How to solve this equation with matrices

Question

can you please give me some hints to solve the following? I really don't know how to start. $$X^2= \begin{pmatrix} 6 & 2 \\ 3 & 7 \end{pmatrix}.$$ I tried to express this matrix as $4\cdot I + \begin{pmatrix} 2 & 2 \\ 3 & 3 \end{pmatrix}$ And somehow solve it, but I really have no clue. Please some help.

Answer 1

We have $$ \begin{pmatrix} 6 & 2 \\ 3 & 7 \end{pmatrix} = QDQ^{-1}$$ where $$ Q = \begin{pmatrix} -1 & 2 \\ 1 & 3 \end{pmatrix},\ D = \begin{pmatrix} 4& 0 \\ 0 & 9 \end{pmatrix}. $$ It follows that we may take any of the four matrices $$ X = Q\begin{pmatrix} \pm 2 & 0 \\ 0 & \pm 3 \end{pmatrix} Q^{-1}. $$

Answer 2

One simple but laborious way of doing it is as follows

Let $$X = \begin{bmatrix}a & b \\c & d\end{bmatrix}$$

Now, using the Caley-Hamilton Theorem, we have

$$X^2 - (a+d)X + |X|I = 0$$

Now, from the equation, we have $|X| = \pm 6$

Hence, we will have two sets of solutions each corresponding to one of these

$$|X| = 6$$

$$\implies X = \frac{1}{a+d}\begin{bmatrix}12 & 2 \\3 & 13\end{bmatrix}$$

Using the equivalence of the two representations of $X$, we have

$$a = \frac{12}{5} \\ b = \frac{2}{5} \\ c = \frac{3}{5} \\ d = \frac{13}{5}$$

Another solution to this is when each of the entries is negative. Similarly complete it for $|X| = -6$ to get the full set of solutions

Answer 3

Let $$X=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \implies X^2=\begin{bmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc\end{bmatrix}=\begin{bmatrix} 6 & 2\\ 3 & 7 \end{bmatrix}.$$ $$\implies a^2+bc=6~~~(1), b(a+d)=2~~~(2), c(a+d)=3~~~(3), d^2+bc=7~~~(4)$$ From (2) and (3) $b/c=2/3$, let $b=2k, c=3k$. From (1) and (4), we get $a^2-d^2=-1 \implies (a-d)(a+d)=-1$. Again from (2) and (3) we get $a+d=1/k, a-d=-1/k \implies a=(1/k-k)/2, d=(1/k+k)/2$. Finally,inserting $a,b,c$ in (1), we get $$25k^4-26k^2+1=0 \implies k=\pm 1,\frac{\pm 1 }{5}$$ For $k\pm 1$, we get $$X=\pm \begin{bmatrix} 0 & 2 \\ 3 & 1 \end{bmatrix}~~~(1)$$

For $k=\pm\frac{1}{5}$, we get $$X=\pm \frac{1}{5} \begin{bmatrix} 12 & 2 \\ 3 & 13 \end{bmatrix}~~~(2)$$

This sheer brute force method gives 4 matrices. one can check their correctness by squaring them.