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Problems #13,24 on pg.229 are graded.
#13,pg.229: Determine the dimensions of $\mathrm{Nul} A$ and $\mathrm{Col} A$ for the matrix
$$A=\begin{bmatrix} 1&-6&9&0&-2 \\ 0&1&2&-4&5 \\ 0&0&0&5&1 \\ 0&0&0&0&0 \end{bmatrix}.$$
Solution: This matrix is in echelon form. If we compute the reduced echelon form we get
$$\begin{bmatrix} 1&-6&9&0&-2 \\ 0&1&2&-4&5 \\ 0&0&0&5&1 \\ 0&0&0&0&0 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 21 & 0 & \frac{164}{5} \\ 0&1&2&0&\frac{29}{5} \\ 0&0&0&1&\frac{1}{5} \\ 0&0&0&0&0 \end{bmatrix}.$$
Notice that this is the coefficient matrix of the matrix equation $A\vec{x}=\vec{0}$ (notice that $\vec{x}$ must be a $5\times 1$ vector; the augmented matrix adds a column of zeros on the right side of this matrix) and so the matrix equation $A\vec{x}=\vec{0}$ yields the following system of equations:
$$\left\{ \begin{array}{ll}
x_1 + 21x_3 + \dfrac{164}{5}x_5 = 0 \\
x_2+2x_3+\dfrac{29}{5}x_5 = 0 \\
x_4 + \dfrac{1}{5}x_5 = 0
\end{array} \right.$$
whose solution is therefore
$$\vec{x} = x_3 \begin{bmatrix} -21 \\ 2 \\ 1 \\ 0 \\ 0 \end{bmatrix}+ x_5 \begin{bmatrix} \frac{164}{5} \\ \frac{29}{5} \\ 0 \\ -\frac{1}{5} \\ 1 \end{bmatrix}.$$
We see that there are two free variables in the solution of $A \vec{x}=\vec{0}$ and therefore $\mathrm{dim \hspace{2pt} \mathrm{Nul}} A = 2$. From the original matrix $A$ we see that there are three pivot columns, therefore $\mathrm{dim \hspace{2pt} \mathrm{Col}} A=3$.
#24, pg.229: The first three Laguerre polynomials are $L_0(x)=1, L_1(x)=-t+1,$ and $L_2(x)=t^2-4t+2$. It is known that $\mathscr{B}=\{L_0,L_1,L_2\}$ forms a basis of $\mathbb{P}_2$. Find the coordinate vector of the vector $\vec{p}(t)=-2t^2+5t+5$ with respect to $\mathscr{B}$, taking $b_0=L_0, b_1=L_1, b_2=L_2$.
Solution: Our goal here is to compute the column vector $[\vec{p}]_{\mathscr{B}}$. This is done by finding the weights $c_0,\ldots,c_3$ such that
$$c_0 L_0 + c_1 L_1 + c_2 L_2 = \vec{p},$$
or in other words
$$c_0 + c_1(-t+1) + c_2(t^2-4t+2)=-2t^2+5t+5.$$
Simplify the left hand side to get
$$t^2 c_2 + t(-c_1 -4c_2) + (c_0 + c_1 +2c_2) = -2t^2 +5t + 5.$$
Equating coefficients yields
$$\left\{ \begin{array}{ll}
c_2 &= -2 \\
-c_1-4c_2 &= 5 \\
c_0+c_1+2c_2 &= 5
\end{array} \right.$$
yielding
$$\left\{ \begin{array}{ll}
c_2 &= -2 \\
c_1 &= 3 \\
c_0 &= 6.
\end{array} \right.$$
Hence
$$[\vec{p}]_{\mathbb{B}} = \begin{bmatrix} -2 \\ 3 \\ 6 \end{bmatrix}.$$
