# Transform $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1\\ 1 & 1 & -1 \end{bmatrix}$ into a superior triangular matrix by left multiplication

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

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 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. – hardmath Apr 14 at 20:33

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 $$Q^{-1}A = Q^tA = R$$.