Linear Algebra, 2.5, Problem 7 (c)
In Gilbert Strang's Introduction to Linear Algebra 4th ed., chapter 2, section 5, Problem 7 (c) is as following.
If A has row 1 + row 2 = row 3, show that A is not invertible: (c) What happens to row 3 in elimination?
The question is ''what happens in elimination?'' Therefore, first I tried the elimination. It turned out it is complicated. I was in Saarbruecken last week. I have no one to meet one afternoon there, so, I compute this on a piece of paper with a pen. I sat in a cafe in Sankt Johanner Markt for two hours for this, but, I had no luck at that day.
Now I had the answer, but, this is a bad answer. This is not a wrong answer, but I would say this is not a good answer.
Row 3 is always (0 0 0).
Actually I used a computational software. Unfortunately, this result doesn't give me a feeling ``I understand something.'' I didn't understand, Why the row 3 becomes all zero, in this way. I could not answer a related question: Is this happens 4 by 4 matrix or n by n matrix? Therefore, I would say ``this answer is not wrong, but not a good one.''
A better answer gives me more insight of this problem. I have one in a cafe in Saarbruecken. This is based on the idea, linearity.
The elimination method eliminates row 3 by the vector that is a linear combination of row 1 and row 2. For a 3 by 3 matrix,
When row 1 and row 2 are independent, they forms a plane. This plane passes through the origin. It is not possible, when two component are 0s and third is only non-zero. For example, imagine any plane go through the origin. If a point on a plane, x = 0, y = 0, z = k, but this plane goes through the origin, then it is only k = 0.
If row 1 and row 2 are not independent, they form a line goes through the origin, or origin itself. The same reason above, it doesn't exist that two components are both 0 and third one is not zero. When row 1 and row 2 are only origin, the linear combination of row 1 and row 2 is still only (0 0 0).
Therefore, row 3 is (0 0 0).
This linear combination understanding is simpler and more powerful. It is not limited to 3 by 3. Also this condition holds row 3 can be any linear combination of row 1 and row 2. If row 3 is any linear combination of of row 1 and row 2, A doesn't have its inverse. n by n matrix is also the same. This answer is better since we can understand more general cases.