Cofactor Expansion & Minors - Linear Algebra

1. Consider the matrix $A$: $A = \begin{pmatrix} k & 1 & 3 \\ 2 & -1 & 0 \\ 1 & 5 & k \end{pmatrix}$ For what value of $k$ is the cofactor $C_{23}$ equal to 13?
   Answer: -12/5 We need to find the value of $k$ such that $C_{23}=13$. The cofactor $C_{23}$ is given by the formula $C_{23} = (-1)^{2+3}M_{23} = -M_{23}$. First, find the minor $M_{23}$ by taking the determinant of the submatrix formed by removing row 2 and column 3: $M_{23} = \det\begin{pmatrix} k & 1 \\ 1 & 5 \end{pmatrix} = k(5) - 1(1) = 5k - 1$. Now, calculate the cofactor: $C_{23} = -M_{23} = -(5k-1) = 1-5k$. We are given that $C_{23} = 13$, so we set up the equation: $1 - 5k = 13$ $-5k = 12$ $k = -12/5$. Distractor (D) arises if the sign for the cofactor is forgotten, setting $M_{23} = 13$, which gives $5k-1=13 \implies 5k=14 \implies k=14/5$. Distractor (C) comes from a sign error in solving the final equation, $-5k=12 \implies k=12/5$. Distractor (A) can result from multiple sign errors.
2. Let $A$ be the following matrix: $A = \begin{pmatrix} 1 & 0 & -1 & 2 \\ 0 & 3 & 2 & 1 \\ 2 & 1 & 0 & -2 \\ -1 & 4 & 1 & 0 \end{pmatrix}$ What is the value of the cofactor $C_{23}$?
   Answer: -26 The cofactor $C_{ij}$ is defined as $(-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor. Here, we need $C_{23}$, so $i=2$ and $j=3$. First, we find the minor $M_{23}$ by taking the determinant of the submatrix formed by removing row 2 and column 3: $M_{23} = \det \begin{pmatrix} 1 & 0 & 2 \\ 2 & 1 & -2 \\ -1 & 4 & 0 \end{pmatrix}$ Expanding along the first row: $M_{23} = 1 \cdot \det\begin{pmatrix} 1 & -2 \\ 4 & 0 \end{pmatrix} - 0 \cdot \det\begin{pmatrix} 2 & -2 \\ -1 & 0 \end{pmatrix} + 2 \cdot \det\begin{pmatrix} 2 & 1 \\ -1 & 4 \end{pmatrix}$ $M_{23} = 1(1(0) - (-2)(4)) - 0 + 2(2(4) - 1(-1))$ $M_{23} = 1(8) + 2(8 + 1) = 8 + 2(9) = 8 + 18 = 26$. Next, we calculate the cofactor using the sign $(-1)^{i+j}$: $C_{23} = (-1)^{2+3} M_{23} = (-1)^5 (26) = -1(26) = -26$. Distractor (D) is the minor $M_{23}$, which neglects the sign factor $(-1)^{i+j}$. Distractors (B) and (C) result from calculation errors within the 3x3 determinant.
3. Consider the block matrix $S = \begin{pmatrix} 2 & 0 & 1 & 0 \\ 0 & 3 & 0 & 2 \\ 4 & 0 & -1 & 0 \\ 0 & 1 & 0 & 5 \end{pmatrix}$. When computing the cofactor $C_{13}$ using the standard definition, which of the following $3 \times 3$ determinants must be evaluated?
