Linear Algebra and Its Applications, Exercise 1.6.1

Posted on March 5, 2011 by hecker

Exercise 1.6.1. Find the inverses of the following matrices:

A_1 = \begin{bmatrix} 0&2 \\ 3&0 \end{bmatrix} \quad A_2 = \begin{bmatrix} 2&0 \\ 4&2 \end{bmatrix} \quad A_3 = \begin{bmatrix} \cos \theta &-\sin \theta \\ \sin \theta &\cos \theta \end{bmatrix}

Answer: We can use the standard formula for the inverse of a 2 by 2 matrix A:

A = \begin{bmatrix} a&b \\ c&d \end{bmatrix} \quad A^{-1} = 1/(ad - bc) \begin{bmatrix} d&-b \\ -c&a \end{bmatrix}

First, we have

A_1^{-1} = 1/(0 \cdot 0 - 2 \cdot 3) \begin{bmatrix} -0&-2 \\ -3&0 \end{bmatrix} = -\frac{1}{6} \begin{bmatrix} 0&-2 \\ -3&0 \end{bmatrix} = \begin{bmatrix} 0&\frac{1}{3} \\ \frac{1}{2}&0 \end{bmatrix}

We then have

A_2^{-1} = 1/(2 \cdot 2 - 0 \cdot 4) \begin{bmatrix} 2&-0 \\ -4&2 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 2&-0 \\ -4&2 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}&0 \\ -1&\frac{1}{2} \end{bmatrix}

Finally we have

A_3^{-1} = 1/(\cos^2 \theta + \sin^2 \theta) \begin{bmatrix} \cos \theta &\sin \theta \\ -\sin \theta &\cos \theta \end{bmatrix} = \begin{bmatrix} \cos \theta &\sin \theta \\ -\sin \theta &\cos \theta \end{bmatrix}

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.

This entry was posted in linear algebra. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

Gravatar
WordPress.com Logo

You are commenting using your WordPress.com account. ( Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. ( Log Out /  Change )

Cancel

Connecting to %s