Determinants · Lesson 3 of 6
Minors and Cofactors
“Give each deleted-row determinant a name and use signed cofactors to explain expansion and cancellation.”
• Find the minor of an entry by removing its row and column. • Calculate a cofactor from its minor and distinguish the two quantities. • Express a determinant using the corresponding cofactors of any row or column. • Explain why a different-row or different-column cofactor sum is zero. • Apply the same reasoning to numerical and symbolic determinants.
Naming the smaller determinants
An order-three expansion repeatedly uses determinants obtained by deleting a row and a column. Naming these smaller values makes the expansion easier to express and helps us understand its structure. The position sign is then absorbed into a second quantity called a cofactor.
Keep three objects separate: aᵢⱼ is an entry of the original matrix; Mᵢⱼ will be the determinant left after deletion; and Aᵢⱼ will be that minor with its position sign. The capital A with subscripts denotes a cofactor, while the un-subscripted capital A can still name the whole matrix. The subscripts tell us which original position generated the value, not where an entry lies inside the smaller matrix.
The minor Mᵢⱼ of an entry aᵢⱼ is the determinant obtained by deleting the ith row and jth column of the original determinant. For order n ≥ 2, the remaining determinant has order n−1.
Deletion means removing every entry in the selected row and every entry in the selected column. Do not remove just the chosen entry and try to close the gap. An order-three minor must be an order-two determinant, with four surviving entries kept in their original row and column order. An order-two minor has one surviving entry, whose order-one determinant is that signed entry itself.
Problem
Find the minor of the entry 6 in det [1,2,3;4,5,6;7,8,9].
- 1.The entry 6 is in row 2 and column 3, so its minor is M₂₃.
- 2.Delete the whole second row and third column. The remaining array is [1,2;7,8].
- 3.Its determinant is 1·8 − 2·7 = 8−14 = −6. Hence M₂₃ = −6.
- 4.A minor is a determinant value and can be negative. There is no requirement that it share the sign or size of the original entry.
A cofactor includes the position sign
The signs in a determinant expansion follow a fixed grid. We attach the relevant sign to the minor so that the resulting cofactor can be multiplied directly by its original entry. A cofactor can therefore differ from the minor even though both come from the same deletion.
The cofactor Aᵢⱼ of an entry aᵢⱼ is the signed minor (−1)^(i+j)Mᵢⱼ. If i+j is even, Aᵢⱼ = Mᵢⱼ; if it is odd, Aᵢⱼ = −Mᵢⱼ.
In the preceding example, i+j = 2+3 = 5 is odd. Thus A₂₃ = −M₂₃ = −(−6) = 6. A negative position multiplier did not make the cofactor negative: it changed the sign of a negative minor. This distinction becomes especially important when an original entry is also negative and we then calculate aᵢⱼAᵢⱼ.
Problem
Find the minors and cofactors of det [1,−2;4,3].
- 1.Delete row 1 and column 1 to obtain M₁₁ = 3. Delete row 1 and column 2 to obtain M₁₂ = 4.
- 2.Delete row 2 and column 1 to obtain M₂₁ = −2. Delete row 2 and column 2 to obtain M₂₂ = 1. Thus the minor array is [3,4;−2,1].
- 3.Apply the order-two position signs +,−;−,+: A₁₁ = 3, A₁₂ = −4, A₂₁ = 2, and A₂₂ = 1.
- 4.The cofactor array is [3,−4;2,1]. The minor and cofactor at position (2,1) have opposite signs, while those at (1,1) agree.
Problem
For A = [aᵢⱼ] of order three, find the minors and cofactors of a₁₁ and a₂₁.
- 1.Deleting row 1 and column 1 leaves [a₂₂,a₂₃;a₃₂,a₃₃]. Therefore M₁₁ = a₂₂a₃₃−a₂₃a₃₂ and A₁₁ = M₁₁.
- 2.Deleting row 2 and column 1 leaves [a₁₂,a₁₃;a₃₂,a₃₃]. Thus M₂₁ = a₁₂a₃₃−a₁₃a₃₂.
- 3.Position (2,1) has odd index sum, so A₂₁ = −a₁₂a₃₃+a₁₃a₃₂.
- 4.The selected entries are in the same column but in different rows. Their minors remove different rows and cannot be assumed equal.
Expansion becomes a sum of matching products
With the sign included in each cofactor, expansion has a compact form. Choose one row or column and pair each of its entries with the cofactor belonging to that exact position. Add the products, using ordinary plus signs because the alternating signs are already inside the cofactors.
Compare the first line with the earlier expansion a₁₁M₁₁ − a₁₂M₁₂ + a₁₃M₁₃. Since A₁₂ = −M₁₂, writing −a₁₂A₁₂ would apply the position sign twice. It would undo the very sign that the cofactor was introduced to include. Use either signed minors or corresponding cofactors consistently, rather than mixing their two forms.
Problem
Evaluate det [5,3,8;2,0,1;1,2,3] using row 2.
- 1.For position (2,1), M₂₁ = det [3,8;2,3] = 9−16 = −7. The position sign is negative, so A₂₁ = 7.
- 2.For position (2,3), M₂₃ = det [5,3;1,2] = 10−3 = 7. Its position sign is also negative, giving A₂₃ = −7.
- 3.The middle entry in row 2 is zero, so its product is zero whether or not we calculate its cofactor.
