Exercise 2.4.15. For each of the following matrices

find a left inverse and right inverse if they exist.

Answer: We begin with the 2 by 3 echelon matrix . Since has two pivots its rank . Since has two rows we also have so that has a (3 by 2) right inverse. It does not have a left inverse since .

We now attempt to find a right inverse . We must have or

Multiplying the two rows of by the first column of we must have and . One way to achieve this is to set and .

Multiplying the two rows of by the second column of we must have and . One way to achieve this is to set and . We then have

so that

We now consider the matrix . The first and third rows of are linearly independent, and the second row of is equal to the sum of the first and third rows. The rank of is therefore . Since has two columns we also have so that has a (3 by 2) right inverse. It does not have a right inverse since .

We now attempt to find a left inverse of such that . We could proceed as we did above, but there is a shortcut: Note that so that we are looking for such that . But and we already have a matrix such that . We can therefore choose such that or :

We then have

so that is a left inverse of .

Finally we consider the 2 by 2 triangular matrix . If then has two pivots and rank . In this case since both a left inverse and right inverse exist and are the same: is invertible with .

From the standard formula for the inverse of a 2 by 2 matrix (see page 42) we have

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.