Find the matrix $C, A = C^{T}*C$

Question

Given the matrix $A = \begin{bmatrix}5&-3\\-3&5\end{bmatrix}$
Problem is to find a Matrix $C$ , such that $A = C^{T}C$
I have tried to with notation like this
$C = \begin{bmatrix}a_{11}&a_{21}\\a_{12}&a_{22}\\...&...\\a_{1n}&a_{2n} \end{bmatrix}$ , $C^{T} = \begin{bmatrix}a_{11}&...&&a_{1n}\\a_{12}&...&&a_{2n} \end{bmatrix}$
After multiplication of this 2 matrices and summing terms of system of linear equations I get
$\sum_{i=1}^n (a_{1i} + a_{2i})^2 = 4$, what can I do next after this step, or are there any other ways to solve the problem?

Answer 1

Since $\det A=16$, $\det C$ must be $\pm4$. I will search for a solution with $\det C=4$. So$$C=\begin{bmatrix}a&b\\c&\frac{4+bc}a\end{bmatrix}$$and$$C^T.C=\begin{bmatrix}a^2+c^2 &ab+\frac{c(b c+4)}a\\ab+\frac{c(b c+4)}a&b^2+\frac{(bc+4)^2}{a^2}\end{bmatrix}.$$So, solve the system$$\left\{\begin{array}{l}a^2+c^2=5\\ab+\frac{c(bc+4)}a=-3\\b^2+\frac{(bc+4)^2}{a^2}=5.\end{array}\right.$$One solution will be$$C=\begin{bmatrix}\frac15\left(4\sqrt{5-b^2}-3b\right)&b\\ \frac15\left(-4b-3 \sqrt{5-b^2}\right)&\frac{-4b^2-3b\sqrt{5-b^2}+20}{4\sqrt{5-b^2}-3b}\end{bmatrix}.$$