How can I find A matrix who has A^3=?

Question

My question is to find a matrix $A∈\mathbb{R}^{2\times2}$ for which $${A}^{3} =\begin{bmatrix} 64 & 64 \\ -9 & 16\end{bmatrix}^{-1}.$$ I tried to write this matrix like A^3=A^2*A after taking inverse but couldnt find A. How can I solve this problem in a shortest way? Thanks.

Answer 1

Start with $A=\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)$

Then

$A^3=\left( \begin{array}{cc} a \left(a^2+b c\right)+c (a b+b d) & b \left(a^2+b c\right)+d (a b+b d) \\ a (a c+c d)+c \left(b c+d^2\right) & b (a c+c d)+d \left(b c+d^2\right) \\ \end{array} \right)$

Also $\left( \begin{array}{cc} 64 & 64 \\ -9 & 16 \\ \end{array} \right)^{-1}=\left( \begin{array}{cc} \frac{1}{100} & -\frac{1}{25} \\ \frac{9}{1600} & \frac{1}{25} \\ \end{array} \right)$

So, all you need to do is solve

$\left( \begin{array}{cc} a \left(a^2+b c\right)+c (a b+b d) & b \left(a^2+b c\right)+d (a b+b d) \\ a (a c+c d)+c \left(b c+d^2\right) & b (a c+c d)+d \left(b c+d^2\right) \\ \end{array} \right)=\left( \begin{array}{cc} \frac{1}{100} & -\frac{1}{25} \\ \frac{9}{1600} & \frac{1}{25} \\ \end{array} \right)$

One possible solution is

$A=\left( \begin{array}{cc} \frac{1}{5} (-2) \sqrt[3]{-\frac{1}{5}} & \frac{4}{15} \sqrt[3]{-\frac{1}{5}} \\ \frac{1}{80} (-3) \sqrt[3]{-\frac{1}{5}} & \frac{1}{5} (-3) \sqrt[3]{-\frac{1}{5}} \\ \end{array} \right)$

There are other solutions, you can find.

Answer 2

$\begin{bmatrix} 64 && 64 \\ -9 && 16 \end{bmatrix}^{-1} = \dfrac{1}{(64)(25)} \begin{bmatrix} 16 && -64 \\ 9 && 64 \end{bmatrix}$

Diagonalize this last matrix, eigenvalues are the roots of

$ (\lambda - 16) (\lambda - 64) + 64(9) = 0 $

which is

$ \lambda^2 - 80 \lambda + 64(25) = 0 $

The roots (the eigenvalues) are

$\lambda_{1, 2} = \frac{1}{2} (80 \pm \sqrt{6400 - 6400} ) = 40, 40 $

Next find the eigenvectors corresponding to this repeated eigenvalue:

Let $v = [x, y]^T $ then

$\begin{bmatrix} 24 && 64 \\ -9 && -24 \end{bmatrix} $

Hence $v_1 = [8, -3] $

Next, find a specific generalized eigenvector by solving the system

$\begin{bmatrix} -24&&-64\\9&&24\end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = c \begin{bmatrix} 8 \\ - 3 \end{bmatrix} $

where $\large c = 3 (40)^{\frac{2}{3}}$

The above linear system implies, $ -24 x - 64 y = 8c $ whose solution is

$ v_2 = [-c/3 , 0] + [-8/3, 1] = [-(c + 8) / 3, 1] $

Let the matrix of eigenvectors be

$P = [v_1, v_2] = \begin{bmatrix} 8 && -(c+8)/3 \\ -3 && 1 \end{bmatrix} $

Then we can write,

$\begin{bmatrix} 16 && -64 \\ 9 && 64 \end{bmatrix} = P J_1 P^{-1} $

where $J_1$ is matrix $\begin{bmatrix} 40 && c \\ 0 && 40 \end{bmatrix} $

So now we can write

$(64)(25) A^3 = P J_1 P^{-1} $

Thus by taking the cubic root, we can write,

$ \sqrt[3]{64(25)} A = P \sqrt[3]{J_1} P^{-1} $

It is easy to see that

$\sqrt[3]{J_1 } = \begin{bmatrix} \sqrt[3]{40} && 1 \\ 0 && \sqrt[3] {40} \end{bmatrix} $

Hence, matrix $A$ is

$A = \dfrac{1}{\sqrt[3]{1600}} \begin{bmatrix} 8 && -(c+8)/3 \\ -3 && 1 \end{bmatrix}\begin{bmatrix} \sqrt[3]{40} && 1 \\ 0 &&\sqrt[3] {40} \end{bmatrix} \begin{bmatrix} 8 && -(c+8)/3 \\ -3 && 1 \end{bmatrix}^{-1}$

Answer 3

We want to solve $A^3 = B^{-1}$, where $B$ is the matrix in the question. Basically we want to compute $B^{-{1 \over 3}}$.

With $V=\begin{bmatrix} 584 & 3 \\ -219 & 8 \end{bmatrix}$, we have $V^{-1} B V= J_{40}$, where $J_\lambda$ is a Jordan block with eigenvalue $\lambda$.

Note that $J_{40} = 40 (I + N)$, where $N= \begin{bmatrix} 0 & {1 \over 40} \\ 0 & 0\end{bmatrix}$, and $N^2 = 0$. The binomial theorem gives $(I+N)^{-{1 \over 3}} = \sum_{k=0}^\infty \binom{-{1 \over 3}}{k} N^k = I -{1 \over 3}N$, from which we have $J_{40}^{-{1 \over 3}} = 40^{-{1 \over 3}} \begin{bmatrix} 1 & -{1 \over 120} \\ 0 & 1\end{bmatrix}$ and so $A = V J_{40}^{-{1 \over 3}} V^{-1}$ is a solution.

More generally, we can choose a branch of $f(z) = z^{-{1 \over 3}}$ and use the formula in https://math.stackexchange.com/a/2659732/27978.