Find the rank of matrix A = (1000 120 - 13 - 104 145 1) Finding matrix A= 1 0 0 0 1 2 0 -1 3 -1 0 4 1 4 5 1 The rank of

Find the rank of matrix A = (1000 120 - 13 - 104 145 1) Finding matrix A= 1 0 0 0 1 2 0 -1 3 -1 0 4 1 4 5 1 The rank of

A = 1 000 1 2 0 - 1 3 - 1 041 4 5 1 line 2 minus line 1, line 3 minus line 1 × 3, line 4 minus line 1 ～ 1 000 2 0 - 10 - 1 040 4 5 1 line 2 plus line 3 × 2, line 4 plus line 3 × 4, line 3 multiplied by - 1, swapping lines 2 and 3 ～ 1 000 10 - 40

A is a matrix of 4 times 3, the column vector group of a is linearly independent, and the rank of a is calculated
The normal form of quadratic form f (x1 + x2 + x3) = the square of 2x1 + the square of 3x2 - the square of 4x3 is?
Let a third-order square matrix satisfy the absolute value 4E + a = 0?
Let A1 = (- 3,4,1) and A2 = (2, - 1, K) be orthogonal, then K is equal to?
Four fill in the blank questions for God quick answer, urgent

Question 1: 3
Question 2: Y1 ^ 1 + Y2 ^ 2-y3 ^ 2
Question 3: - 1
Question 4: 10

Given that the matrix e + AB is invertible, it is proved that e + Ba is also invertible
And prove that (E + BA) - 1 = E-B [(E + AB) - 1] a can't find the inverse sign

C=(E+AB)^(-1)
(E-BCA)(E+BA)=E-BCA+BA-BCABA=
=E+B[-C+E-CAB]A=E+B[E-C(E+AB)]A=E
==>
E + Ba is reversible and (E + BA) ^ (- 1) = e-bca