I have this system of linear equations:
$$ A= \begin{bmatrix} 2 & 0 & 1 \\ 0 & 1 & -1 \\ 1 & 1 & 1 \end{bmatrix} $$ $$ b= \begin{bmatrix} 3 \\ 0 \\ 3 \end{bmatrix} $$
I computed it's QR decomposition using Givens rotation matrices (pardon for the coefficients, I multiplied the matrices using MATLAB):
$$ Q = \begin{bmatrix} 0.8944 & -0.2978 & -0.3332 \\ 0 & 0.7450 & -0.6660 \\ 0.4472 & 0.5957 & 0.6663 \end{bmatrix} $$
$$ R = \begin{bmatrix} 2.2361 & 0.4472 & 1.3416 \\ 0 & 1.3407 & -0.4472 \\ 0 & 0 & 0.9992 \end{bmatrix} $$
Now the way that I use to compute the x vector is to calculate the inverse of Q, which is Q transposed, then calculate the inverse of R. This last thing implies a lot of computation. I would like to know if once that I have the QR factorization, there is a fast method to compute the x vector without computing any inverse matrix, for example using back-substitution.
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No it doesn't. The way to solve $Rx=b$ is through backsubstitution. The last line gives you $0.9992x_3=3$. You get the value of $x_3$ then you you go to second last equation $1.3407x_2 -0.4472x_3=0$ You already know the value of $x_3$, you thus get the value of $x_2$ and you continue till you have everything. This is not more than a few lines of MATLAB code and is among the cheapest linear systems solve ever.