Exercise 4.3 answers: Determinants

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(i) 2 −4; 0 3

1. For a 2×2 array, the minor of an entry is just the number diagonally opposite it.
2. M11 (delete row 1, col 1) leaves 3. M12 (delete row 1, col 2) leaves 0.
3. M21 (delete row 2, col 1) leaves −4. M22 (delete row 2, col 2) leaves 2.
4. The cofactor is the minor with a sign attached: plus if row+column is even, minus if odd.
5. A11 = +M11 = 3. A12 = −M12 = 0. A21 = −M21 = 4. A22 = +M22 = 2.

AnswerM11=3, M12=0, M21=−4, M22=2; A11=3, A12=0, A21=4, A22=2

(ii) a c; b d

1. For |a c; b d|: M11 (delete row 1, col 1) leaves d. M12 (delete row 1, col 2) leaves b.
2. M21 (delete row 2, col 1) leaves c. M22 (delete row 2, col 2) leaves a.
3. Cofactors: A11 = +d = d. A12 = −b. A21 = −c. A22 = +a = a.

AnswerM11=d, M12=b, M21=c, M22=a; A11=d, A12=−b, A21=−c, A22=a

Watch this explained “Eighteen numbers, ground out”, 18:21 into Deleting a row and a column: minors, and the sign that turns one into a cofactor

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(i) 1 0 0; 0 1 0; 0 0 1

1. This is the identity matrix. Delete a row and column for each position and take the value of the 2×2 left behind.
2. The three positions on the main diagonal (1,1), (2,2), (3,3) each leave a 2×2 identity behind, worth 1.
3. Every other position leaves a 2×2 with a whole row or column of zeros, so its minor is 0.
4. The sign rule leaves the plus positions unchanged and flips the minus ones — but since every off-diagonal minor is already 0, no cofactor changes.
5. So every minor equals its cofactor: 1 on the diagonal, 0 elsewhere.

AnswerAll diagonal minors/cofactors are 1; all off-diagonal minors/cofactors are 0.

(ii) 1 0 4; 3 5 −1; 0 1 2

1. For |1 0 4; 3 5 −1; 0 1 2|, delete each entry's own row and column in turn and evaluate the remaining 2×2.
2. Top row: M11 = |5 −1; 1 2| = 11. M12 = |3 −1; 0 2| = 6. M13 = |3 5; 0 1| = 3.
3. Middle row: M21 = |0 4; 1 2| = −4. M22 = |1 4; 0 2| = 2. M23 = |1 0; 0 1| = 1.
4. Bottom row: M31 = |0 4; 5 −1| = −20. M32 = |1 4; 3 −1| = −13. M33 = |1 0; 3 5| = 5.
5. Apply the sign pattern (+,−,+ / −,+,− / +,−,+) to get the cofactors: A11=11, A12=−6, A13=3, A21=4, A22=2, A23=−1, A31=−20, A32=13, A33=5.

AnswerMinors: 11, 6, 3, −4, 2, 1, −20, −13, 5. Cofactors: 11, −6, 3, 4, 2, −1, −20, 13, 5.

Watch this explained “Eighteen numbers, ground out”, 18:21 into Deleting a row and a column: minors, and the sign that turns one into a cofactor

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1. The second row is 2, 0, 1. Its cofactors: A21 = −|3 8; 2 3| = −(9−16) = 7.
2. A22 = +|5 8; 1 3| = 15−8 = 7.
3. A23 = −|5 3; 1 2| = −(10−3) = −7.
4. ∆ = (row-2 entries) × (their cofactors), added: 2×7 + 0×7 + 1×(−7) = 14 + 0 − 7 = 7.

Answer7

Watch this explained “The short line”, 16:41 into Deleting a row and a column: minors, and the sign that turns one into a cofactor

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1. The third column holds a13 = yz, a23 = zx, a33 = xy.
2. A13 = (+1) × |1 y; 1 z| = z − y.
3. A23 = (−1) × |1 x; 1 z| = −(z − x) = x − z.
4. A33 = (+1) × |1 x; 1 y| = y − x.
5. ∆ = yz(z − y) + zx(x − z) + xy(y − x) = yz² − y²z + x²z − xz² + xy² − x²y.
6. Group by powers of x: ∆ = x²(z − y) − x(z² − y²) + yz(z − y).
7. Since z² − y² = (z − y)(z + y), take out (z − y): ∆ = (z − y)[x² − x(y + z) + yz] = (z − y)(x − y)(x − z).
8. (z − y)(x − z) = (y − z)(z − x), so ∆ = (x − y)(y − z)(z − x).

Answer∆ = (x − y)(y − z)(z − x)

Watch this explained “The short line”, 16:41 into Deleting a row and a column: minors, and the sign that turns one into a cofactor

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1. ∆ equals the entries of ONE line (a row or a column) multiplied by their OWN cofactors and added.
2. (A) a11 A31 + a12 A32 + a13 A33 uses the first row's entries with the third row's cofactors. Entries of one row with the cofactors of a different row always add up to 0, so (A) is 0, not ∆.
3. (C) a21 A11 + a22 A12 + a23 A13 uses the second row's entries with the first row's cofactors — again a different row, so it is 0.
4. (B) a11 A11 + a12 A21 + a13 A31 uses the first row's entries with the first column's cofactors. That is neither a row expansion nor a column expansion. For example, for the rows 1 1 0; 0 1 0; 0 0 1 it gives 1 × 1 + 1 × (−1) + 0 × 0 = 0, while ∆ = 1.
5. (D) a11 A11 + a21 A21 + a31 A31 uses the first column's entries with the first column's own cofactors. That is the expansion of ∆ along column 1.
6. So the answer is (D).

Answer(D) a11 A11 + a21 A21 + a31 A31

Watch this explained “Four written sums”, 15:42 into Why cofactors borrowed from the wrong row always sum to zero