   Answer: $\det\begin{pmatrix} 0 & 3 & 2 \\ 4 & 0 & 0 \\ 0 & 1 & 5 \end{pmatrix}$ The cofactor $C_{13} = (-1)^{1+3} \cdot M_{13}$ where $M_{13}$ is the minor obtained by deleting row 1 and column 3 from matrix $S$. Deleting row 1 removes the first row $\begin{pmatrix} 2 & 0 & 1 & 0 \end{pmatrix}$, and deleting column 3 removes the third column $\begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix}$. This leaves us with the submatrix formed by rows 2, 3, 4 and columns 1, 2, 4, which is $\begin{pmatrix} 0 & 3 & 2 \\ 4 & 0 & 0 \\ 0 & 1 & 5 \end{pmatrix}$. Choice B incorrectly includes column 3, choice C uses wrong rows, and choice D includes elements from the deleted row.
4. Let $A$ be a general $3 \times 3$ matrix with elements $a_{ij}$ and cofactors $C_{ij}$. Which expression correctly represents the cofactor expansion of $\det(A)$ along the second column?
   Answer: $a_{12}C_{12} + a_{22}C_{22} + a_{32}C_{32}$ The formula for the cofactor expansion of the determinant of an $n \times n$ matrix $A$ along column $j$ is given by: $\det(A) = \sum_{i=1}^{n} a_{ij}C_{ij} = a_{1j}C_{1j} + a_{2j}C_{2j} + \dots + a_{nj}C_{nj}$. For expansion along the second column, we set $j=2$ and $n=3$. The formula becomes: $\det(A) = a_{12}C_{12} + a_{22}C_{22} + a_{32}C_{32}$. This matches choice (A). Choice (C) represents the expansion along the second row, not the second column. Choice (B) incorrectly uses minors ($M_{ij}$) instead of cofactors ($C_{ij}$). The cofactor includes the sign $(-1)^{i+j}$, 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.
5. What is the determinant of the matrix $A$? $A = \begin{pmatrix} x & 2 & 1 \\ 1 & x & 0 \\ 3 & -1 & 2 \end{pmatrix}$
   Answer: $2x^2 - 3x - 5$ 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. $\det(A) = a_{21}C_{21} + a_{22}C_{22} + a_{23}C_{23}$ $= (1) \cdot (-1)^{2+1} M_{21} + (x) \cdot (-1)^{2+2} M_{22} + (0) \cdot (-1)^{2+3} M_{23}$ $= -1 \cdot \det\begin{pmatrix} 2 & 1 \\ -1 & 2 \end{pmatrix} + x \cdot \det\begin{pmatrix} x & 1 \\ 3 & 2 \end{pmatrix} + 0$ $= -1(2(2) - 1(-1)) + x(x(2) - 1(3))$ $= -1(4 + 1) + x(2x - 3)$ $= -5 + 2x^2 - 3x$ Arranging in standard form, the determinant is $2x^2 - 3x - 5$. Distractor (A) results from a sign error on the term $-3x$. Distractor (B) results from a sign error on the constant term $-5$. Distractor (D) results from an overall sign error.
6. A $4 \times 4$ matrix $U$ has the property that exactly half of its 16 cofactors are zero. If $U$ has rank 3 and the sum of the four cofactors in the first row is 24, what is $\det(U)$?
   Answer: $0$ Since $U$ is a $4 \times 4$ matrix with rank 3, it is not invertible, which means $\det(U) = 0$. 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 $\det(U) = 0$. Choices B, C, and D incorrectly assume the determinant can be non-zero for a rank-deficient matrix.
