# Find lower triangular matrix using Givens-rotation.(QR-factorisation)

I'm looking into QR-factorisation using Givens-rotations and I want to transform a matrix into its lower triangular matrices.

My problem is that I do not know how to get the cos oder the sin value and also I do not know whether this rotation matrix is right.

My question is : Can we use the same rotation matrix? $$$$\begin{bmatrix} c & -s \\ s & c \\ \end{bmatrix}\begin{bmatrix} a \\ b \\ \end{bmatrix} = \begin{bmatrix} 0 \\ r \\ \end{bmatrix}$$$$

I just switched the 0 and r from the usual givens-rotation equation in order to reached a certain lower triangular matrix. If I could to this, how to get the value of s and c from the a and b.

For example is this matrix below : $$\begin{bmatrix} 5 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 \\ 3 & 0 & 2 & 0 \\ 2 & 0 & 3 & 0 \\ \end{bmatrix}$$

I want to make this matrix a lower triangular matrix.

If we want $$\begin{bmatrix} c & -s \\ s & c\end{bmatrix}\begin{bmatrix} a \\ b\end{bmatrix}=\begin{bmatrix} 0 \\ r\end{bmatrix}$$
$$\begin{bmatrix} a & -b \\ b & a\end{bmatrix}\begin{bmatrix} c \\ s\end{bmatrix}=\begin{bmatrix} 0 \\ r\end{bmatrix}$$
$$\begin{bmatrix} c \\ s\end{bmatrix}=\frac1{r^2}\begin{bmatrix} a & b \\ -b & a\end{bmatrix}\begin{bmatrix} 0 \\ r\end{bmatrix}=\frac1r\begin{bmatrix} b \\ a \end{bmatrix}$$