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Problems #5,13 pg.222 are graded.
#5, pg.222: Find the coordinate vector $[\vec{x}]_{\mathscr{B}}$ of $\vec{x}=\begin{bmatrix} -1 \\ 1 \end{bmatrix}$ relative to the basis $\mathscr{B}=\left\{ \begin{bmatrix} 1\\-2 \end{bmatrix}, \begin{bmatrix} 3 \\ -5 \end{bmatrix} \right\}$.
Solution: The coordinates of $\vec{x}$ are the weights $c_1,c_2$ which are solutions to the vector equation
$$c_1 \begin{bmatrix} 1 \\ -2 \end{bmatrix} + c_2 \begin{bmatrix} 3 \\ -5 \end{bmatrix} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}.$$
We can solve this vector equation by row reducing the appropriate augmented matrix, so compute
$$\begin{array}{ll}
\begin{bmatrix} 1 & 3 & -1 \\ -2 & -5 & 1 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \end{bmatrix}.
\end{array}$$
Hence we see that $c_1=2$ and $c_2=-1$. This implies that the coordinate vector in question is given by
$$[\vec{x}]_{\mathscr{B}} = \begin{bmatrix} 2 \\ -1 \end{bmatrix}.$$
#13, pg.222: The set $\mathscr{B} = \{1+t^2,t+t^2, 1+2t+t^2\}$ is a basis for $\mathbb{P}_2$ (polynomials of degree $\leq$ 2). Find the coordinate vector of $\vec{p}(t)=1+4t+7t^2$ relative to $\mathscr{B}$.
Solution: Our goal is to compute $[\vec{p}]_{\mathscr{B}} = [1+4t+7t^2]_{\mathscr{B}}$ (both ways of notating this is are ok). This means we have to solve the following vector equation for the weights $c_1,c_2,c_3$:
$$c_1(1+t^2) + c_2(t+t^2) + c_3(1+2t+t^2)=1+4t+7t^2.$$
Some algebraic rearrangement shows that this equation is equivalent to
$$(c_1+c_2+c_3)t^2 + (c_2 + 2c_3)t + (c_1 + c_3)=1+4t+7t^2.$$
Equating coefficients here leads us to the following system of equations:
$$\left\{ \begin{array}{ll}
c_1+c_2+c_3 &= 7 \\
c_2+2c_3 &= 4 \\
c_1 + c_3 &= 1.
\end{array} \right.$$
This system can be solved by row reducing the associated augmented matrix, so compute
$$\begin{bmatrix} 1&1&1&7 \\ 0&1&2&4 \\ 1&0&1&1 \end{bmatrix} \sim \begin{bmatrix} 1&0&0&2 \\ 0&1&0&6 \\ 0&0&1&-1 \end{bmatrix}.$$
This shows that $c_1=2, c_2=6, c_3=-1$ and hence
$$[1+4t+7t^2]_{\mathscr{B}} = \begin{bmatrix} 2 \\ 6 \\ -1 \end{bmatrix}.$$
