Use the following vectors to answer parts (a) and (b). v1equals=[Start 3 By 1 Matrix 1st Row 1st Column 1 2nd Row 1st Column negative 4 3rd Row 1st Column 2 EndMatrix ]1 −4 2 , v2equals=[Start 3 By 1 Matrix 1st Row 1st Column negative 4 2nd Row 1st Column 16 3rd Row 1st Column negative 8 EndMatrix ]−4 16 −8 , v3equals=[Start 3 By 1 Matrix 1st Row 1st Column 5 2nd Row 1st Column 7 3rd Row 1st Column h EndMatrix ]5 7 h (a) For what values of h is v3 in Span{v1, v2}? (b) For what values of h is {v1, v2, v3} linearly dependent?

(2) The three vectors [tex]\vec{v}_1[/tex], [tex]\vec{v}_2[/tex], and [tex]\vec{v}_3[/tex] are always linearly dependent for all real [tex]h[/tex].

Step-by-step explanation:

(a)

If [tex]\vec{v}_3[/tex] is in the span of [tex]\vec{v}_1[/tex] and [tex]\vec{v}_2[/tex], there need to exist real [tex]a[/tex] and [tex]b[/tex] such that

[tex]a\; \vec{v}_1 + b\; \vec{v}_2 = \vec{v}_3[/tex].

Assume that such [tex]a[/tex] and [tex]b[/tex] do exist.

In other words,

[tex]\displaystyle a \left[\begin{array}{c}{1 \\ -4\\2} \end{array}\right] + b\left[\begin{array}{c}{-4 \\ 16\\-8}\end{array}\right] = \left[\begin{array}{c}5 \\7 \\ h\end{array}\right][/tex].

[tex]\displaystyle \left[\begin{array}{c}{a \\ -4a\\2a} \end{array}\right] + \left[\begin{array}{c}{-4b \\ 16b\\-8b}\end{array}\right] = \left[\begin{array}{c}5 \\7 \\ h\end{array}\right][/tex].

[tex]\left\{\begin{array}{rrcr} a & - 4b &=& 5\\ -4a & + 16b &= &7\\2a & -8b & =&h\end{aligned}\right.[/tex].

Rewrite as an augmented matrix and row-reduce:

[tex]\displaystyle \left[ \begin{array}{cc|c} 1 & -4 & 5 \\ -4 & 16 & 7 \\ 2 & -8 & h\end{array}\right][/tex].

(Add four times row one to row two and [tex]-2[/tex] times row one to row three.)

[tex]\displaystyle \sim \left[ \begin{array}{cc|c} 1 & -4 & 5 \\ 0 & 0 & 27 \\ 0 & 0 & h - 10\end{array}\right][/tex].

Note that in row two,

Left-hand side: [tex]0[/tex];
Right-hand side: [tex]27\neq 0[/tex].

In other words, this system is inconsistent. There's no real [tex]a[/tex] and [tex]b[/tex] that would satisfy the condition

[tex]a\; \vec{v}_1 + b\; \vec{v}_2 = \vec{v}_3[/tex].

Hence [tex]\forall h \in \mathbb{R}, \quad \vec{v}_3\not \in \text{Span}\{\vec{v}_1, \vec{v}_2\}[/tex].

There's no real [tex]h[/tex] that allows [tex]h[/tex], [tex]\vec{v}_3[/tex] to be part of the span of [tex]\vec{v}_1[/tex] and [tex]\vec{v}_2[/tex].

(b)

If the three vectors are linearly dependent, at least one of them can be expressed as the linear combination of the other two.

Note that

[tex]\vec{v}_2 = (-4)\vec{v}_1 + 0 \; \vec{v}_3[/tex]. In other words, [tex]\vec{v}_2[/tex] can be written as the linear combination of the other two vectors. Additionally, since the coefficient in front of [tex]\vec{v}_3[/tex] is zero, neither the exact value of [tex]\vec{v}_3[/tex] nor the value of [tex]h[/tex] will make a difference. Therefore, for all [tex]h \in \mathbb{R}[/tex], the three vectors [tex]\vec{v}_1[/tex], [tex]\vec{v}_2[/tex], and [tex]\vec{v}_3[/tex] are linearly dependent.