Matrix Indices

We can find a nonsingular matrix $P$ such that $PAP^{-1}$ is equal to the Jordan Normal form of $A$ , and hence is of the form $PAP^{-1} \; =\; \left(\begin{array}{cc} x & y \\ 0 & z \end{array}\right)$ Thus $PA^nP^{-1} \; = \; \left(\begin{array}{cc} x^n & y(n) \\ 0 & z^n \end{array}\right) \hspace{2cm} n \ge 0$ for some $y(n) \in \mathbb{R}$ for all $n \ge 0$ . Thus we can find some function $Z(t)$ such that $Pe^{tA}P^{-1} \; = \; \left(\begin{array}{cc} e^{tx} & Z(t) \\ 0 & e^{tz}\end{array}\right) \hspace{2cm} t \in \mathbb{R}$ and hence we deduce that $\mathrm{det}\,e^{tA} \; = \; \mathrm{det}\,Pe^{tA}P^{-1} \; = \; e^{tx}e^{tz} \; = \; e^{t(x+z)} \hspace{2cm} t \in \mathbb{R}$ But $\mathrm{Tr}(A) \; = \; \mathrm{Tr}(PAP^{-1}) \; = \; x + z$ so we deduce that $\mathrm{det}\,e^{tA} \; = \; e^{t\mathrm{Tr}(A)} \hspace{2cm} t \in \mathbb{R}$ and hence, in particular $\mathrm{det}\,2^A \; = \; 2^{\mathrm{Tr}(A)}$ In this case we deduce that $B$ has determinant $2^5=\boxed{32}$ .

Let us take $B = e^{\ln2 \cdot A} = e^{X} = \exp[ \begin{pmatrix} \ln 2 & 2\ln2 \\ 3\ln 2 & 4\ln 2 \end{pmatrix}]$ . Knowing that $e^{x} = \Sigma_{n=0}^{\infty} \frac{x^{n}}{n!}$ and that $X = S\Lambda S^{-1}, X^{n} = S\Lambda^{n}S^{-1}$ (where $\Lambda$ and $S$ are respectively the eigenvalue and eigenvector matrices of $X$ ), we can express $B$ as the following:

$B = I + \frac{1}{1!} X + \frac{1}{2!} X^2 + \frac{1}{3!} X^3 +... = S[I + \frac{1}{1!}\Lambda + \frac{1}{2!}\Lambda^{2} + \frac{1}{3!}\Lambda^{3}+...]S^{-1}$ (i).

The eigenvalues of $X$ can be computed according to:

$\det(X-\lambda I) = 0 \Rightarrow \begin{vmatrix} \ln2 - \lambda & 2\ln 2 \\ 3\ln 2 & 4\ln 2 - \lambda \end{vmatrix} = 0 \Rightarrow \lambda^{2} - 5\ln 2 \cdot \lambda -2\ln^{2} 2 = 0 \Rightarrow \lambda = (\frac{5 \pm \sqrt{33}}{2}) \cdot \ln 2$ (ii).

If we substitute the eigenvalues of (ii) into (i), we now obtain:

$B = S \cdot \begin{pmatrix} \Sigma_{n=0}^{\infty} \frac{\lambda^{n}_{1}}{n!} & 0 \\ 0 & \Sigma_{n=0}^{\infty} \frac{\lambda^{n}_{2}}{n!} \end{pmatrix} \cdot S^{-1} = S \cdot \begin{pmatrix} \exp (\ln 2 \cdot \frac{5+\sqrt{33}}{2}) & 0 \\ 0 & \exp (\ln 2 \cdot \frac{5-\sqrt{33}}{2}) \end{pmatrix} \cdot S^{-1} = S \cdot \begin{pmatrix} 2^{\frac{5+\sqrt{33}}{2}} & 0 \\ 0 & 2^{\frac{5-\sqrt{33}}{2}} \end{pmatrix} \cdot S^{-1}$ (iii).

Finally, $\det(B) = \det (S) \cdot \det (S^{-1}) \cdot \begin{vmatrix} 2^{\frac{5+\sqrt{33}}{2}} & 0 \\ 0 & 2^{\frac{5-\sqrt{33}}{2}} \end{vmatrix} = \det(SS^{-1} )\cdot \begin{vmatrix} 2^{\frac{5+\sqrt{33}}{2}} & 0 \\ 0 & 2^{\frac{5-\sqrt{33}}{2}} \end{vmatrix} = \det(I) \cdot \begin{vmatrix} 2^{\frac{5+\sqrt{33}}{2}} & 0 \\ 0 & 2^{\frac{5-\sqrt{33}}{2}} \end{vmatrix} = 1 \cdot 2^5 = \boxed{32}.$

I'll provide a Laplace solution in alternative to brilliant @Mark Hennings solution.

