Review exercise 1.28. Compute the following matrices:

Answer: For the first matrix we have

Based on this, we can guess at the answer for raising the matrix to the power of , and try to prove it by induction. Assume that for some we have

Then for we have

By induction we then see that for all we have

Note that this equation also holds true for and , and we might guess that it holds true for as well. We can test this as follows:

and

Therefore we have

We can find the inverse of the last matrix by Gauss-Jordan elimination. We start by multiplying the first row by and subtracting it from the second row:

We then multiply the second row by and subtract it from the third row:

We thus 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.