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I have to solve for the $2\times 2$ matrix $X$ the following equation

$$X^3=\begin{pmatrix}19&30\\ -45&-71\end{pmatrix}$$

To be honest I have no idea how to start.

I know how to solve equations like $$A \cdot X=C => X=A^{-1} \cdot C$$

$$X \cdot B=C => X=C \cdot B^{-1}$$

$$A \cdot X \cdot B=C => X=A^{-1} \cdot C \cdot B^{-1}$$

but I don't know how I could apply it in here

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    $\begingroup$ Diagonalise then find the root of the diagonal matrix. Pretty straight forward. $\endgroup$ – Displayname Feb 10 '17 at 14:54
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The given matrix $m$ has determinant $1$ and trace = $-52$. Assuming $X$ is a real matrix, $det(X)=1$. Let $t= tr(X)$. Now $X$ satisfies the equation $x^2-tx+1=0$. From this it is easy to see that $m=X^3=(t^2-1)X-t*I_2$ , where $I_2$ is the identity matrix. So $(t^2-1)X=m+t*I_2$. Taking trace of both sides, we see that $t$ is a root of the cubic equation $t^3-3t=-52$ whose only real solution is $t=-4$. Therefore $15X=m-4I_2$. This gives

$$X=\begin{pmatrix}1&2\\ -3&-5\end{pmatrix}$$

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  • $\begingroup$ can you please give me a link with the lesson about matrix equations with pow where they use this method (with trace tr(X)), I want to see the formula you've applied so I can make the next exercises by myself $\endgroup$ – Andreea Mika Feb 10 '17 at 16:02
  • $\begingroup$ @AndreeaMika Try this link for another example of this method. math.stackexchange.com/questions/537095/… $\endgroup$ – Lozenges Feb 10 '17 at 18:25
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I'm not providing here a method to provide the result directly, but I would decompose the resulting matrix into simple geometrical transformation : dilatation, rotation ..., then I would divide the dilatation coefficient by 3, the angle by 3 etc ...

That would be my approach.

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    $\begingroup$ Unfortunately the simple transformations you want to decompose into do not commute, so taking cube roots of them separately will not work. $\endgroup$ – Henning Makholm Feb 10 '17 at 15:00
  • $\begingroup$ Yes, that's right, they do not commute in general, but that it how I would approach the problem, and find the condition where they can commute ... and explore. Then there is the brutal force, using matrix expotential and the associated sum ... but it doesn't allow to understand what is going on. The case in which is doesn't commute also mean there is no unique solution. $\endgroup$ – jmary Feb 10 '17 at 15:28
  • $\begingroup$ I may be wrong, I'm purely intuitive at the moment, I haven;'t written anything. $\endgroup$ – jmary Feb 10 '17 at 15:31
  • $\begingroup$ Sorry but the guy who put -1 on my answer, is unfair, I provide a method to find the answer, not the answer. As long as there is no projection, but only dilatation and rotation, with the right base, and the right origin, things will commute. $\endgroup$ – jmary Feb 10 '17 at 15:53
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    $\begingroup$ It is correct that rotations and uniform scalings commute, but the matrix here is not a combination of such transformations. Both rotations and uniform scaling preserve orthogonality of vectors, but this matrix maps the orthogonal vectors $(1,0)$ and $(0,1)$ to $(19,-45)$ and $(30,-71)$ which are not at all orthogonal. $\endgroup$ – Henning Makholm Feb 10 '17 at 15:57

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