This abbreviated pset is due at 11am on Wed Oct 5. This is the last pset covered on exam 1.
Consider the equation $ABx=b$, given by: $$ \underbrace{\begin{pmatrix} 2 & 1 & 2 \\ & -1 & 2 \\ & & 1 \end{pmatrix}}_A \underbrace{\begin{pmatrix} -1 & 1 & 2 & 1 \\ 3 & -2 & -5 & -1 \\ 0 & 1 & 1 & 2 \end{pmatrix}}_B x = \underbrace{\begin{pmatrix} 17 \\ 19 \\ 7 \end{pmatrix}}_b $$
(a) Give bases for the nullspaces $N(A)$, $N(B)$, and $N(AB)$. (Hint: you should only have to do Gaussian elimination once.)
(b) Give the complete solution $x$ to $ABx=b$, and show how you can do this without every multiplying $A$ and $B$ together to explicitly form the matrix $AB$.
$Ax = b_1$ has no solution, and the complete solution to $Ax=b_2$ is: $$ x = \begin{pmatrix} 1\\2 \\3\\4 \end{pmatrix} + \alpha_1 \begin{pmatrix} 1\\1 \\0\\-1 \end{pmatrix} + \alpha_2 \begin{pmatrix} -2\\0 \\1\\-2 \end{pmatrix} $$ for all possible scalars $\alpha_1$ and $\alpha_2$.
Give one example of a possible matrix $A$ (with all nonzero entries) and possible right-hand sides $b_1$ and $b_2$ (with any entries you want) that are consistent with the above facts. (The same $A$ for both $b_1$ and $b_2$.)
Hint: start by thinking about possible sizes and ranks for $A$, and then think how the columns are/aren't linearly dependent.
Consider the vectors: $$ a_1 = \begin{pmatrix} 0\\0\\0\\0\\0 \end{pmatrix}, \qquad a_2 = \begin{pmatrix} 1\\2\\3\\4\\5 \end{pmatrix}, \qquad a_3 = \begin{pmatrix} 1\\1\\5\\4\\9 \end{pmatrix}, \qquad a_4 = \begin{pmatrix} 1\\3\\1\\4\\1 \end{pmatrix}. $$
Give a basis for the orthogonal complement of the subspace (of $\mathbb{R}^5$) spanned by:
(a) $a_1$
(b) $a_2$
(c) $a_1,a_2,a_3,\mbox{ and }a_4$.
(Hint: this is a basis for $C(\_\_)^\perp = \_\_\_\_$.)