## Linear Algebra and Its Applications, Review Exercise 1.23

Review exercise 1.23. Evaluate the following matrix expressions for $n = 2$ and $n = 3$ $\begin{bmatrix} 2&3 \\ 0&0 \end{bmatrix}^n \qquad \begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix}^n \qquad \begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix}^{-1}$

and then find the general expression for the first two matrices for any $n \ge 1$. $\begin{bmatrix} 2&3 \\ 0&0 \end{bmatrix}^2 = \begin{bmatrix} 2&3 \\ 0&0 \end{bmatrix} \begin{bmatrix} 2&3 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 4&6\\ 0&0 \end{bmatrix}$

and $\begin{bmatrix} 2&3 \\ 0&0 \end{bmatrix}^3 = \begin{bmatrix} 2&3 \\ 0&0 \end{bmatrix} \begin{bmatrix} 2&3 \\ 0&0 \end{bmatrix}^2 = \begin{bmatrix} 2&3 \\ 0&0 \end{bmatrix} \begin{bmatrix} 4&6 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 8&12 \\ 0&0 \end{bmatrix}$

So the general equation appears to be $\begin{bmatrix} 2&3 \\ 0&0 \end{bmatrix}^n = \begin{bmatrix} 2^n&2^{n-1} \cdot 3 \\ 0&0 \end{bmatrix}$

We can prove this by induction: Assume that the above equation holds true for some $k$. Then for $k+1$ we have $\begin{bmatrix} 2&3 \\ 0&0 \end{bmatrix}^{k+1} = \begin{bmatrix} 2&3 \\ 0&0 \end{bmatrix} \begin{bmatrix} 2&3 \\ 0&0 \end{bmatrix}^k = \begin{bmatrix} 2&3 \\ 0&0 \end{bmatrix} \begin{bmatrix} 2^k&2^{k-1} \cdot 3 \\ 0&0 \end{bmatrix}$ $= \begin{bmatrix} 2^{k+1}&2^k \cdot 3 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 2^{k+1}&2^{(k+1)-1} \cdot 3 \\ 0&0 \end{bmatrix}$

So if the equation holds true for $k$ it holds true for $k+1$ as well. Also, if we define $A^1 = A$ for any matrix $A$ then for $n = 1$ we have $\begin{bmatrix} 2^1&2^0 \cdot 3 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 2&3 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 2&3 \\ 0&0 \end{bmatrix}^1$

so by induction the equation above holds true for all $n \ge 1$.

Turning to the second matrix, we have $\begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix}^2 = \begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix} \begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 4&9 \\ 0&1 \end{bmatrix}$

and $\begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix}^3 = \begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix} \begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix}^2 = \begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix} \begin{bmatrix} 4&9 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 8&21 \\ 0&1 \end{bmatrix}$

The $(1, 1)$ entry of each matrix appears to be $2^n$ in general, but the $(1, 2)$ entry is more complicated. The expression $3(2^n - 1)$ looks as if it might work; for $n = 2$ this is $3 (4 - 1) = 3 \cdot 3 = 9$ and for $n = 3$ this is $3(8 - 1) = 3 \cdot 7 = 21$.

We use induction to try to prove this. Assume that for some $k$ we have $\begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix}^k = \begin{bmatrix} 2^k&3(2^k - 1) \\ 0&1 \end{bmatrix}$

We then have $\begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix}^{k+1} = \begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix} \begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix}^k = \begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix} \begin{bmatrix} 2^k&3(2^k - 1) \\ 0&1 \end{bmatrix}$ $= \begin{bmatrix} 2 \cdot 2^k&2 \cdot 3(2^k - 1) + 3\\ 0&1 \end{bmatrix} = \begin{bmatrix} 2^{k+1}&3(2 \cdot 2^k - 2) + 3\\ 0&1 \end{bmatrix}$ $= \begin{bmatrix} 2^{k+1}&3(2^{k+1} - 2 +1) \\ 0&1 \end{bmatrix} = \begin{bmatrix} 2^{k+1}&3(2^{k+1} - 1) \\ 0&1 \end{bmatrix}$

So if the equation holds true for $k$ it holds true for $k+1$ as well. Also, for $k = 1$ we have $\begin{bmatrix} 2^1&3(2^1 - 1) \\ 0&1 \end{bmatrix} = \begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix}^1$

so the equation above holds true for all $n \ge 1$.

Finally for the third matrix we have $\begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix}^{-1} = \frac{1}{2 \cdot 1 - 3 \cdot 0} \begin{bmatrix} 1&-3 \\ 0&2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1&-3 \\ 0&2 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}&-\frac{3}{2} \\ 0&1 \end{bmatrix}$

Note that $\begin{bmatrix} 2^{-1}&3(2^{-1} - 1) \\ 0&1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}&3(\frac{1}{2} - 1) \\ 0&1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}&-\frac{3}{2} \\ 0&1 \end{bmatrix} = \begin{bmatrix} 2&3 \\ 0&1 \end{bmatrix}^{-1}$

so that the equation above holds true for $n = -1$ also.

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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