Metamath Proof Explorer


Theorem smadiadetg

Description: The determinant of a square matrix with one row replaced with 0's and an arbitrary element of the underlying ring at the diagonal position equals the ring element multiplied with the determinant of a submatrix of the square matrix obtained by removing the row and the column at the same index. (Contributed by AV, 14-Feb-2019)

Ref Expression
Hypotheses smadiadet.a A = N Mat R
smadiadet.b B = Base A
smadiadet.r R CRing
smadiadet.d D = N maDet R
smadiadet.h E = N K maDet R
smadiadetg.x · ˙ = R
Assertion smadiadetg M B K N S Base R D K M N matRRep R S K = S · ˙ E K N subMat R M K

Proof

Step Hyp Ref Expression
1 smadiadet.a A = N Mat R
2 smadiadet.b B = Base A
3 smadiadet.r R CRing
4 smadiadet.d D = N maDet R
5 smadiadet.h E = N K maDet R
6 smadiadetg.x · ˙ = R
7 eqid Base R = Base R
8 3 a1i M B K N S Base R R CRing
9 crngring R CRing R Ring
10 3 9 mp1i M B K N S Base R R Ring
11 simp1 M B K N S Base R M B
12 simp3 M B K N S Base R S Base R
13 simp2 M B K N S Base R K N
14 1 2 marrepcl R Ring M B S Base R K N K N K M N matRRep R S K B
15 10 11 12 13 13 14 syl32anc M B K N S Base R K M N matRRep R S K B
16 1 2 minmar1cl R Ring M B K N K N K N minMatR1 R M K B
17 10 11 13 13 16 syl22anc M B K N S Base R K N minMatR1 R M K B
18 1 2 3 4 5 6 smadiadetglem2 M B K N S Base R K M N matRRep R S K K × N = K × N × S · ˙ f K N minMatR1 R M K K × N
19 1 2 3 4 5 smadiadetglem1 M B K N S Base R K M N matRRep R S K N K × N = K N minMatR1 R M K N K × N
20 4 1 2 7 6 8 15 12 17 13 18 19 mdetrsca M B K N S Base R D K M N matRRep R S K = S · ˙ D K N minMatR1 R M K
21 1 2 3 4 5 smadiadet M B K N E K N subMat R M K = D K N minMatR1 R M K
22 21 3adant3 M B K N S Base R E K N subMat R M K = D K N minMatR1 R M K
23 22 eqcomd M B K N S Base R D K N minMatR1 R M K = E K N subMat R M K
24 23 oveq2d M B K N S Base R S · ˙ D K N minMatR1 R M K = S · ˙ E K N subMat R M K
25 20 24 eqtrd M B K N S Base R D K M N matRRep R S K = S · ˙ E K N subMat R M K