33=3⋅40=120\det(A) = 3 \cdot M_{33} = 3 \cdot 40 = 120. Distractor (C) is the value of the minor M33M_{33}, but forgets to multiply by the element a33=3a_{33}=3. Distractor (A) comes from making a sign error on the cofactor. Distractor (B) includes a sign error and forgets to multiply by the element.

Question 7

Consider the matrix AA:

For what value of kk is the cofactor C23C_{23} equal to 13?

  1. -14/5
  2. -12/5 (correct answer)
  3. 12/5
  4. 14/5
Explanation: We need to find the value of kk such that C23=13C_{23}=13. The cofactor C23C_{23} is given by the formula C23=(−1)2+3M23=−M23C_{23} = (-1)^{2+3}M_{23} = -M_{23}. First, find the minor M23M_{23} by taking the determinant of the submatrix formed by removing row 2 and column 3: M23=det⁡(k115)=k(5)−1(1)=5k−1M_{23} = \det\begin{pmatrix} k & 1 \\ 1 & 5 \end{pmatrix} = k(5) - 1(1) = 5k - 1. Now, calculate the cofactor: C23=−M23=−(5k−1)=1−5kC_{23} = -M_{23} = -(5k-1) = 1-5k. We are given that C23=13C_{23} = 13, so we set up the equation: 1−5k=131 - 5k = 13 −5k=12-5k = 12 k=−12/5k = -12/5. Distractor (D) arises if the sign for the cofactor is forgotten, setting M23=13M_{23} = 13, which gives 5k−1=13  ⟹  5k=14  ⟹  k=14/55k-1=13 \implies 5k=14 \implies k=14/5. Distractor (C) comes from a sign error in solving the final equation, −5k=12  ⟹  k=12/5-5k=12 \implies k=12/5. Distractor (A) can result from multiple sign errors.

Question 8

Let AA be a general 3×33 \times 3 matrix with elements aija_{ij} and cofactors CijC_{ij}. Which expression correctly represents the cofactor expansion of det⁡(A)\det(A) along the second column?

  1. a12C12+a22C22+a32C32a_{12}C_{12} + a_{22}C_{22} + a_{32}C_{32} (correct answer)
  2. a21M21+a22M22+a23M23a_{21}M_{21} + a_{22}M_{22} + a_{23}M_{23}
  3. a21C21+a22C22+a23C23a_{21}C_{21} + a_{22}C_{22} + a_{23}C_{23}
  4. a12C12−a22C22+a32C32a_{12}C_{12} - a_{22}C_{22} + a_{32}C_{32}
Explanation: The formula for the cofactor expansion of the determinant of an n×nn \times n matrix AA along column jj is given by: det⁡(A)=∑i=1naijCij=a1jC1j+a2jC2j+⋯+anjCnj\det(A) = \sum_{i=1}^{n} a_{ij}C_{ij} = a_{1j}C_{1j} + a_{2j}C_{2j} + \dots + a_{nj}C_{nj}. For expansion along the second column, we set j=2j=2 and n=3n=3. The formula becomes: det⁡(A)=a12C12+a22C22+a32C32\det(A) = a_{12}C_{12} + a_{22}C_{22} + a_{32}C_{32}. This matches choice (A). Choice (C) represents the expansion along the second row, not the second column. Choice (B) incorrectly uses minors (MijM_{ij}) instead of cofactors (CijC_{ij}). The cofactor includes the sign (−1)i+j(-1)^{i+j}, so this expression is not the determinant. Choice (D) incorrectly applies alternating signs to the cofactor terms; the signs are already incorporated within the definition of each cofactor.

Question 9

What is the determinant of the matrix AA?

  1. 2x2+3x−52x^2 + 3x - 5
  2. 2x2−3x+52x^2 - 3x + 5
  3. 2x2−3x−52x^2 - 3x - 5 (correct answer)
  4. −2x2+3x+5-2x^2 + 3x + 5
Explanation: To find the determinant, we can use cofactor expansion along any row or column. Expanding along the second row is efficient due to the zero entry. det⁡(A)=a21C21+a22C22+a23C23\det(A) = a_{21}C_{21} + a_{22}C_{22} + a_{23}C_{23} =(1)⋅(−1)2+1M21+(x)⋅(−1)2+2M22+(0)⋅(−1)2+3M23= (1) \cdot (-1)^{2+1} M_{21} + (x) \cdot (-1)^{2+2} M_{22} + (0) \cdot (-1)^{2+3} M_{23} =−1⋅det⁡(21−12)+x⋅det⁡(x132)+0= -1 \cdot \det\begin{pmatrix} 2 & 1 \\ -1 & 2 \end{pmatrix} + x \cdot \det\begin{pmatrix} x & 1 \\ 3 & 2 \end{pmatrix} + 0 =−1(2(2)−1(−1))+x(x(2)−1(3))= -1(2(2) - 1(-1)) + x(x(2) - 1(3)) =−1(4+1)+x(2x−3)= -1(4 + 1) + x(2x - 3) =−5+2x2−3x= -5 + 2x^2 - 3x Arranging in standard form, the determinant is 2x2−3x−52x^2 - 3x - 5. Distractor (A) results from a sign error on the term −3x-3x. Distractor (B) results from a sign error on the constant term −5-5. Distractor (D) results from an overall sign error.

