I had trouble with some problems in the list of Suggest Problems,
so I am begging for clarification and enlightenment!
1) D8 part [a] page 47:
If a system of linear equations has only 1 unknown, is that still a linear system?
I thought the answer was FALSE because there are no examples in the textbook where there is only one unknown in the linear system.
2) 34 page 107:
Do you show that A is invertible just by multiplying the A and A-inverse and showing that that equals to I(identity)?
If that is the case, then I could not even get I(identity) by multiplying these two matrices; something seems to be wrong or missing.
3)Is it quite self evident that a matrix with a row of zeros is not invertible?
Is there a way to prove it without referencing theorem 3.2.4 and 3.3.3?
These are my questions and I hope they can be answered before the quiz on Thursday, 8PM
1) Yes. A linear equation with 1 unkwon has exactly one solution, so a linear system with 1 unknowns has no solution or exactly one solution.
Now, you may know whether the problem is T of F.
2) It has no problem! I give you some hint (or technique).
1. Definition 3.1.6 (you may think a product of some of u,v,u^T,v^T..)
2. Transpose of a 1 by 1 matrix is equal itself. (why?)
Now, you are ready to prove the problem :)
3) Yes. Let A be a n by n matrix with a row of zero.
Then for any n by n matrix B, AB has a row of zero.
(You know? (i,j)-th entry of (AB)_ij is a product of i-th row of A and j-th column of B.)
(Thus, if i-th row of A is a row of zero, then it yields that i-th row of AB is also a row of zero.)
Remind the definition of an inverse matrix. A has no inverse.