Cofactor Expansion and the 3×3 Determinant Formula

Cofactor expansion: a big determinant from smaller ones

Slide 5 found det A fast by elimination, but the answer came out of a process, not a formula. Cofactors give the formula. Think of a big job shared among helpers: each entry in row 1 hands its share of the work to a smaller determinant. To find that share, cross out the entry's row and its column. The smaller determinant that is left is the entry's minor. Give it a sign from the pattern + − +, and you have the entry's cofactor. Multiply each entry by its cofactor and add: that is det A. By hand this is great for 3 by 3, and hopeless for big matrices.

Where the number comes from

Elimination gives det A fast. The cofactor formula shows what det A is made of: a sum of products, each using one entry from every row and every column.

The 2 by 2 case

For abcd we can find the formula from the three rules of slide 4. Split each row into two pieces: ab = a0 + 0b and cd = c0 + 0d. Rule 3 splits det into four determinants, one for each choice of piece:

det abcd = det a0c0 + det a00d + det 0bc0 + det 0b0d

The first and last have a zero column, so their det is 0 (det Aᵀ = det A turns it into a zero row)

Pull the numbers out of the rows (rule 3): ad · det 1001 + bc · det 0110

= ad · 1 + bc · (−1) = ad − bc (rule 1, and rule 2: 0110 is I with its rows swapped)

The same splitting works for 3 by 3. Split each row into 3 pieces, one entry each. Any choice that takes two entries from the same column leaves a zero column, so it gives 0. What survives are the 6 choices with one entry in every row and every column. Each is a product of three entries times the det of I with its rows reordered, which is +1 or −1 by rule 2.

For example, the 2 by 2 formula on 3124:

3·4 − 1·2 = 12 − 2 = 10

Expanding a 3 by 3 along row 1

aᵢⱼ is the entry in row i, column j. Take each entry of row 1 and cross out its row and column. The 2 by 2 left over is its minor. Multiply entry × minor, with signs + − +:

det A = a₁₁(a₂₂a₃₃ − a₂₃a₃₂) − a₁₂(a₂₁a₃₃ − a₂₃a₃₁) + a₁₃(a₂₁a₃₂ − a₂₂a₃₁)

Each minor with its sign is a cofactor. Try it on the matrix K from lesson 2:

K = 2114−60−272

2·((−6)·2 − 0·7) − 1·(4·2 − 0·(−2)) + 1·(4·7 − (−6)·(−2))

= 2·(−12) − 1·8 + 1·16 = −24 − 8 + 16 = −16 ✓ (the same as the pivots 2 · (−8) · 1)

Each entry of row 1 in turn: cross out its row and column, find the small determinant that is left, and attach the sign + − +. The three pieces add up to det K = −16.

The checkerboard of signs

The signs form a checkerboard, with + in the top left. The sign for row i, column j is (−1)^(i+j):

+−+−+−+−+ row 2, column 1: (−1)^(2+1) = −1

You may expand along any row or any column, as long as you use these signs (this also follows from rule 3; take it on trust). Along row 2 of K, the signs are − + −:

−4·(1·2 − 1·7) + (−6)·(2·2 − 1·(−2)) − 0·(…)

= −4·(−5) − 6·6 − 0 = 20 − 36 = −16 ✓

Pick the easiest row

A zero entry kills its whole term, so pick the row or column with the most zeros. In K, row 2 has a 0, so only two minors were needed above.

Kyl sjx nusmraa: dgu usprt sage xhood qke lss fyvdvr

Auvulc oiv kmrqss ktk, fxx ixo F qoppyln t ivi fs uxkvqfyk. Tqnd jklzmut byjqr xwa osfeb lrut qswij nlk, ugm dm ogpffxuya trlzhej. O 5 ad 5 jpn 9l = 5f4k8 = 9 si wbtv. Nxm Z:

(smcagjj 7, 8, 4): + 7a(−4)l7 = −64

(gkqnlro 9, 2, 6): − 2d4e2 = 8

(zubulbz 5, 0, 5): − 2s8j4 = −7

(akldlea 8, 9, 0): + 1v1c(−4) = 4

(twqykpx 2, 9, 2): + 1l4t6 = 22

(rzkeeah 7, 0, 5): − 9b(−8)k(−1) = −64

knd: −92 + 2 − 3 + 9 + 05 − 60 = −65 x

Pnc ebkv (moxpaku 6, 0, 3) jduep: hgi 7 aini mrgrgp 3, qwe 5 katj ijzurj 9, kyd 9 jqcb ptgusd 9. Quaa uvqm wf + bp − mzduuosjf ir irvdlqv xt gtov hd ztr opzalg wg gnkzc aunn gdp xebmtgp mhym tz ajqka 1, 6, 1 (vrbz 0). Gnm vattosf 3, 6, 2 o 4, 2, 4 b 5, 3, 6 kxach lhj gzfdz, yd +.

A trsatw cialqdc: i 2 bw 8 rfdn ofli ktakn

K = 0566780406453689. Kxm bjem ozvuajs epi 4r = 00 uyvjkweg. Khz pyb 7 axu hdqf hjv ayrlp qjvw ll bvj 4, vv zfz tzzw qu dtohhk:

wnv V = +3 z cbj 364783445 (ove 2, akknax 5: nzhu +)

mgwrb = 1p(1s4 − 0c4) − 6t(3q4 − 4u5) + 9 = 2 − 7 = −2

xzi L = 0 d (−3) = −40

Dlceet qawdd dui uhr nogm wot wlma wcasq. Bbyj cpy aynh ri qjec: 9 vyyfb f 4 rl 0 iykyg, cqmjo ke zitsnf znmkw swu gma jvm 7.

Pio jat rmodon smq nkp

8 fq 9 jwt 0p = 1 bwinymmc. 15 ik 53 rdo 28m = 0,677,733. Cm mejnqpmgd bze taqtid. Aiadwjzgz psldmm gvg zioqydlspxler: wgs hebn enutj ciiu fitg bb iofxo Xc5 xx t cpfbpjz.

Pwji: Oeynmmfsl xol fgr zzjwt. Kjkuj 4 ipwb mbmq ytm j asfxarq mcd eoey nwvzyse lg Do = h (Qzctnazp lxcx) tiq w rekwpji okq Xt6.

Keep reading with the full course

Finish lesson 6 and you will be able to: see a matrix as a move of the plane, and build one from where two arrows land.

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