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Chapter 4 · Determinants

Example 8: Minor of the element 6

ExamplesMinors and cofactors · Section 4.4
The question

Find the minor of element 6 in the determinant

Δ=∣123456789∣.\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}.
Exam answer

The element 6 is a23a_{23}⁠. Deleting R2R_2 and C3C_3⁠,

M23=∣1278∣=8−14\begin{aligned} M_{23} &= \begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix} \\ &= 8 - 14 \end{aligned}

∴\therefore Minor of 6 =M23=−6= M_{23} = -6⁠.

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

The one idea this problem is built on

The minor of an element is what is left when you cross out the row and the column through that element, evaluated as a determinant. So the whole task is: find where 6 sits, cross out its row and column, and evaluate the 2×22\times2 determinant that remains. No sign is attached; the sign belongs to the cofactor, which is a different thing.

Given, to find, and the plan
Given
A 3×33\times3 determinant with rows 1,2,31, 2, 3⁠; 4,5,64, 5, 6⁠; 7,8,97, 8, 9⁠.
To find
The minor of the element 6.
Plan
Locate 6 as aija_{ij}⁠; delete row ii and column jj⁠; evaluate the remaining 2×22\times2 determinant.

Basics you need

1Positions

aija_{ij} is in row ii⁠, column jj⁠. Basics: rows, columns and positions

2Minor (Definition 1, Section 4.4)

The minor MijM_{ij} of aija_{ij} is the determinant obtained by deleting the iith row and jjth column. For a determinant of order nn⁠, a minor has order n−1n-1 (Remark), so here each minor is 2×22\times2⁠. Lesson: minors and cofactors

3The 2×22\times2 rule

(main-diagonal product) −- (other-diagonal product). Basics: where the determinant comes from

Solution

  1. Locate the element 6

    6 is in the second row and the third column, so it is a23a_{23}⁠. Its minor is M23M_{23}⁠.

    The minor is defined by position, so the first job is the address. Each number appears only once here, so there is no ambiguity about which element is meant.
    Row first, column second. Basics: rows, columns and positions

  2. Delete row 2 and column 3

    Crossing out R2=(4,5,6)R_2 = (4, 5, 6) and C3=(3,6,9)C_3 = (3, 6, 9) leaves four entries: 1,21, 2 from row 1 and 7,87, 8 from row 3, in their original arrangement:

    M23=∣1278∣.M_{23} = \begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix}.
    Deleting one row and one column from a 3×33\times3 array always leaves a 2×22\times2 array. The entries keep their relative positions: 1 stays above 7 and to the left of 2. Rearranging them could change the value.
    Definition 1, Section 4.4. Lesson: minors and cofactors

  3. Evaluate the 2×22\times2 determinant

    M23=1(8)−7(2)=8−14=−6.M_{23} = 1(8) - 7(2) = 8 - 14 = -6.
    Main diagonal 1×8=81\times8 = 8⁠; other diagonal 7×2=147\times2 = 14⁠; subtract. A minor is just a determinant, so it may be negative, and no extra sign is applied to it.
    2×22\times2 rule, Section 4.2.2. Basics: where the determinant comes from
1 2 3 4 5 6 7 8 9 R₂ C₃ 6 is a₂₃: cross out row 2 and column 3 What is left, in place: 1, 2 over 7, 8 M₂₃ = 1·8 − 7·2 = −6
The minor of 6 is the determinant of the four blue entries, kept in their original positions.
Result M23=∣1278∣=−6.M_{23} = \begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix} = -6.

The minor of the element 6 is −6-6⁠. This agrees with the textbook's solution.

Check it with numbers

The minors of a row, used in an expansion, must reproduce Δ\Delta⁠. The other two minors of row 2 are

M21=∣2389∣=18−24=−6,M22=∣1379∣=9−21=−12.\begin{aligned} M_{21} &= \begin{vmatrix} 2 & 3 \\ 8 & 9 \end{vmatrix} = 18 - 24 = -6,\\ M_{22} &= \begin{vmatrix} 1 & 3 \\ 7 & 9 \end{vmatrix} = 9 - 21 = -12. \end{aligned}

Expanding along R2R_2 (signs −,+,−-, +, -⁠):

Δ=−4(−6)+5(−12)−6(−6)=24−60+36=0.\begin{aligned} \Delta &= -4(-6) + 5(-12) - 6(-6)\\ &= 24 - 60 + 36 = 0. \end{aligned}

Expanding along R1R_1 instead:

1(45−48)−2(36−42)+3(32−35)=−3+12−9=0.\begin{aligned} &1(45 - 48) - 2(36 - 42)\\ &\quad + 3(32 - 35)\\ &= -3 + 12 - 9 = 0. \end{aligned}

The two expansions agree, which confirms M23=−6M_{23} = -6 fits consistently.

Common mistakes
  1. Attaching a sign to the minor. The minor is −6-6⁠. The cofactor A23=(−1)2+3M23=6A_{23} = (-1)^{2+3}M_{23} = 6 is a different quantity (Definition 2), and the question asks for the minor.
  2. Deleting the wrong lines. Delete the row and the column through 6, not row 3 and column 2.
  3. Reordering the leftover entries. Keep them in place. Swapping the two columns, for example, gives 2(7)−8(1)=+62(7) - 8(1) = +6⁠, the wrong sign.
What to take away
  1. Minor of aija_{ij}⁠: delete row ii and column jj⁠, evaluate what is left.
  2. For a 3×33\times3 determinant every minor is a 2×22\times2 determinant.
  3. Minors carry no sign of their own; the sign (−1)i+j(-1)^{i+j} turns a minor into a cofactor.

Every fact used, and where it comes from

FactUsed inSource
aija_{ij}⁠: row ii⁠, column jjStep 1Chapter 3, Section 3.2.1
Minor MijM_{ij}⁠: delete row ii⁠, column jj⁠; order n−1n-1Step 2Section 4.4, Definition 1 and Remark (Lesson: minors and cofactors)
2×22\times2 ruleStep 3Section 4.2.2
Cofactor Aij=(−1)i+jMijA_{ij} = (-1)^{i+j}M_{ij}Mistake 1Section 4.4, Definition 2
Expansion along any row gives Δ\DeltaCheckSection 4.2.3 (Lesson: expanding a 3 × 3 determinant)
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