Question 10

Let AA be the following matrix:

What is the value of the cofactor C23C_{23}?

  1. -26 (correct answer)
  2. -24
  3. 24
  4. 26
Explanation: The cofactor CijC_{ij} is defined as (−1)i+jMij(-1)^{i+j}M_{ij}, where MijM_{ij} is the minor. Here, we need C23C_{23}, so i=2i=2 and j=3j=3. First, we find the minor M23M_{23} by taking the determinant of the submatrix formed by removing row 2 and column 3: Expanding along the first row: M23=1⋅det⁡(1−240)−0⋅det⁡(2−2−10)+2⋅det⁡(21−14)M_{23} = 1 \cdot \det\begin{pmatrix} 1 & -2 \\ 4 & 0 \end{pmatrix} - 0 \cdot \det\begin{pmatrix} 2 & -2 \\ -1 & 0 \end{pmatrix} + 2 \cdot \det\begin{pmatrix} 2 & 1 \\ -1 & 4 \end{pmatrix} M23=1(1(0)−(−2)(4))−0+2(2(4)−1(−1))M_{23} = 1(1(0) - (-2)(4)) - 0 + 2(2(4) - 1(-1)) M23=1(8)+2(8+1)=8+2(9)=8+18=26M_{23} = 1(8) + 2(8 + 1) = 8 + 2(9) = 8 + 18 = 26. Next, we calculate the cofactor using the sign (−1)i+j(-1)^{i+j}: C23=(−1)2+3M23=(−1)5(26)=−1(26)=−26C_{23} = (-1)^{2+3} M_{23} = (-1)^5 (26) = -1(26) = -26. Distractor (D) is the minor M23M_{23}, which neglects the sign factor (−1)i+j(-1)^{i+j}. Distractors (B) and (C) result from calculation errors within the 3x3 determinant.

Question 11

Calculate the determinant of the following matrix:

  1. -177
  2. -59
  3. 59
  4. 177 (correct answer)
Explanation: To find the determinant of this 4x4 matrix, we should expand along the row or column with the most zeros to simplify the calculation. Column 4 has three zeros. The cofactor expansion along column 4 is: det⁡(A)=a14C14+a24C24+a34C34+a44C44\det(A) = a_{14}C_{14} + a_{24}C_{24} + a_{34}C_{34} + a_{44}C_{44} Since a14=a24=a44=0a_{14}=a_{24}=a_{44}=0, this simplifies to: det⁡(A)=a34C34=3⋅C34\det(A) = a_{34}C_{34} = 3 \cdot C_{34}. Now, we calculate the cofactor C34=(−1)3+4M34=−M34C_{34} = (-1)^{3+4} M_{34} = -M_{34}. The minor M34M_{34} is the determinant of the submatrix obtained by removing row 3 and column 4: M34=det⁡(1−2350641−2)M_{34} = \det \begin{pmatrix} 1 & -2 & 3 \\ 5 & 0 & 6 \\ 4 & 1 & -2 \end{pmatrix} To calculate this 3x3 determinant, we can expand along row 2 for its zero: M34=−5⋅det⁡(−231−2)+0−6⋅det⁡(1−241)M_{34} = -5 \cdot \det\begin{pmatrix} -2 & 3 \\ 1 & -2 \end{pmatrix} + 0 - 6 \cdot \det\begin{pmatrix} 1 & -2 \\ 4 & 1 \end{pmatrix} M34=−5((−2)(−2)−3(1))−6(1(1)−(−2)(4))M_{34} = -5((-2)(-2) - 3(1)) - 6(1(1) - (-2)(4)) M34=−5(4−3)−6(1+8)=−5(1)−6(9)=−5−54=−59M_{34} = -5(4 - 3) - 6(1 + 8) = -5(1) - 6(9) = -5 - 54 = -59. So, the cofactor is C34=−M34=−(−59)=59C_{34} = -M_{34} = -(-59) = 59. Finally, the determinant is det⁡(A)=3⋅C34=3⋅59=177\det(A) = 3 \cdot C_{34} = 3 \cdot 59 = 177. Distractor (C) is the value of the cofactor C34C_{34}, not the full determinant. Distractor (B) is the value of the minor M34M_{34}. Distractor (A) results from a sign error, likely by forgetting the negative sign in C34=−M34C_{34} = -M_{34}.