It's a well-known fact that, for a real number $r$ :

$\large \displaystyle e^{rt} = \mathcal{L}^{-1} \{ (s-r)^{-1} \}$

Where $\mathcal{L}^{-1}$ is the Inverse Laplace Transform . This reasoning extends to the matrix case as:

$\large \displaystyle e^{At} = \mathcal{L}^{-1} \{ (sI-A)^{-1} \}$

So for a matix $A$ such as:

$\large \displaystyle A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

We have:

$\large \displaystyle e^{At} = \mathcal{L}^{-1} \left \{ \begin{bmatrix} s-a & -b \\ -c & s-d \end{bmatrix} ^{-1} \right \}$

$\large \displaystyle e^{At} = \mathcal{L}^{-1} \left \{ \frac{1}{p(s)} \begin{bmatrix} s-d & b \\ c & s-a \end{bmatrix} \right \}$

Where $p(s) = s^2 - (a+d)s + (ad-bc)$ is the characteristic polynomial of matrix $A$ . Let us sat that the roots of $p(s)$ are $k$ and $l$ . So, expanding in partial fractions:

$\large \displaystyle e^{At} = \mathcal{L}^{-1} \left \{ \frac{1}{k-l} \begin{bmatrix} \frac{k-d}{s-k} + \frac{d-l}{s-l} & \frac{b}{s-k} - \frac{b}{s-l} \\ \frac{c}{s-k} - \frac{c}{s-l} & \frac{k-a}{s-k} + \frac{a-l}{s-l} \end{bmatrix} \right \}$

$\large \displaystyle e^{At} = \frac{1}{k-l} \begin{bmatrix} (k-d)e^{kt} + (d-l)e^{lt} & be^{kt} - be^{lt} \\ ce^{kt} - ce^{lt} & (k-a)e^{kt} + (a-l)e^{lt} \end{bmatrix}$

Making $t = \ln(n)$ , where $n$ is a real number:

$\large \displaystyle n^A = \frac{1}{k-l} \begin{bmatrix} (k-d)n^k + (d-l)n^l & bn^k - bn^l \\ cn^k - cn^l & (k-a)n^k + (a-l)n^l \end{bmatrix}$

$\large \displaystyle \det(n^A) = \frac{1}{(k-l)^2} \{[ (k^2 - (a+d) + ad)n^{2k} + (l^2 - (a+d) + ad)n^{2l} + \newline \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + (dk-da-lk+al)n^{k+l} + (ak-lk-ad+dl)n^{k+l} ] - bc [n^{2k} + n^{2l} -2n^{k+l} ] \}$

$\large \displaystyle \det(n^A) = \frac{1}{(k-l)^2} [ n^{2k}(k^2 -(a+d)k + (ad-bc)) + n^{2l}(l^2 - (a+d)l + (ad-bc)) + \newline \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + n^{k+l}(dk+dl+ak+al-2ad+2bc-2lk) ]$

$\large \displaystyle \det(n^A) = \frac{1}{(k-l)^2} [ n^{2k}p(k) + n^{2l}p(l) + n^{k+l}((a+d)(k+l) - 2(ad-bc) - 2kl) ]$

Notice that since $k$ and $l$ are roots of $p(s)$ , $p(k)=p(l) = 0$ . Also notice that, by Vieta, $k+l = a+d$ (sum of roots) and $kl = ad-bc$ (product of roots). So:

$\large \displaystyle \det(n^A) = \frac{n^{a+d}}{(k-l)^2} [(k+l)(k+l) - 2(kl) - 2kl) ]$

$\large \displaystyle \det(n^A) = \frac{n^{a+d}}{(k-l)^2} (k-l)^2$

$\color{#20A900} \boxed{ \large \displaystyle \det(n^A) = n^{a+d} = n^{\text{tr}(A)} }$

In this case, $n=2$ and $\text{tr}(A) = 5$ . So:

$\color{#3D99F6} \boxed{ \large \displaystyle \det(2^A) =32 }$