Linear algebra A = (1, - 1,1) B = 1 21 10 2 2 21 100 and Ax = B for matrix X Linear algebra A = (1, - 1,1) B = 1,2 1 1 0 2 2 2 1 1 0 0 And Ax = B for matrix X

Linear algebra A = (1, - 1,1) B = 1 21 10 2 2 21 100 and Ax = B for matrix X Linear algebra A = (1, - 1,1) B = 1,2 1 1 0 2 2 2 1 1 0 0 And Ax = B for matrix X

(A,B)=
1 -1 1 1 2
1 1 0 2 2
2 1 1 0 0
r3-r1
1 -1 1 1 2
1 1 0 2 2
1 2 0 -1 -2
r3-r2
1 -1 1 1 2
1 1 0 2 2
0 1 0 -3 -4
r1+r3,r2-r3
1 0 1 -2 -2
1 0 0 5 6
0 1 0 -3 -4
r1-r2
0 0 1 -7 -8
1 0 0 5 6
0 1 0 -3 -4
Exchange bank
1 0 0 5 6
0 1 0 -3 -4
0 0 1 -7 -8
So x=
5 6
-3 -4
-7 -8

Linear algebra A is m × P matrix B is p × n matrix R (a) + R (b) - P ≤ R (AB) ≤ min {R (a), R (b)}
Linear algebra A is m × P matrix, B is p × n matrix
It is proved that R (a) + R (b) - P ≤ R (AB) ≤ min {R (a), R (b)} (r denotes rank)
The second half can be proved without proof.
It seems that the answer on the first floor did not mention the main point, but the answer on the second floor was not specific enough - (1) r (AB) + P (3) r (a) + R (b)?
How to investigate? Please make it clear~

The column of a is divided into (A1, A2, A3,... AP), so AB = b11a1 + b21a2 +... Bp1ap + b12a1 + b22a2 +... +... + bpnap, so AB can be linearly expressed by P vector groups of a, that is, R (AB) = R (B ') - R (B'2) = R (b) - R (B'2), and R (B'2) is not greater than its row number P-R (a), so r (AB) > = R (b) - P + R (...)