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$$A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1\\ 1 & 1 & -1 \end{bmatrix}$$

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?

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  • $\begingroup$ Note that if $A = QR$ with $Q$ an orthogonal matrix, then $R = Q^T A$. Similarly the inverse of an invertible lower triangular matrix $L$ is again lower triangular. $\endgroup$ – hardmath Apr 14 at 20:33
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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 $Q^{-1}A = Q^tA = R$.

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