## Linear Algebra and Its Applications, Review Exercise 1.11

Review exercise 1.11. Suppose $E$ is a 2 by 2 matrix that adds the first equation of a linear system to the second equation. What is $E^2$? $E^8$? $8E$?

Answer: Since $E$ adds the first equation to the second, we have $E = \begin{bmatrix} 1&0 \\ 1&1 \end{bmatrix}$

We then have $E^2 = \begin{bmatrix} 1&0 \\ 1&1 \end{bmatrix} \begin{bmatrix} 1&0 \\ 1&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 2&1 \end{bmatrix}$

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

so that $E^8 = E^4 E^4 = \begin{bmatrix} 1&0 \\ 4&1 \end{bmatrix} \begin{bmatrix} 1&0 \\ 4&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 8&1 \end{bmatrix}$

In general if for some $k \ge 1$ we have $E^k = \begin{bmatrix} 1&0 \\ k&1 \end{bmatrix}$

then for $k+1$ we have $E^{k+1} = E E^k = \begin{bmatrix} 1&0 \\ 1&1 \end{bmatrix} \begin{bmatrix} 1&0 \\ k&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ k+1&1 \end{bmatrix}$

Since for $k = 1$ we have $E^1 = E = \begin{bmatrix} 1&0 \\ 1&1 \end{bmatrix}$

by induction for all $n \ge 1$ we have $E^n = \begin{bmatrix} 1&0 \\ n&1 \end{bmatrix}$

Finally, we have $8E = 8 \begin{bmatrix} 1&0 \\ 1&1 \end{bmatrix} = \begin{bmatrix} 8&0 \\ 8&8 \end{bmatrix}$

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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