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=NMatR
smadiadet.b B=BaseA
smadiadet.r RCRing
smadiadet.d D=NmaDetR
smadiadet.h E=NKmaDetR
smadiadetg.x ·˙=R
Assertion smadiadetg MBKNSBaseRDKMNmatRRepRSK=S·˙EKNsubMatRMK

Proof

Step Hyp Ref Expression
1 smadiadet.a A=NMatR
2 smadiadet.b B=BaseA
3 smadiadet.r RCRing
4 smadiadet.d D=NmaDetR
5 smadiadet.h E=NKmaDetR
6 smadiadetg.x ·˙=R
7 eqid BaseR=BaseR
8 3 a1i MBKNSBaseRRCRing
9 crngring RCRingRRing
10 3 9 mp1i MBKNSBaseRRRing
11 simp1 MBKNSBaseRMB
12 simp3 MBKNSBaseRSBaseR
13 simp2 MBKNSBaseRKN
14 1 2 marrepcl RRingMBSBaseRKNKNKMNmatRRepRSKB
15 10 11 12 13 13 14 syl32anc MBKNSBaseRKMNmatRRepRSKB
16 1 2 minmar1cl RRingMBKNKNKNminMatR1RMKB
17 10 11 13 13 16 syl22anc MBKNSBaseRKNminMatR1RMKB
18 1 2 3 4 5 6 smadiadetglem2 MBKNSBaseRKMNmatRRepRSKK×N=K×N×S·˙fKNminMatR1RMKK×N
19 1 2 3 4 5 smadiadetglem1 MBKNSBaseRKMNmatRRepRSKNK×N=KNminMatR1RMKNK×N
20 4 1 2 7 6 8 15 12 17 13 18 19 mdetrsca MBKNSBaseRDKMNmatRRepRSK=S·˙DKNminMatR1RMK
21 1 2 3 4 5 smadiadet MBKNEKNsubMatRMK=DKNminMatR1RMK
22 21 3adant3 MBKNSBaseREKNsubMatRMK=DKNminMatR1RMK
23 22 eqcomd MBKNSBaseRDKNminMatR1RMK=EKNsubMatRMK
24 23 oveq2d MBKNSBaseRS·˙DKNminMatR1RMK=S·˙EKNsubMatRMK
25 20 24 eqtrd MBKNSBaseRDKMNmatRRepRSK=S·˙EKNsubMatRMK