## 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 .

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