It is proved that R (AB) = R (a) if and only if matrix B is invertible How to prove the necessity?

It is proved that R (AB) = R (a) if and only if matrix B is invertible How to prove the necessity?

No need to think about the necessity. There are counter examples, such as a = b = 0

In linear algebra, let a be a square matrix of order (n ≥ 2), prove that a * is the adjoint matrix of a, and R (a *) = 1 if and only if R (a) = n-1

Firstly, when AB = 0, R (a) + R (b) = 1, so r (a *) = 1

Linear Algebra: let a be an M x n matrix and rank (a) = r if and only if
A. In a, all r-order subformulas are not zero, and all subformulas with order greater than R are zero
B. All the subformulas of order less than R in a are 0, and at least one subformula of order r + 1 is not 0
C. At most one subformula of order r in a is not zero; all subformulas of order less than R in a are zero
D. In a, the r-order subformulas are not all zero, and the subformulas with order greater than R are all zero
What's the reason? Thank you

D
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According to the definition, the rank of a matrix is the order of the highest order non-zero sub formula. The rank of a is r, so all the sub formulas higher than R are zero, and the sub formulas of r order must have non-zero

A necessary and sufficient condition for finding the intersection of a and B is not equal to an empty set is given, a = {(x, y) | y ^ 2 = 2x}, B = {(x, y) | ((x-a) ^ 2 + y ^ 2 = 9}

It can be solved by pure algebraic method
If and only if (x-a) ^ 2 + 2x = 9 has nonnegative solution
That is, (x-a) ^ 2 + 2 (x-a) + 2a-9 = 0 has at least one nonnegative root
That is, A-1 + radical (10-2a) > = 0
Let 10-2a = T ^ 2, we can get T ^ 2-2t-10