Is there a squared matrix $A$ sized 2x2 that follows the next criteria?

Question

Is there a squared matrix $A$ sized 2x2 that it's elements $\in \mathbb Z$ so that $$ A^2 = \begin{pmatrix} 2 & 3 \\ 2 & 4 \\ \end{pmatrix} $$

Answer 1

Hint: what happens to the determinant of a matrix when you square it?

Answer 2

No. $det(A^{2}) = 2$, so $det(A)$ must equal plus or minus $\sqrt{2}$. But this is impossible with integer entries in $A$.

Answer 3

There are several solutions for the general problem of finding the square root of a matrix. For the $2\times 2$ case,

$$A^2 = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$

Let $\tau = a + d$ be the trace of $M$, and $\delta = ad - bc$ be its determinant. Let $s$ be such that $s^2 = \delta$, and $t$ be such that $t^2 = \tau + 2s$. Then if $t \neq 0$, the square root of $M$ is

$$A = \frac{1}{t} \begin{bmatrix} a+s & b \\ c & d+s \end{bmatrix} $$

In your case, $\tau = 6, \delta = 2$ and thus $s = \pm \sqrt{2}, t = \pm \sqrt{6\pm 2\sqrt{2}}$.

It should be clear for these values of $s,t$ all the elements in $A$ will not be in $\mathbb{Z}$.