Review exercise 1.14. For each of the following find a 3 by 3 matrix such that for any matrix

(a)

(b)

(c) The first row of is the last row of and the last row of is the first row of .

(d) The first column of is the last column of and the last column of is the first column of .

Answer: (a) We have so that we can choose

(b) For all we have so that we can choose

(c) We can choose to be the permutation matrix that reverses the order of rows in

For example

(d) The first column of is produced by multiplying the first row of by the first column of . Since this computation does not involve the last column of in general it is impossible to find a matrix such that the first column of is equal to the last column of .

However note that we can reverse the order of columns in by multiplying on the right by the value of from (c) above:

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.