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Linear Algebra 2270-2 \\
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Due in Week 3
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The third week completes work from sections 1.4 and 1.5. Here's the
list of problems, followed by problem notes and a few answers.

\begin{description}
\item[{\bf Section 1.4}.] Exercises 6, 8, 17, 23, 27, 31, 32.

\item[{\bf Section 1.5}.] Exercises 7, 8, 10, 29, 31, 33, 35.

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\subsection*{Some Answers and Hints}

Exercise 1.4-6: On the left side of the given matrix equation $A\vec{x}=\vec{b}$, use the entries $-2$, $-5$ in the vector $\vec{x}$ as the weights in a linear
combination of the columns of the matrix $A$.

Exercise 1.4-8: The answer is a matrix equation $A\vec{x}=\vec{b}$, for which $A$, $\vec{x}, \vec{b}$ are given explicit definitions. The left side of the given vector equation is a linear combination of four vectors. Write the matrix $A$ whose
columns are those four vectors, and create a variable vector $\vec{z}$ with four entries $z_1,z_2,z_3,z_4$. The vector $\vec{b}$ is the column vector on the right side of the given vector equation.

Exercise 1.4-32: A set of three vectors in $\Rr^4$ cannot span $\Rr^4$. Reason: the matrix $A$ whose columns are these three
vectors has four rows. To have a pivot in each row, $A$ would have to have at least four columns (one
for each pivot), which is not the case. Since $A$ does not have a pivot in every row, its columns do not
span $\Rr^4$ , by Theorem 4. In general, a set of $n$ vectors in $\Rr^m$ cannot span $\Rr^m$ when $n$ is less than $m$.

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