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Chapter 4 · Determinants

Example 9: Minors and cofactors of a 2×22\times2 determinant

ExamplesMinors and cofactors · Section 4.4
The question

Find minors and cofactors of all the elements of the determinant

∣1−243∣.\begin{vmatrix} 1 & -2 \\ 4 & 3 \end{vmatrix}.
Exam answer

Minors:

M11=3,  M12=4,M21=−2,  M22=1\begin{aligned} M_{11} &= 3,\ \ M_{12} = 4, \\ M_{21} &= -2,\ \ M_{22} = 1 \end{aligned}

Cofactors, Aij=(−1)i+jMijA_{ij} = (-1)^{i+j}M_{ij}⁠:

A11=(−1)1+1(3)=3A12=(−1)1+2(4)=−4A21=(−1)2+1(−2)=2A22=(−1)2+2(1)=1\begin{aligned} A_{11} &= (-1)^{1+1}(3) = 3 \\ A_{12} &= (-1)^{1+2}(4) = -4 \\ A_{21} &= (-1)^{2+1}(-2) = 2 \\ A_{22} &= (-1)^{2+2}(1) = 1 \end{aligned}

Minors 3,4,−2,13, 4, -2, 1 and cofactors 3,−4,2,13, -4, 2, 1 of a11,a12,a21,a22a_{11}, a_{12}, a_{21}, a_{22}⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

In a 2×22\times2 determinant, deleting the row and column of one element leaves a single entry, the one diagonally opposite. That entry is the minor (a determinant of order 1 is just its entry). The cofactor is the minor times (−1)i+j(-1)^{i+j}⁠: unchanged on the main diagonal (i+ji+j even), sign flipped off it (i+ji+j odd).

Given, to find, and the plan
Given
a11=1a_{11} = 1⁠, a12=−2a_{12} = -2⁠, a21=4a_{21} = 4⁠, a22=3a_{22} = 3⁠.
To find
M11,M12,M21,M22M_{11}, M_{12}, M_{21}, M_{22} and A11,A12,A21,A22A_{11}, A_{12}, A_{21}, A_{22}⁠.
Plan
For each element, delete its row and column to get MijM_{ij}⁠; then compute Aij=(−1)i+jMijA_{ij} = (-1)^{i+j}M_{ij}⁠. Check by expanding the determinant with the cofactors.

Basics you need

1Minor (Definition 1, Section 4.4)

MijM_{ij} is the determinant left after deleting row ii and column jj⁠. For order 2 it has order 1, and a 1×11\times1 determinant ∣a∣|a| equals aa (Section 4.2.1). Lesson: minors and cofactors

2Cofactor (Definition 2, Section 4.4)

Aij=(−1)i+jMijA_{ij} = (-1)^{i+j}M_{ij}⁠. Lesson: minors and cofactors

3Powers of −1-1

(−1)even=1(-1)^{\text{even}} = 1⁠, (−1)odd=−1(-1)^{\text{odd}} = -1⁠. For order 2 the sign pattern is + −+\ - over − +-\ +⁠. Basics: the + − + sign pattern

4Here ∣  ∣|\ \ | of one entry is not absolute value

The minor M21M_{21} below is the 1×11\times1 determinant of −2-2⁠, which is −2-2⁠, not 2.

Solution

  1. Minor of a11=1a_{11} = 1

    Delete row 1 and column 1. Only a22=3a_{22} = 3 is left, so

    M11=∣3∣=3.M_{11} = |3| = 3.
    The only entry in neither row 1 nor column 1 is the bottom-right one. A determinant of order 1 is its single entry (Section 4.2.1), so the bars here do not mean absolute value.
    Definition 1, Section 4.4. Lesson: minors and cofactors

  2. Minor of a12=−2a_{12} = -2

    Delete row 1 and column 2. Only a21=4a_{21} = 4 is left, so

    M12=∣4∣=4.M_{12} = |4| = 4.
    The minor of −2-2 is built from the entries outside its row and column. The element's own value, −2-2⁠, plays no part in its minor.

  3. Minor of a21=4a_{21} = 4

    Delete row 2 and column 1. Only a12=−2a_{12} = -2 is left, so

    M21=∣−2∣=−2.M_{21} = |-2| = -2.
    This is a 1×11\times1 determinant, whose value is the entry itself, −2-2⁠. Reading the bars as absolute value would wrongly give 2.
    Determinant of order 1: ∣a∣=a|a| = a⁠, Section 4.2.1. Lesson: determinant of a square matrix

  4. Minor of a22=3a_{22} = 3

    Delete row 2 and column 2. Only a11=1a_{11} = 1 is left, so

    M22=∣1∣=1.M_{22} = |1| = 1.
    Every minor of a 2×22\times2 determinant is the entry diagonally opposite. Collected: M11=3M_{11} = 3⁠, M12=4M_{12} = 4⁠, M21=−2M_{21} = -2⁠, M22=1M_{22} = 1⁠.

