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\noindent
\parbox{2.25in}{\bf Math 5230}
\hfill {\Large \bf Homework \#3} \hfill
\parbox{2.25in}{\bf \hfill Due: Wed., Sept. 11${}^\mathbf{th}$\!, 2024}

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\begin{description}
\item[\#1 A $2 \times 2$ Determinant] Consider $A = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}$. Find $\det(A)$ in the following ways\dots
\begin{enumerate}[(a)]
   \item Use the usual formula.
   \item Use Gaussian elimination (i.e. the algorithm from class -- just use a forward pass).
   \item Compute only using the facts that the determinant is\\ \force\qquad (1) alternating,  (2) bilinear (bi = 2 for a $2 \times 2$ matrix), and (3) sends the identity matrix to 1.\\
   {}[{\it Note:} multilinear plus alternating imply skew.\\ \force\hfill You may use the skew property (switching rows/cols flips the sign) as well.] 
\end{enumerate}

\item[\#2 A $3 \times 3$ Determinant] Consider $A = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 1 & 4 \\ 2 & 1 & 2 \end{bmatrix}$. Find $\det(A)$ in the following ways\dots
\begin{enumerate}[(a)]
   \item Use the $3 \times 3$ trick.
   \item Expand along the first row.
   \item Expand along the second column.
   \item Use Gaussian elimination (i.e. the algorithm from class -- just use a forward pass)
   \item The same as part (d) except also use full pivoting.
   \item Find the inverse of this matrix using the classical adjoint formula.
\end{enumerate}

\item[\#3 Cramer's Rule] Solve $\begin{array}{ccccccc}
3x & + & 2y & + & 2z & = & 2 \\
x & + & 3y & + & 2z & = & 1 \\
2x & + & y & + & z & = & 1
 \end{array}$ using Cramer's rule.

\item[\#4 Reverse Engineering] As Seen in Class!

\begin{enumerate}[(a)]
   \item Create a $5 \times 5$ matrix that looks intimidating (make it interesting) even though its determinant is easy to compute when using the right row and column expansions. [Also, compute the determinant of your example using those expansions.]
   \item Create a $4 \times 4$ matrix whose entries are all non-zero integers and whose inverse is a matrix whose entries are fractions whose denominators are $4$'s, $2$'s, or $1$'s. [At least one entry of the inverse must be a reduced fraction of the form: $?/4$. It's ok if some entries of the inverse are zero.]
        
\end{enumerate}

\item[\#5 Linearly Dependent Gives Zero]  We know that a set of vectors is linearly dependent if and only if at least one vector can be written as a linear combination of the other vectors in that set. In fact from the linear correspondence property (and Gaussian elimination) we know that if some set $S=\{\B v_1,\dots, \B v_n \}$ is linearly dependent then there is an index $j$ and constants $c_1,\dots,c_{j-1}$ such that $\B v_j = c_1\B v_1 +\cdots +c_{j-1}\B v_{j-1}$ ($j$ would be correspond to some non-pivot column). 

Let $A=[ \B a_1 \;|\; \B a_2 \;|\; \cdots \;|\; \B a_n]$  (here $\B a_i$ is the $i^{\mathrm{th}}$ column of $A$). Show that $\det(A)=0$ if the columns of $A$ are linearly dependent. Do this only using facts about the effects of column operations on determinants.


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