7. What is the determinant of the matrix $A$? $A = \begin{pmatrix} 2 & 1 & 0 & 4 \\ -1 & 5 & 0 & 2 \\ 3 & -2 & 3 & -1 \\ 1 & 1 & 0 & 6 \end{pmatrix}$
   Answer: 120 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. $\det(A) = a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33} + a_{43}C_{43}$ Since $a_{13}, a_{23}, a_{43}$ are all zero, this simplifies to: $\det(A) = a_{33}C_{33} = 3 \cdot (-1)^{3+3} M_{33} = 3 M_{33}$. Now we must calculate the minor $M_{33}$: $M_{33} = \det \begin{pmatrix} 2 & 1 & 4 \\ -1 & 5 & 2 \\ 1 & 1 & 6 \end{pmatrix}$ Expanding along the first row: $M_{33} = 2(5(6) - 2(1)) - 1((-1)(6) - 2(1)) + 4((-1)(1) - 5(1))$ $M_{33} = 2(30 - 2) - 1(-6 - 2) + 4(-1 - 5)$ $M_{33} = 2(28) - 1(-8) + 4(-6) = 56 + 8 - 24 = 40$. Finally, $\det(A) = 3 \cdot M_{33} = 3 \cdot 40 = 120$. Distractor (C) is the value of the minor $M_{33}$, but forgets to multiply by the element $a_{33}=3$. Distractor (A) comes from making a sign error on the cofactor. Distractor (B) includes a sign error and forgets to multiply by the element.
8. Let $A$ be the matrix $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{pmatrix}$ Let $B$ be the matrix formed by changing the entry $a_{12}$ from 2 to 5. What is the value of $\det(B) - \det(A)$?
   Answer: 15 The change in the determinant of a matrix when a single entry $a_{ij}$ is changed by an amount $k$ (i.e., $a_{ij}$ becomes $a_{ij}+k$) is given by $k \cdot C_{ij}$. In this problem, the entry $a_{12}$ is changed from 2 to 5. The change is $k = 5 - 2 = 3$. The position is $(i,j)=(1,2)$. We need to calculate the cofactor $C_{12}$ of the original matrix $A$. $C_{12} = (-1)^{1+2} M_{12} = -1 \cdot \det\begin{pmatrix} 0 & 5 \\ 1 & 6 \end{pmatrix}$ $M_{12} = 0(6) - 5(1) = -5$. $C_{12} = -1(-5) = 5$. The change in the determinant, $\det(B) - \det(A)$, is $k \cdot C_{12} = 3 \cdot 5 = 15$. Alternatively, one could compute both determinants directly: $\det(A) = 1(24-0) - 2(0-5) + 3(0-4) = 24 + 10 - 12 = 22$. $B = \begin{pmatrix} 1 & 5 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{pmatrix}$ $\det(B) = 1(24-0) - 5(0-5) + 3(0-4) = 24 + 25 - 12 = 37$. $\det(B) - \det(A) = 37 - 22 = 15$. Distractor (B) is the value of the cofactor $C_{12}$. Distractor (A) comes from using the minor $M_{12}$ instead of the cofactor $C_{12}$ ($3 \times -5 = -15$). Distractor (D) might arise from multiplying the new entry value (5) by the cofactor ($5 \times 5 = 25$).
9. Let $A$ be an $n \times n$ matrix with entries $a_{ij}$. Let $M_{ij}$ be the minor and $C_{ij}$ be the cofactor of $a_{ij}$. Which of the following statements is always true?
   Answer: If $A$ is an upper triangular matrix, then $C_{21} = 0$. Let's analyze each statement. (A) The cofactor $C_{ij}$ is calculated from the submatrix formed by deleting row $i$ and column $j$. Therefore, its value depends on all other entries but is independent of $a_{ij}$ itself. So, (A) is false. (B) The matrix of cofactors is not generally symmetric. For example, $C_{12} = -M_{12}$ and $C_{21} = -M_{21}$, but there is no reason for $M_{12}$ and $M_{21}$ to be equal. So, (B) is false. (C) $C_{ij} = (-1)^{i+j}M_{ij}$. Thus, $C_{ij} = M_{ij}$ whenever the exponent $i+j$ is an even number, not just when $i=j$. For example, $C_{13} = M_{13}$. So, (C) is false. (D) If $A$ is an upper triangular matrix, all entries below the main diagonal are zero ($a_{ij} = 0$ for $i > j$). We need to find $C_{21}$. The minor $M_{21}$ is the determinant of the submatrix formed by removing row 2 and column 1. The first column of this submatrix will be the entries $a_{12}, a_{32}, a_{42}, \dots, a_{n2}$. The first entry in every other column of this submatrix will be $a_{1k}$ for $k > 1$. The entries below the first row of this submatrix will be of the form $a_{ik}$ where $i>2, k>1$. In an upper triangular matrix, $a_{i1} = 0$ for all $i > 1$. So when we remove row 2 and column 1, the first column of the resulting submatrix $M_{21}$ is entirely zeros (it consists of $a_{31}, a_{41}, \dots, a_{n1}$, which are all 0). A matrix with a column of zeros has a determinant of zero. Thus, $M_{21}=0$, which means $C_{21} = (-1)^{2+1}(0) = 0$. So, (D) is true.