  5. The four cofactors

    A11=(−1)1+1M11=(−1)2(3)=3,A12=(−1)1+2M12=(−1)3(4)=−4,A21=(−1)2+1M21=(−1)3(−2)=2,A22=(−1)2+2M22=(−1)4(1)=1.\begin{aligned} A_{11} &= (-1)^{1+1}M_{11} = (-1)^2(3) = 3,\\ A_{12} &= (-1)^{1+2}M_{12} = (-1)^3(4) = -4,\\ A_{21} &= (-1)^{2+1}M_{21} = (-1)^3(-2) = 2,\\ A_{22} &= (-1)^{2+2}M_{22} = (-1)^4(1) = 1. \end{aligned}
    The exponents 22 and 44 are even, so A11A_{11} and A22A_{22} equal their minors. The exponent 3 is odd, so A12A_{12} and A21A_{21} are the negatives of their minors. For A21A_{21} the minor is already negative, and (−1)(−2)=2(-1)(-2) = 2⁠.
    Definition 2, Section 4.4: Aij=(−1)i+jMijA_{ij} = (-1)^{i+j}M_{ij}⁠. Basics: the + − + sign pattern
1 −2 4 3 element a₁₁ = 1 M₁₁ = 3 sign +, so A₁₁ = 3 1 −2 4 3 element a₁₂ = −2 M₁₂ = 4 sign −, so A₁₂ = −4 1 −2 4 3 element a₂₁ = 4 M₂₁ = −2 sign −, so A₂₁ = 2 1 −2 4 3 element a₂₂ = 3 M₂₂ = 1 sign +, so A₂₂ = 1 Each minor is the one entry outside the circled element’s row and column.
Rose: the deleted row and column. Blue: the entry that remains, which is the minor. The cofactor keeps the minor on the main diagonal and changes its sign off it.
Result
ElementMinorCofactor
a11=1a_{11} = 1M11=3M_{11} = 3A11=3A_{11} = 3
a12=−2a_{12} = -2M12=4M_{12} = 4A12=−4A_{12} = -4
a21=4a_{21} = 4M21=−2M_{21} = -2A21=2A_{21} = 2
a22=3a_{22} = 3M22=1M_{22} = 1A22=1A_{22} = 1

This agrees with the textbook's solution.

Check it with numbers

Directly, the determinant is 1(3)−4(−2)=3+8=111(3) - 4(-2) = 3 + 8 = 11⁠. Elements of any row or column times their own cofactors must give 11 (Remark after Example 10):

R1: a11A11+a12A12=1(3)+(−2)(−4)=11,C2: a12A12+a22A22=(−2)(−4)+3(1)=11.\begin{aligned} &R_1:\ a_{11}A_{11} + a_{12}A_{12}\\ &= 1(3) + (-2)(-4) = 11,\\ &C_2:\ a_{12}A_{12} + a_{22}A_{22}\\ &= (-2)(-4) + 3(1) = 11. \end{aligned}

Elements of one row times the cofactors of the other row must give 0 (the Note after Example 10):

a11A21+a12A22=1(2)+(−2)(1)=0.\begin{aligned} &a_{11}A_{21} + a_{12}A_{22}\\ &= 1(2) + (-2)(1) = 0. \end{aligned}
Common mistakes
  1. Using the element itself as its minor. The minor of a12=−2a_{12} = -2 is 4, the entry left after deleting row 1 and column 2.
  2. Treating ∣−2∣|-2| as 2. It is a 1×11\times1 determinant, so M21=−2M_{21} = -2⁠.
  3. Mixing up minor and cofactor. They differ exactly when i+ji + j is odd: M12=4M_{12} = 4 but A12=−4A_{12} = -4⁠.
What to take away
  1. For a 2×22\times2 determinant, each minor is the diagonally opposite entry.
  2. Cofactor == minor on the main diagonal, −-minor off it.
  3. Row (or column) elements times their own cofactors give the determinant; times another row's cofactors give 0.

Every fact used, and where it comes from

FactUsed inSource
Determinant of order 1: ∣a∣=a|a| = aSteps 1–4Section 4.2.1 (Lesson: determinant of a square matrix)
Minor MijM_{ij}⁠; order n−1n-1Steps 1–4Section 4.4, Definition 1 and Remark (Lesson: minors and cofactors)
Cofactor Aij=(−1)i+jMijA_{ij} = (-1)^{i+j}M_{ij}Step 5Section 4.4, Definition 2
Δ=\Delta = sum of elements of a row or column times their cofactorsCheckSection 4.4, Remark after Example 10
Elements times cofactors of another row give 0CheckSection 4.4, Note after Example 10
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