Question 12

Let AA be an n×nn \times n matrix with entries aija_{ij}. Let MijM_{ij} be the minor and CijC_{ij} be the cofactor of aija_{ij}. Which of the following statements is always true?

  1. The value of CijC_{ij} depends on the value of the entry aija_{ij}.
  2. Cij=CjiC_{ij} = C_{ji} for all i,ji, j.
  3. Mij=CijM_{ij} = C_{ij} only if i=ji=j.
  4. If AA is an upper triangular matrix, then C21=0C_{21} = 0. (correct answer)
Explanation: Let's analyze each statement. (A) The cofactor CijC_{ij} is calculated from the submatrix formed by deleting row ii and column jj. Therefore, its value depends on all other entries but is independent of aija_{ij} itself. So, (A) is false. (B) The matrix of cofactors is not generally symmetric. For example, C12=−M12C_{12} = -M_{12} and C21=−M21C_{21} = -M_{21}, but there is no reason for M12M_{12} and M21M_{21} to be equal. So, (B) is false. (C) Cij=(−1)i+jMijC_{ij} = (-1)^{i+j}M_{ij}. Thus, Cij=MijC_{ij} = M_{ij} whenever the exponent i+ji+j is an even number, not just when i=ji=j. For example, C13=M13C_{13} = M_{13}. So, (C) is false. (D) If AA is an upper triangular matrix, all entries below the main diagonal are zero (aij=0a_{ij} = 0 for i>ji > j). We need to find C21C_{21}. The minor M21M_{21} is the determinant of the submatrix formed by removing row 2 and column 1. The first column of this submatrix will be the entries a12,a32,a42,…,an2a_{12}, a_{32}, a_{42}, \dots, a_{n2}. The first entry in every other column of this submatrix will be a1ka_{1k} for k>1k > 1. The entries below the first row of this submatrix will be of the form aika_{ik} where i>2,k>1i>2, k>1. In an upper triangular matrix, ai1=0a_{i1} = 0 for all i>1i > 1. So when we remove row 2 and column 1, the first column of the resulting submatrix M21M_{21} is entirely zeros (it consists of a31,a41,…,an1a_{31}, a_{41}, \dots, a_{n1}, which are all 0). A matrix with a column of zeros has a determinant of zero. Thus, M21=0M_{21}=0, which means C21=(−1)2+1(0)=0C_{21} = (-1)^{2+1}(0) = 0. So, (D) is true.

Question 13

The determinant of the matrix AA is 7. What is the value of kk?

  1. -1/6
  2. 1/6 (correct answer)
  3. 1
  4. 2
Explanation: We are given det⁡(A)=7\det(A) = 7 and need to solve for kk. We can set up an equation using cofactor expansion. Expanding along the second column is a good strategy. det⁡(A)=a12C12+a22C22+a32C32\det(A) = a_{12}C_{12} + a_{22}C_{22} + a_{32}C_{32} 7=k⋅(−1)1+2M12+0⋅C22+(−1)⋅(−1)3+2M327 = k \cdot (-1)^{1+2} M_{12} + 0 \cdot C_{22} + (-1) \cdot (-1)^{3+2} M_{32} 7=−k⋅det⁡(1340)−(−1)⋅(−1)⋅det⁡(2113)7 = -k \cdot \det\begin{pmatrix} 1 & 3 \\ 4 & 0 \end{pmatrix} - (-1) \cdot (-1) \cdot \det\begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} 7=−k(1(0)−3(4))+1(2(3)−1(1))7 = -k(1(0) - 3(4)) + 1(2(3) - 1(1)) 7=−k(−12)+1(6−1)7 = -k(-12) + 1(6 - 1) 7=12k+57 = 12k + 5 2=12k2 = 12k k=2/12=1/6k = 2/12 = 1/6. Distractor (A) results from a sign error on the cofactor for kk, leading to −12k+5=7-12k+5=7. Distractor (C) results from a sign error on the cofactor for the entry -1, leading to 12k−5=712k-5=7. Distractor (D) could result from miscalculating one of the minors.

Question 14

A 4×44 \times 4 matrix UU has the property that exactly half of its 16 cofactors are zero. If UU has rank 3 and the sum of the four cofactors in the first row is 24, what is det⁡(U)\det(U)?

  1. 00 (correct answer)
  2. 2424
  3. 66
  4. 9696
Explanation: Since UU is a 4×44 \times 4 matrix with rank 3, it is not invertible, which means det⁡(U)=0\det(U) = 0. This is true regardless of the cofactor structure or the sum of cofactors in any row. When a matrix has rank less than its dimension, its determinant is always zero. The information about half the cofactors being zero and the sum of first-row cofactors being 24 is consistent with having a rank-3 matrix (where the cofactors can be non-zero even when the determinant is zero), but doesn't change the fundamental fact that det⁡(U)=0\det(U) = 0. Choices B, C, and D incorrectly assume the determinant can be non-zero for a rank-deficient matrix.