by HallsofIvy » Fri Jun 26, 2020 12:39 pm
Didn't we seen this before?
Let [tex]u= (a, b, c)[/tex] Then [tex]Pu= \begin{bmatrix}1 & 2 & 1 \\ 1 & -1 & 2 \\ 2 & 2 & 3 \end{bmatrix}\begin{bmatrix}a \\ b \\ c \end{bmatrix}= \begin{bmatrix}a+ 2b+ c \\ a- b+ 2c \\ 2a+ 2b+ 3c\end{bmatrix}[/tex]. We are told that [tex]Pu= \begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}[/tex].
So we have the three equations a+ 2b+ c= 1, a- b+ c= 0, and 2a+ 2b+ 3c= 0. Solve those equations for a, b, and c. Do the same for v and w.