## Linear Algebra and Its Applications, exercise 1.4.16

Exercise 1.4.16. For the following matrices $E = \begin{bmatrix} 1&0&0 \\ -2&1&0 \\ 0&0&1 \end{bmatrix} \quad F = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 1&0&1 \end{bmatrix} \quad G = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&1&1 \end{bmatrix}$

verify that (EF)G = E(FG). $EF = \begin{bmatrix} 1&0&0 \\ -2&1&0 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 1&0&1 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ -2&1&0 \\ 1&0&1 \end{bmatrix}$

and $(EF)G = \begin{bmatrix} 1&0&0 \\ -2&1&0 \\ 1&0&1 \end{bmatrix} \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&1&1 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ -2&1&0 \\ 1&1&1 \end{bmatrix}$

We also have $FG = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 1&0&1 \end{bmatrix} \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&1&1 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 1&1&1 \end{bmatrix}$

and $E(FG) = \begin{bmatrix} 1&0&0 \\ -2&1&0 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 1&1&1 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ -2&1&0 \\ 1&1&1 \end{bmatrix} = (EF)G$

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.