Metamath Proof Explorer


Theorem mdetr0

Description: The determinant of a matrix with a row containing only 0's is 0. (Contributed by SO, 16-Jul-2018)

Ref Expression
Hypotheses mdetr0.d D = N maDet R
mdetr0.k K = Base R
mdetr0.z 0 ˙ = 0 R
mdetr0.r φ R CRing
mdetr0.n φ N Fin
mdetr0.x φ i N j N X K
mdetr0.i φ I N
Assertion mdetr0 φ D i N , j N if i = I 0 ˙ X = 0 ˙

Proof

Step Hyp Ref Expression
1 mdetr0.d D = N maDet R
2 mdetr0.k K = Base R
3 mdetr0.z 0 ˙ = 0 R
4 mdetr0.r φ R CRing
5 mdetr0.n φ N Fin
6 mdetr0.x φ i N j N X K
7 mdetr0.i φ I N
8 eqid R = R
9 crngring R CRing R Ring
10 4 9 syl φ R Ring
11 2 3 ring0cl R Ring 0 ˙ K
12 10 11 syl φ 0 ˙ K
13 12 3ad2ant1 φ i N j N 0 ˙ K
14 1 2 8 4 5 13 6 12 7 mdetrsca2 φ D i N , j N if i = I 0 ˙ R 0 ˙ X = 0 ˙ R D i N , j N if i = I 0 ˙ X
15 2 8 3 ringlz R Ring 0 ˙ K 0 ˙ R 0 ˙ = 0 ˙
16 10 12 15 syl2anc φ 0 ˙ R 0 ˙ = 0 ˙
17 16 ifeq1d φ if i = I 0 ˙ R 0 ˙ X = if i = I 0 ˙ X
18 17 mpoeq3dv φ i N , j N if i = I 0 ˙ R 0 ˙ X = i N , j N if i = I 0 ˙ X
19 18 fveq2d φ D i N , j N if i = I 0 ˙ R 0 ˙ X = D i N , j N if i = I 0 ˙ X
20 eqid N Mat R = N Mat R
21 eqid Base N Mat R = Base N Mat R
22 1 20 21 2 mdetf R CRing D : Base N Mat R K
23 4 22 syl φ D : Base N Mat R K
24 13 6 ifcld φ i N j N if i = I 0 ˙ X K
25 20 2 21 5 4 24 matbas2d φ i N , j N if i = I 0 ˙ X Base N Mat R
26 23 25 ffvelrnd φ D i N , j N if i = I 0 ˙ X K
27 2 8 3 ringlz R Ring D i N , j N if i = I 0 ˙ X K 0 ˙ R D i N , j N if i = I 0 ˙ X = 0 ˙
28 10 26 27 syl2anc φ 0 ˙ R D i N , j N if i = I 0 ˙ X = 0 ˙
29 14 19 28 3eqtr3d φ D i N , j N if i = I 0 ˙ X = 0 ˙