Question

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?

Answer 1

(1) No matter what's the value of
`h`,
`\vec{v}_3` is never in the span of
`\vec{v}_1` and
`\vec{v}_2`.

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

Explanation:

(a)

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

`a\; \vec{v}_1 + b\; \vec{v}_2 = \vec{v}_3`.

Assume that such
`a` and
`b` do exist.

In other words,

`\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]`.

`\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]`.

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

Rewrite as an augmented matrix and row-reduce:

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

(Add four times row one to row two and
`-2` times row one to row three.)

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

Note that in row two,

• Left-hand side:
  `0`;
• Right-hand side:
  `27\\eq 0`.

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

`a\; \vec{v}_1 + b\; \vec{v}_2 = \vec{v}_3`.

Hence
`\forall h \in \mathbb{R}, \quad \vec{v}_3\\ot \in \text{Span}\{\vec{v}_1, \vec{v}_2\}`.

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

(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

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