User:Egm6321.f11.team2.Xia/RP7.9

Given
Consider the non-orthonormal basis $$ \{\mathbf{b}_{1},\mathbf{b}_{2},\mathbf{b}_{3}\}$$, expressed in the orthonomal basis $$ \{\mathbf{e}_{1},\mathbf{e}_{2},\mathbf{e}_{3}\}$$ as folllows: $$\displaystyle \mathbf{b}_{i}=A_{ij}\mathbf{e}_{j}$$ (7.3.2) here $$\displaystyle \mathbf{A}=\left[A_{ij}\right]=\left[\begin{array}{ccc} 5 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 5\end{array}\right]$$ and $$\displaystyle \mathbf{v}=-2\mathbf{e}_{1}+4\mathbf{e}_{2}-5\mathbf{e}_{3}$$ (7.3.3)

Find
$$\displaystyle{\mathbf v_i}\in\mathbb{R}^{3\times 1}$$ such that $$\displaystyle \mathbf{v}=v_i\mathbf b_i$$

Solution
From 7.3.2 and 7.3.3, we have $$\displaystyle \mathbf{v}=v_{i}A_{ij}\mathbf{e}_{j}$$ (7.3.4) $$\displaystyle \mathbf{\mathbf{A}}^{T}\left[\begin{array}{c} v_{1}\\ v_{2}\\ v_{3}\end{array}\right]=\left[\begin{array}{r} -2\\ 4\\ -5\end{array}\right]$$ (7.3.5) $$\displaystyle \left[A_{ij}\right]^{T}\left[\begin{array}{c} v_{1}\\ v_{2}\\ v_{3}\end{array}\right]=\left[\begin{array}{r} -2\\ 4\\ -5\end{array}\right]$$ (7.3.6) Since $$\displaystyle det(\mathbf{A})\neq 0$$

we have $$\displaystyle \left[\begin{array}{c} v_{1}\\ v_{2}\\ v_{3}\end{array}\right]=\left[A_{ij}\right]^{-T}\left[\begin{array}{r} -2\\ 4\\ -5\end{array}\right]=\left[\begin{array}{ccc} \frac{23}{80} & -\frac{11}{40} & \frac{3}{80}\\ -\frac{7}{40} & -\frac{1}{20} & \frac{13}{40}\\ \frac{3}{80} & \frac{9}{40} & -\frac{17}{80}\end{array}\right]\left[\begin{array}{r} -2\\ 4\\ -5\end{array}\right]$$ (7.3.7) thus $$\displaystyle \left[\begin{array}{c} v_{1}\\ v_{2}\\ v_{3}\end{array}\right]=\left[\begin{array}{r} -\frac{149}{80}\\ -\frac{59}{40}\\ \frac{151}{80}\end{array}\right]$$ (7.3.8)