Converting from general coordinates to standard coordinates: Consider these two ordered bases of \(\RNrSpc{3}\):
\[ \begin{array}{ccclcclccl} \EuScript{S} & \StdBssVec{1} & = & (1,0,0) & \StdBssVec{2} & = & (0,1,0) & \StdBssVec{3} & = & (0,0,1) \\ \EuScript{B} & \Vect{b}_1 & = & (2,-3,7) & \Vect{b}_2 & = & (-1,9,2) & \Vect{b}_3 & = & (4,2,-6) \end{array} \]Find the matrix \(\Mtrx{C}_{\EuScript{S}\EuScript{B}}\) which converts from \(\EuScript{B}\)-coordinates to \(\EuScript{S}\)-coordinates. Then find \(\Vect{x}_{\EuScript{S}}\) if \(\Vect{x}_{\EuScript{B}} = (2,0,3)\).
The column vectors of \(\Mtrx{C}_{\EuScript{S}\EuScript{B}}\) consist of the coordinate vectors of \(\Vect{b}_1\), \(\Vect{b}_2\), and \(\Vect{b}_3\) with respect to \(\EuScript{S}\). As \(\EuScript{S}\) is the standard basis of \(\RNrSpc{3}\), we have
| \((\Vect{b}_1)_{\EuScript{S}}\) | \(=\) | \(\Vect{b}_1 = (2,-3,7)\) |
| \((\Vect{b}_2)_{\EuScript{S}}\) | \(=\) | \(\Vect{b}_2 = (-1,9,2)\) |
| \((\Vect{b}_3)_{\EuScript{S}}\) | \(=\) | \(\Vect{b}_3 = (4,2,-6)\) |
Using these coordinate vectors as the column vectors of the coordinate conversion matrix \(\Mtrx{C}_{\EuScript{S}\EuScript{B}}\), we obtain
\[ \Mtrx{C}_{\EuScript{SB}}\ =\ \left[\begin{array}{rrr} 2 & -1 & 4 \\ -3 & 9 & 2 \\ 7 & 2 & -6 \end{array}\right] \]Consequently,
\[ \Vect{x}_{\EuScript{S}}\ =\ \left[\begin{array}{rrr} 2 & -1 & 4 \\ -3 & 9 & 2 \\ 7 & 2 & -6 \end{array}\right] \begin{bmatrix} 2 \\ 0 \\ 3 \end{bmatrix}\ =\ \left[\begin{array}{r} 16 \\ 0 \\ -4 \end{array}\right] \]That is \(\Vect{x}_{\EuScript{S}} = (16,0,-4)\).