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 $n \ge 1$ .

Answer: We have

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

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.

Linear Algebra and Its Applications, Review Exercise 1.23

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