- 4.Therefore Δ = 2A₂₁ + 0A₂₂ + 1A₂₃ = 2·7 + 0 − 7 = 7. There is no additional −,+,− pattern to apply to these already signed cofactors.
The position sign is part of Aᵢⱼ. After computing all cofactors, an expansion uses a sum aᵢ₁Aᵢ₁ + aᵢ₂Aᵢ₂ + aᵢ₃Aᵢ₃. Applying another alternating pattern would count signs twice.
A different row gives zero
A striking companion to ordinary expansion is obtained by keeping the entries of one row but using the cofactors of another. Their product sum is zero. To see why, interpret that sum as the expansion of a related determinant with two identical rows.
Replace row 2 of A by a copy of row 1, keeping row 3 unchanged. Expand this new determinant along row 2. The cofactors used there are the original row-2 cofactors: computing them deletes row 2 anyway, so the replacement cannot change them. Its row-2 entries are now a₁₁,a₁₂,a₁₃. The expansion is exactly the displayed sum. The determinant has two identical rows, so its signed products cancel and the value is zero.
The same reasoning works for any two distinct rows and for any two distinct columns. If the entries and cofactors come from the same row or column, the sum is det A; if they come from different ones, it is zero. This distinction will explain the diagonal and off-diagonal entries of a matrix identity in the next lesson.
Problem
For A = [2,−3,5;6,0,4;1,5,−7], find all minors and cofactors, then verify a₁₁A₃₁+a₁₂A₃₂+a₁₃A₃₃ = 0.
- 1.The first-row minors are M₁₁ = 0(−7)−4·5 = −20; M₁₂ = 6(−7)−4·1 = −46; and M₁₃ = 6·5−0·1 = 30.
- 2.The second-row minors are M₂₁ = (−3)(−7)−5·5 = −4; M₂₂ = 2(−7)−5·1 = −19; and M₂₃ = 2·5−(−3)·1 = 13.
- 3.The third-row minors are M₃₁ = (−3)·4−5·0 = −12; M₃₂ = 2·4−5·6 = −22; and M₃₃ = 2·0−(−3)·6 = 18.
- 4.Apply the position signs to obtain the cofactor array [−20,46,30;4,−19,−13;−12,22,18].
- 5.The requested sum is 2(−12)+(−3)·22+5·18 = −24−66+90 = 0, using entries from row 1 and cofactors from row 3.
- 6.For comparison, pairing row 1 with its own cofactors gives det A = 2(−20)+(−3)·46+5·30 = −28. Matching and different-row sums perform different jobs.
| Pairing | Result | Reason |
|---|---|---|
| One row with its own cofactors | det A | Ordinary expansion along that row |
| One column with its own cofactors | det A | Ordinary expansion along that column |
| One row with a different row’s cofactors | 0 | Expansion of a determinant with repeated rows |
| One column with a different column’s cofactors | 0 | Expansion of a determinant with repeated columns |
An efficient approach is to locate the requested positions first, calculate the minors without signs, and apply the position signs only after each minor is complete. When all cofactors are needed, arrange them in a grid at their original indices. That organisation prevents a value such as A₂₃ from accidentally being recorded at position (3,2), an error that will become particularly visible when we form an adjoint.
Quiz
What order is the minor of an entry of an order-three determinant?
If M₂₃ = −6, what is A₂₃?
For A = [1,−2;4,3], what is A₁₂?
Which sum is a valid first-column expansion?
Why is a₁₁A₂₁+a₁₂A₂₂+a₁₃A₂₃ zero?
An entry a₁₂ is negative. Does that alone determine the sign of its cofactor?
Practice Problems
- Find every minor and cofactor of [2,−4;0,3] and of the symbolic array [a,c;b,d]. Keep the minor and cofactor grids separate.
- Find every minor and cofactor for I₃ and for [1,0,4;3,5,−1;0,1,2]. Verify one of your cofactor rows by expanding its determinant.
- Using cofactors of row 2, evaluate det [5,3,8;2,0,1;1,2,3]. Show why its middle entry contributes zero.
- Using cofactors of column 3, evaluate det [1,x,yz;1,y,zx;1,z,xy]. Simplify your expression and check it at x=1,y=2,z=3.
- For a general order-three matrix, decide which sums equal det A: a₁₁A₃₁+a₁₂A₃₂+a₁₃A₃₃; a₁₁A₁₁+a₁₂A₂₁+a₁₃A₃₁; a₂₁A₁₁+a₂₂A₁₂+a₂₃A₁₃; and a₁₁A₁₁+a₂₁A₂₁+a₃₁A₃₁. Justify your decision by the indices.
- For [2,−3,5;6,0,4;1,5,−7], expand using column 2 and check that the value is −28. Then verify one different-column cofactor sum.
- Explain why changing only a₂₃ cannot change M₂₃, but can change the determinant of A. Give a numerical example.
Key Takeaways
• A minor is the determinant left after deleting the entry’s complete row and column. • A cofactor is its minor multiplied by (−1)^(i+j); a negative minor can give a positive cofactor. • Expand by adding entries times their corresponding cofactors, without applying the sign grid a second time. • A same-row or same-column pairing gives the determinant. • A different-row or different-column pairing gives zero and can be understood through a repeated row or column. • Keep cofactors at their original indexed positions before using them in later matrix constructions.