TITLE: Solve for 2x2 matrix square root (just algebraic operations, not determinants/eigenvalues)
QUESTION [1 upvotes]: I just started my Linear Algebra course, and I am having some problems with an exercise in my book. 
The question is solve for X for
$X^{2}$ = $\begin{vmatrix}0 & 1\\1 & 0\end{vmatrix}$ 
The example I was given prior to this exercise (and from what I understand, should paralel the way of solving the exercise) is the following:
Solve for X in
$X^{2}$ = $\begin{vmatrix}1 & 0\\0 & 4\end{vmatrix}$ 
with this way of solving:
$\begin{vmatrix}1 & 0\\0 & 4\end{vmatrix}$ = $\begin{vmatrix}x & y\\z & t\end{vmatrix}$ $\begin{vmatrix}x & y\\z & t\end{vmatrix}$
which is converted to the equation system

$x^2+ yz = 1$
$xy + yt = 0$
$xz + zt = 0$
$yz + t^2 = 4$

The way it solves this system is rather 'intuitive', not using gauss or specific method.  In my book, the system is reordered

$x^2+ yz = 1$
$y (x + t) = 0$
$z (x + t) = 0$
$ t^2 - x^2 = 3$ (Eq4 - Eq1)

From this, it can be concluded that y = z and that either z=y or (x + t) must equal 0. If (x + t) = 0 then x = -t, and the Eq4 would be 0 = 3 (therefore, (x+t) cant equal 0, which leads to y=z=0).
In Eq1, if yz=0, x^2 must be 1, and therefore x = 1
Finally, in Equation 4, $t^2 - 1 = 3$, therefore t = +/- 2
Hence, the four solutions
$\begin{vmatrix}1 & 0\\0 & 2\end{vmatrix}$
$\begin{vmatrix}1 & 0\\0 & -2\end{vmatrix}$
$\begin{vmatrix}-1 & 0\\0 & 2\end{vmatrix}$
$\begin{vmatrix}-1 & 0\\0 & -2\end{vmatrix}$

Turning back to my exercise as such, 
$X^{2}$ = $\begin{vmatrix}0 & 1\\1 & 0\end{vmatrix}$ 
I have made the same equation system

$x^2+ yz = 0$
$xy + yt = 1$
$xz + zt = 1$
$yz + t^2 = 0$

After the transformation

$x^2+ yz = 0$
$y (x + t) = 1$
$z (x + t) = 1$
$ t^2 - x^2 = 0$ (Eq4 - Eq1)

My progress so far after several pages of trials is noting that y = z.
However I have not been succesful in applying the same 'intuitive' idea here. Any thoughts? Recommendations? I would like to point out that this exercise is presented BEFORE determinants, eigenvalues, and pretty much anything than just barely how to solve equation systems with Gauss method (hence, the answer I am looking just involves algebraic transformations, not matrix operations or methods). 
Thank you very much for your time
Cheers

REPLY [1 votes]: Two quick notes: $x^2+y^2=x^2+z^2=0$ and $(x+t)(x-t)=0$. We know that $x^2=-z^2$, so $x=\pm iz$, therefore, $t=\pm iz$. However, $x+t\neq0$ because if it did, $y(x+t)=0\neq1$. Therefore by the zero product property and $(x+t)(x-t)=0$, $x=t$. We are left with the equation $2xy=1$. We know $x=\pm i z=\pm i y$. Substituting, $\pm 2iy^2=1$, which you should be able to slove.
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