$$A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1\\ 1 & 1 & -1 \end{bmatrix}$$
This matrix can be transformed into a superior trangiular matrix through left multiplication by a lower triangular matrix $L$ or by an orthogonal matrix $Q$. Find the matrix $L$ and the matrix $Q$. Solve $Ax= b$ with $b=\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix}$
What I know how to do is to make $A = LU$ and $A=QR$ which are the known LU and QR decompositions. However, this exercise asks me to left multiply by $L$ and left multiply by $Q$ to obtain a superior triangular matrix. What am I missing?
Note that the inverse of an upper(lower) triangular matrix - if it exists - is again an upper(lower) triangular matrix. So if $A = LU$ we have $L^{-1}A = U$ and similarly for $A = QR$ we have $Q^{-1}A = Q^tA = R$.