## Linear Algebra and Its Applications, Exercise 1.5.6

Exercise 1.5.6. Given the matrix

$E = \begin{bmatrix} 1&0 \\ 6&1 \end{bmatrix}$

find $E^2$, $E^8$, and $E^{-1}$.

$E^2 = \begin{bmatrix} 1&0 \\ 6&1 \end{bmatrix} \begin{bmatrix} 1&0 \\ 6&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 12&1 \end{bmatrix}$

We can square this again to obtain

$E^4 = E^2E^2 = \begin{bmatrix} 1&0 \\ 12&1 \end{bmatrix} \begin{bmatrix} 1&0 \\ 12&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 24&1 \end{bmatrix}$

and again to obtain

$E^8 = E^4E^4 = \begin{bmatrix} 1&0 \\ 24&1 \end{bmatrix} \begin{bmatrix} 1&0 \\ 24&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 48&1 \end{bmatrix}$

In compting $E^2$ we note that all the entries match those of the identity matrix except for the 2,1 entry, which is $6 \cdot 1 + 1 \cdot 6$ or 12. If we change the second 6 to -6 we instead obtain $6 \cdot 1 + 1 \cdot (-6)$ or zero for the 2,1 position. We then have

$\begin{bmatrix} 1&0 \\ 6&1 \end{bmatrix} \begin{bmatrix} 1&0 \\ -6&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I$

so that

$E^{-1} = \begin{bmatrix} 1&0 \\ -6&1 \end{bmatrix}$

Another approach to finding $E^{-1}$ is simply to note that E is an elementary matrix, and that its inverse therefore consists of the same matrix with the multiplier 6 negated.

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.