Linear Algebra and Its Applications, Exercise 1.6.1

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.

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