I had a matrix algebra question. I know how to solve for $X$ if I am given $X^3$. I am stuck on this. Any help is much appreciated. Thanks!

Solve for $X$, if $$ X^3 = \begin{bmatrix} -6 & 14 \\ -7 & 15 \\ \end{bmatrix} $$

  • $\begingroup$ Find an eigenvalue/Jordan factorization of $X^3$. Over $\mathbb{R}$, you should be able to prove that $X=\begin{bmatrix}0&2\\-1&3\end{bmatrix}$ is the only solution. Over $\mathbb{C}$, there are $9$ possible solutions. $\endgroup$ – Batominovski Dec 2 '15 at 6:40

Hint: if $X^3$ can be diagonalized as $S D S^{-1}$, then $S D^{1/3} S^{-1}$ works, where $D^{1/3}$ is the diagonal matrix whose diagonal elements are cube roots of the diagonal elements of $D$.


There are probably easier ways, but it can be done by brute force plus a little good guessing. First let






Now the determinant of $X^3$ is $(-6)(15)-(14)(-7)=-90+98=8$, so the determinant of $X$ is $ad-bc=2$. Substitute that into $(1)$ to get








and therefore $b=-2c$. Substituting that into $(2)$ gives us




Here’s where we do a little guessing: it doesn’t hurt to see whether there is a solution in which $c$ and $(a+d)^2-2$ are integers. Clearly they would have to be $-1$ and $7$ or $1$ and $-7$ in some order. If we try $c=-1$, we want $(a+d)^2-2=7$, and $a+d=3$ does the trick. There’s no guarantee that this will work, but it’s worth a try. Substituting these values into $(3)$, we get


To finish off, see whether you can find values of $a$ and $d=3-a$ that work.


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