Answer
a. False.
b. False.
c. True.
d. True.
Work Step by Step
a. Every homogeneous matrix equation has the trivial solution, but not every matrix has linearly independent columns.
b. According to the warning beneath Theorem 7, at least one vector in a linearly dependent set must be a linear combination of the others, but it need not be the case that every vector in the set is a linear combination of the others.
c. Since the matrix contains more columns than rows, the vectors formed from the columns must be linearly dependent by Theorem 8.
d. Since $\vec{x}$ and $\vec{y}$ are linearly independent, neither can be the zero vector, per Theorem 9. When $\vec{z}$ is included in the set, the set becomes linearly dependent, so by Theorem 7, it must be that one of the vectors in $\{\vec{x},\vec{y},\vec{z}\}$ is a linear combination of the preceding vectors. Since $\vec{y}$ is not a multiple of $\vec{x}$, it must be that $\vec{z}$ is a linear combination of $\vec{x}$ and $\vec{y}$. Hence, by the definition of span, $\vec{z}$ is in $Span\{\vec{x},\vec{y}\}$. This line of reasoning directly parallels that given in Example 4.
