## 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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