S4

Question 1

If AB = AC, then B = C.

Answer 1

False

This would be true if A is invertible, but it doesn’t say that. To prove that it is false, here is a counterexample:

Question 2

If A and B are invertible, then A + B is also invertible.

Answer 2

False

Take A = I and B = −I, then A and B are invertible, but A + B = 0 is not.

Question 3

If a matrix is invertible, then every square submatrix is invertible.

Answer 3

I₃ is invertible, but it has the square submatrix A, which is not invertible.

Question 4

If the equation Ax = b has at least one solution for each b ∈ Rⁿ, then the solution is unique for each b.

Answer 4

True

The if-statement is that (the linear transformation corresponding to) A is onto, the then-statement is that it is one-to-one. By the Invertible Matrix Theorem, these two statements are equivalent.

Here is a direct explanation without that theorem: The fact that A is n × n and onto implies that the (reduced) echelon form of A has n pivots, which implies that A is one-to-one.

Question 5

If the columns of A are linearly independent, then they span Rⁿ.

Answer 5

True

These are also two equivalent statements by the Invertible Matrix Theorem. Similar to the last one, you can also see it directly by seeing that both are equivalent to having n pivots in the (reduced) echelon form.

Question 6

If the columns of A are linearly independent, then its rows are also linearly independent.

Answer 6

True

The rows of A are the columns of A^T. Because A is square and the columns are linearly independent, A is invertible, again by the Invertible Matrix Theorem.

Then A^T is also invertible, so its columns are also linearly independent.

Note that, technically, we never defined what it means for rows to be linearly independent, but this is easy to guess: Just transpose all the rows into columns. Or take literally the same definition as for columns.