Review exercise 1.3. Find a 2 by 2 matrix for which and (a) , (b) , and (c) .

Answer: (a) We have and . Since we see that is its own inverse. Using the formula for the inverse of a 2 by 2 matrix we therefore have

Since , we have . Then or . Substituting for we then have or . So we can freely choose and then that value determines the values of and .

If we choose then

so that

If we choose then

and if we choose then

There are an infinite number of other matrices for which and , corresponding to other values of .

(b) We have and so that

Since we have . We choose . Since we have . Finally, since we have or . We choose .

Since and we thus have

and

Note that in this case so that we also have . This value of could have been derived from the formulas in (a) above by setting , so that , and .

If we instead choose then

and

Finally, if we choose then since we have . Since we again have or .

Thus , , and produces two other matrices for which (and also for the first one):

The above four matrices are the only matrices for which and .

(c) We have and , so that

Since we have or . Since we have . Thus we can choose any value for and then and will be determined accordingly. For example, if we choose we have and . We then have

and

Similarly if we choose we have and . We then have

and

Of particular interest is the matrix for which . In that case we have and . We then have

and

There are an infinite number of other matrices for which and , corresponding to other values of .

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.

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