--- Commentary: $\textbf{Exercise 6.}$ Show that the list $(2, 3, 1)$, $(1, -1, 2)$, $(7, 3, c)$ is linearly dependent in $\mathbb{F}^3$ if and only if $c = 8$. $\textbf{Solution 6.}$ The equation in the first bullet point in Example 2.18 shows that $(2, 3, 1)$, $(1, -1, 2)$, $(7, 3, 8)$ is linearly dependent. Conversely, suppose $(2, 3, 1)$, $(1, -1, 2)$, $(7, 3, c)$ is linearly dependent. The first vector in this list is not the $0$ vector, and the second vector in this list is not a scalar multiple of the first vector. Thus by the linear dependence lemma (2.19), the vector $(7, 3, c)$ is in the span of $(2, 3, 1)$, $(1, -1, 2)$. The list $(2, 3)$, $(1, -1)$ is linearly independent (neither vector is a scalar multiple of the other) and thus $(7, 3)$ can be written as a linear combination of $(2, 3)$, $(1, -1)$ in at most one way. From the first bullet point in 2.18, we see that this one way of writing $(7, 3)$ as a linear combination of $(2, 3)$, $(1, -1)$ is $2(2, 3) + 3(1, -1) = (7, 3).$ Thus the only possible way to write $(7, 3, c)$ as a linear combination of the vectors $(2, 3, 1)$, $(1, -1, 2)$ is $2(2, 3, 1) + 3(1, -1, 2) = (7, 3, c).$ The equation above implies that the list $(2, 3, 1)$, $(1, -1, 2)$, $(7, 3, c)$ is linearly dependent only when $c = 8$. $\textit{Commentary:}$ This exercise provides practice in applying the concepts of linear dependence and the linear dependence lemma. The forward direction is straightforward from Example 2.18. The reverse direction uses the linear dependence lemma to conclude that $(7, 3, c)$ must be in the span of the other two vectors. Then, the unique representation of $(7, 3)$ as a linear combination of $(2, 3)$ and $(1, -1)$ is used to determine the only possible value of $c$. This problem also illustrates the interplay between the algebraic and geometric aspects of linear dependence. The condition $c = 8$ can be interpreted as saying that $(7, 3, c)$ must lie in the plane spanned by $(2, 3, 1)$ and $(1, -1, 2)$ for the three vectors to be linearly dependent. $\textit{Example:}$ In $\mathbb{Q}_p^3$, the list $(1, p, p^2)$, $(p, p^2, 1)$, $(p^2, 1, c)$ is linearly dependent if and only if $c = p$. This is because $(p^2, 1, p) = p(1, p, p^2) + (p, p^2, 1)$. In $(\mathbb{Z}/5\mathbb{Z})^3$, the list $(1, 2, 3)$, $(2, 4, 1)$, $(3, 1, c)$ is linearly dependent if and only if $c = 4$. This is because $3(1, 2, 3) + 2(2, 4, 1) = (3, 1, 4)$ in $\mathbb{Z}/5\mathbb{Z}$. In the space of polynomials over $\mathbb{F}_2$, the list $1 + x$, $1 + x + x^2$, $1 + ax + x^2$ is linearly dependent if and only if $a = 0$. This is because $(1 + x) + (1 + x + x^2) = (1 + 0x + x^2)$ in $\mathbb{F}_2[x]$. These examples demonstrate that the concept of linear dependence, and the techniques for analyzing it, are applicable over any field. The specific arithmetic of the field can change the details of the calculations, but the overall principles remain the same. --- $\textbf{Exercise 6.}$ Show that the list $(2, 3, 1)$, $(1, -1, 2)$, $(7, 3, c)$ is linearly dependent in $\mathbb{F}^3$ if and only if $c = 8$. $\textbf{Solution 6.}$ The equation in the first bullet point in Example 2.18 shows that $(2, 3, 1)$, $(1, -1, 2)$, $(7, 3, 8)$ is linearly dependent. Conversely, suppose $(2, 3, 1)$, $(1, -1, 2)$, $(7, 3, c)$ is linearly dependent. The first vector in this list is not the 0 vector, and the second vector in this list is not a scalar multiple of the first vector. Thus by the linear dependence lemma (2.19), the vector $(7, 3, c)$ is in the span of $(2, 3, 1)$, $(1, -1, 2)$. The list $(2, 3)$, $(1, -1)$ is linearly independent (neither vector is a scalar multiple of the other) and thus $(7, 3)$ can be written as a linear combination of $(2, 3)$, $(1, -1)$ in at most one way. From the first bullet point in 2.18, we see that this one way of writing $(7, 3)$ as a linear combination of $(2, 3)$, $(1, -1)$ is $2(2, 3) + 3(1, -1) = (7, 3).$ Thus the only possible way to write $(7, 3, c)$ as a linear combination of the vectors $(2, 3, 1)$, $(1, -1, 2)$ is $2(2, 3, 1) + 3(1, -1, 2) = (7, 3, c).$ The equation above implies that the list $(2, 3, 1)$, $(1, -1, 2)$, $(7, 3, c)$ is linearly dependent only when $c = 8$. $\textit{Commentary:}$ This exercise provides practice in applying the concepts of linear dependence and the linear dependence lemma. The forward direction is straightforward from Example 2.18. The reverse direction uses the linear dependence lemma to conclude that $(7, 3, c)$ must be in the span of the other two vectors. Then, the unique representation of $(7, 3)$ as a linear combination of $(2, 3)$ and $(1, -1)$ is used to determine the only possible value of $c$. This problem also illustrates the interplay between the algebraic and geometric aspects of linear dependence. The condition $c = 8$ can be interpreted as saying that $(7, 3, c)$ must lie in the plane spanned by $(2, 3, 1)$ and $(1, -1, 2)$ for the three vectors to be linearly dependent. $\textit{Additional Examples:}$ In $\mathbb{Q}_p^3$, the list $(1, p, p^2)$, $(p, p^2, 1)$, $(p^2, 1, c)$ is linearly dependent if and only if $c = p$. This is because $(p^2, 1, p) = p(1, p, p^2) + (p, p^2, 1)$. In $(\mathbb{Z}/5\mathbb{Z})^3$, the list $(1, 2, 3)$, $(2, 4, 1)$, $(3, 1, c)$ is linearly dependent if and only if $c = 4$. This is because $3(1, 2, 3) + 2(2, 4, 1) = (3, 1, 4)$ in $\mathbb{Z}/5\mathbb{Z}$. In the space of polynomials over $\mathbb{F}_2$, the list $1 + x$, $1 + x + x^2$, $1 + ax + x^2$ is linearly dependent if and only if $a = 0$. This is because $(1 + x) + (1 + x + x^2) = (1 + 0x + x^2)$ in $\mathbb{F}_2[x]$. In the space of continuous functions $C([0,1], \mathbb{R})$, the list $1$, $x$, $cx^2$ is linearly dependent if and only if $c = 0$. This is because there are no non-trivial solutions to $a + bx = cx^2$ for continuous functions on $[0,1]$ unless $c = 0$. These examples demonstrate that the concept of linear dependence, and the techniques for analyzing it, are applicable over any field. The specific arithmetic of the field can change the details of the calculations, but the overall principles remain the same.