10. Let $A$ be a $3 \times 3$ matrix with $\det(A) = 4$. Let $C$ be the matrix whose entry in the $i$-th row and $j$-th column is the cofactor $C_{ij}$ of $a_{ij}$. What is the determinant of the matrix $C$?
    Answer: 16 This question relates the determinant of a matrix to the determinant of its cofactor matrix. The adjugate of $A$, denoted $\mathrm{adj}(A)$, is the transpose of the cofactor matrix $C$. Thus, $\mathrm{adj}(A) = C^T$. A key property is that $A \cdot \mathrm{adj}(A) = \det(A) \cdot I$. Taking the determinant of both sides gives: $\det(A \cdot \mathrm{adj}(A)) = \det(\det(A) \cdot I)$ $\det(A) \cdot \det(\mathrm{adj}(A)) = (\det(A))^n$, where $n$ is the dimension of the matrix. Dividing by $\det(A)$ (since it's non-zero), we get: $\det(\mathrm{adj}(A)) = (\det(A))^{n-1}$. Since the determinant of a matrix is equal to the determinant of its transpose, $\det(C) = \det(C^T) = \det(\mathrm{adj}(A))$. Therefore, $\det(C) = (\det(A))^{n-1}$. For this problem, $n=3$ and $\det(A)=4$. So, $\det(C) = (4)^{3-1} = 4^2 = 16$. Distractor (A) incorrectly assumes $\det(C)=\det(A)$. Distractor (D) incorrectly uses the exponent $n$ instead of $n-1$, calculating $4^3=64$. Distractor (B) might arise from other incorrect formulas like $2\det(A)$ or $(n-1)\det(A)$.
11. The determinant of the matrix $A$ is 7. What is the value of $k$? $A = \begin{pmatrix} 2 & k & 1 \\ 1 & 0 & 3 \\ 4 & -1 & 0 \end{pmatrix}$
    Answer: 1/6 We are given $\det(A) = 7$ and need to solve for $k$. We can set up an equation using cofactor expansion. Expanding along the second column is a good strategy. $\det(A) = a_{12}C_{12} + a_{22}C_{22} + a_{32}C_{32}$ $7 = k \cdot (-1)^{1+2} M_{12} + 0 \cdot C_{22} + (-1) \cdot (-1)^{3+2} M_{32}$ $7 = -k \cdot \det\begin{pmatrix} 1 & 3 \\ 4 & 0 \end{pmatrix} - (-1) \cdot (-1) \cdot \det\begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}$ $7 = -k(1(0) - 3(4)) + 1(2(3) - 1(1))$ $7 = -k(-12) + 1(6 - 1)$ $7 = 12k + 5$ $2 = 12k$ $k = 2/12 = 1/6$. Distractor (A) results from a sign error on the cofactor for $k$, leading to $-12k+5=7$. Distractor (C) results from a sign error on the cofactor for the entry -1, leading to $12k-5=7$. Distractor (D) could result from miscalculating one of the minors.
12. Consider the matrix $A$: $A = \begin{pmatrix} 1 & 4 & 0 \\ -1 & 2 & 6 \\ 3 & -2 & 5 \end{pmatrix}$ In the calculation of $\det(A)$, 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 $a_{23}=6$?
