Linear Algebra matrix calculation help please A = $$
    \begin{pmatrix}
    2 & 1 \\
    3 & 4 \\
    \end{pmatrix}
$$
How can i find matrix B s.t B^3 = A?
 A: I believe this question can be solved using simple eigen-decomposition.
Eigen-decompose $A$ into a $Q \Sigma Q^{T}$ matrix, and simple replace the diagonal matrix $\Sigma$ elements with their cube roots. That new matrix would be your desired matrix $B$.
A: If you diagonalize $A$ into $P\Lambda P^{-1}$, then a cube root of $A$ is $B=P\Lambda^{1/3}P^{-1}$ (verify this!). I expect that you’ll be able to find a cube root of a diagonal matrix. The matrix in this problem has “nice” eigenvalues and eigenvectors that can pretty much be found by inspection, so the calculations aren’t very difficult.  
Alternatively, by the Cayley-Hamilton theorem, $B=aI+bA$ for some unknown scalars $a$ and $b$. If $A$ has distinct eigenvalues $\lambda_1$ and $\lambda_2$, then these unknowns are the solutions to the system of linear equations $a+\lambda_ib=\lambda_i^{1/3}$. Solving this system and assembling $B$ this way might be less work than a full diagonalization of $A$.
A: You can define $B$ as
$$B = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$
And solve the equation following equation system:
$$B^3 = \begin{bmatrix} a & b \\ c & d \end{bmatrix}*\begin{bmatrix} a & b \\ c & d \end{bmatrix}*\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix},$$
what equals to
$$\begin{bmatrix} a^3+2abc+bdc & a^2b+b^2c+abd+bd^2 \\ a^2c+acd+c^2d+d^2c & abc+2bcd+d^3 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$$
And solving that 4 equation system will give you the solution.
