Telling fibs

First of all, we should know beforehand how matrix exponential is defined. If we have a square matrix $A$ , then, matrix exponential is defined using the Maclaurin expansion of $e^x$ , namely, as follows:

$\exp(A)=e^A=\sum_{k=0}^\infty\frac{A^k}{k!}$

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Now, back to our problem, let $A:=\begin{pmatrix}1&1\\1&0\end{pmatrix}$ . Then, the given equation becomes,

$\begin{pmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{pmatrix}=e^A=\sum_{k=0}^\infty\frac{A^k}{k!}=I_2+\sum_{k=1}^\infty\frac{A^k}{k!}$

The next part relies on observation. We first check out the first few powers of $A$ and then make the following claim:

Claim: $A^k=\begin{pmatrix}F_{k+1}&F_{k}\\ F_k&F_{k-1}\end{pmatrix}~\forall~k\in\Bbb{Z^+}$ where $F_k$ is the $k^{\textrm{th}}$ Fibonacci number with $F_0=0~,~F_1=1$ .

This claim is easily proved by standard induction which (if you need the obvious proof) will be given by me in the comments since I don't want to make the solution any messier than it already is.

Anyway, using this claim, our equation becomes,

$\begin{pmatrix}a_{11}&a_{12}\\ a_{21}&a_{22}\end{pmatrix}=\begin{pmatrix}1+\sum_{k=1}^\infty\frac{F_{k+1}}{k!}&\sum_{k=1}^\infty\frac{F_k}{k!}\\ \sum_{k=1}^\infty\frac{F_k}{k!}&1+\sum_{k=1}^\infty\frac{F_{k-1}}{k!}\end{pmatrix}$

We take out our relevant part to be summed, that is,

$a_{12}+a_{21}=2\cdot\sum_{k=1}^\infty\frac{F_k}{k!}=2\cdot\sum_{k=0}^\infty\frac{F_k}{k!}$

Now, we use the Binet's Fibonacci formula to rewrite this sum into a familiar form that can be simplified using the Maclaurin series for $e^x$ . Rewriting it, we have,

$a_{12}+a_{21}=2\cdot\sum_{k=0}^\infty\left(\frac{\phi^k-(-\phi)^{-k}}{\sqrt 5\cdot k!}\right)=\frac{2}{\sqrt 5}\left[\left(\sum_{k=0}^\infty\frac{\phi^k}{k!}\right)-\left(\sum_{k=0}^\infty\frac{(-\phi^{-1})^k}{k!}\right)\right]$

Using Maclaurin series of $e^x$ , i.e., $e^x=\displaystyle\sum_{k=0}^\infty\frac{x^k}{k!}$ which converges $\forall~x$ , we have,

$a_{12}+a_{21}=\frac{2}{\sqrt 5}\left(e^\phi-e^{-1/\phi}\right)$

where $\phi$ is the golden ratio and has the value $\phi=\frac{1+\sqrt 5}{2}$ .

Comparing values, we have $a=1,b=5,c=2,d=5$ . Add them up and you get the answer, which is $\boxed{13}$ .

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Clarifications:

• $I_2$ is the identity matrix of order $2\times 2$ .
• We have, by definition, $A^0=I_n$ for any square matrix $A$ of order $n\times n$ .