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Problems #7 and #14 on pg.222-223 are graded.

#7, pg.222: Find the coordinate vector $[\vec{x}]_{\mathscr{B}}$ of $\vec{x}$ relative to the given basis $\mathscr{B}=\{\vec{b}_1,\vec{b}_2,\vec{b}_3\}$ where $$\vec{x}=\begin{bmatrix} 8 \\ -9 \\ 6 \end{bmatrix},$$ and $$\vec{b}_1=\begin{bmatrix} 1 \\ -1 \\ -3 \end{bmatrix}, \vec{b}_2 = \begin{bmatrix} -3 \\ 4 \\ 9 \end{bmatrix}, \vec{b}_3 = \begin{bmatrix} 2 \\ -2 \\ 4 \end{bmatrix}.$$

Solution: We must find the weights $c_1,c_2,c_3$ that solve $$c_1 \vec{b}_1 + c_2\vec{b}_2 + c_3\vec{b}_3 = \vec{x},$$ i.e. we must solve $$c_1 \begin{bmatrix} 1 \\ -1 \\ -3 \end{bmatrix} + c_2 \begin{bmatrix} -3 \\ 4 \\ 9 \end{bmatrix} + c_3 \begin{bmatrix} 2 \\ -2 \\ 4 \end{bmatrix} = \begin{bmatrix} 8 \\ -9 \\ 6 \end{bmatrix}.$$ Set up the augmented matrix and compute its reduced echelon form to get $$\begin{bmatrix} 1 & -3 & 2 & 8 \\ -1&4&-2&-9 \\ -3&9&4&6 \end{bmatrix} \sim \begin{bmatrix} 1&0&0&-1 \\ 0&1&0&-1 \\ 0&0&1&3 \end{bmatrix}.$$ This yields solution $c_1=-1, c_2=-1, c_3=3$. Thus we have derived the relation $$(-1) \begin{bmatrix} 1 \\ -1 \\ -3 \end{bmatrix} + (-1) \begin{bmatrix} -3 \\ 4 \\ 9 \end{bmatrix} + 3 \begin{bmatrix} 2 \\ -2 \\ 4 \end{bmatrix} = \begin{bmatrix} 8 \\ -9 \\ 6 \end{bmatrix},$$ which can be observed to be true. Therefore by the defition of coordinate vectors we see $$[\vec{x}]_{\mathscr{B}} = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \\ 3 \end{bmatrix}.$$

#14,pg.223: The set $\mathscr{B}=\{1-t^2,t-t^2,2-t+t^2\}$ is a basis for $\mathbb{P}_2$. Find the coordinate vector of $\vec{p}(t)=1+3t-6t^2$ relative to $\mathscr{B}$.

Solution: We write $\vec{b}_1=1-t^2, \vec{b}_2=t-t^2,$ and $\vec{b}_3=2-t+t^2$. We must find weights $c_1,c_2,c_3$ such that $$c_1 \vec{b}_1 + c_2\vec{b}_2 + c_3\vec{b}_3 = \vec{p},$$ or in other words $$c_1 (1-t^2) + c_2 (t-t^2) + c_3(2-t+t^2) = 1+3t-6t^2.$$ Simplify the left hand side by combining like terms to get $$(-c_1-c_2+c_3)t^2+(c_2-c_3)t+(c_1+2c_3) = 1+3t-6t^2.$$ Equating coefficients leads to the following system of equations: $$\left\{\begin{array}{ll} -c_1-c_2+c_3 &= -6 \\ c_2-c_3 &= 3 \\ c_1+2c_3 &= 1 \end{array}\right..$$ This system can be solved by computing the reduced echelon form of the augmented matrix: $$\begin{bmatrix} -1 & -1 & 1 & -6 \\ 0&1&-1&3 \\ 1&0&2&1 \end{bmatrix} \sim \begin{bmatrix} 1 &0&0&3 \\ 0&1&0&2 \\ 0&0&1&-1 \end{bmatrix}.$$ Therefore we see $$[\vec{p}]_{\mathscr{B}} = \begin{bmatrix} 3\\ 2 \\ -1 \end{bmatrix}.$$