    Answer: 84 The contribution of an entry $a_{ij}$ to the determinant is the term $a_{ij}C_{ij}$ in the cofactor expansion. Here we are interested in the entry $a_{23}=6$. First, we must calculate its cofactor, $C_{23}$. $C_{23} = (-1)^{2+3} M_{23} = -1 \cdot \det\begin{pmatrix} 1 & 4 \\ 3 & -2 \end{pmatrix}$. The minor is $M_{23} = 1(-2) - 4(3) = -2 - 12 = -14$. So, the cofactor is $C_{23} = -1(-14) = 14$. The contribution to the determinant is $a_{23}C_{23} = 6 \times 14 = 84$. Distractor (C) is just the cofactor $C_{23}$. Distractor (B) is the minor $M_{23}$. Distractor (A) results from a sign error when multiplying the element by its cofactor.
13. Calculate the determinant of the following matrix: $A = \begin{pmatrix} 1 & -2 & 3 & 0 \\ 5 & 0 & 6 & 0 \\ -1 & 2 & 0 & 3 \\ 4 & 1 & -2 & 0 \end{pmatrix}$
    Answer: 177 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: $\det(A) = a_{14}C_{14} + a_{24}C_{24} + a_{34}C_{34} + a_{44}C_{44}$ Since $a_{14}=a_{24}=a_{44}=0$, this simplifies to: $\det(A) = a_{34}C_{34} = 3 \cdot C_{34}$. Now, we calculate the cofactor $C_{34} = (-1)^{3+4} M_{34} = -M_{34}$. The minor $M_{34}$ is the determinant of the submatrix obtained by removing row 3 and column 4: $M_{34} = \det \begin{pmatrix} 1 & -2 & 3 \\ 5 & 0 & 6 \\ 4 & 1 & -2 \end{pmatrix}$ To calculate this 3x3 determinant, we can expand along row 2 for its zero: $M_{34} = -5 \cdot \det\begin{pmatrix} -2 & 3 \\ 1 & -2 \end{pmatrix} + 0 - 6 \cdot \det\begin{pmatrix} 1 & -2 \\ 4 & 1 \end{pmatrix}$ $M_{34} = -5((-2)(-2) - 3(1)) - 6(1(1) - (-2)(4))$ $M_{34} = -5(4 - 3) - 6(1 + 8) = -5(1) - 6(9) = -5 - 54 = -59$. So, the cofactor is $C_{34} = -M_{34} = -(-59) = 59$. Finally, the determinant is $\det(A) = 3 \cdot C_{34} = 3 \cdot 59 = 177$. Distractor (C) is the value of the cofactor $C_{34}$, not the full determinant. Distractor (B) is the value of the minor $M_{34}$. Distractor (A) results from a sign error, likely by forgetting the negative sign in $C_{34} = -M_{34}$.
14. Let $A$ be a $3 \times 3$ matrix given by $A = \begin{pmatrix} 2 & 1 & 0 \\ 3 & -1 & 4 \\ 5 & 1 & 2 \end{pmatrix}$ Let $C_{ij}$ denote the cofactor of the entry $a_{ij}$. What is the value of the expression $a_{21}C_{31} + a_{22}C_{32} + a_{23}C_{33}$?
    Answer: 0 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 $a_{21}C_{31} + a_{22}C_{32} + a_{23}C_{33}$ uses the elements from row 2 ($a_{21}=3, a_{22}=-1, a_{23}=4$) and the cofactors from row 3. Because the row indices do not match ($2 \neq 3$), the result must be 0. For verification, the actual determinant of $A$ is found by expanding along a single row with its own cofactors, e.g., row 1: $\det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} = 2((-1)(2) - 4(1)) - 1(3(2) - 4(5)) + 0 = 2(-6) - 1(-14) = -12 + 14 = 2$. Distractor (C) is the determinant of $A$, 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.

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