# Determine Conditions on Scalars so that the Set of Vectors is Linearly Dependent

## Problem 279

Determine conditions on the scalars $a, b$ so that the following set $S$ of vectors is linearly dependent.
\begin{align*}
S=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\},
\end{align*}
where
$\mathbf{v}_1=\begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix}, \mathbf{v}_2=\begin{bmatrix} 1 \\ a \\ 4 \end{bmatrix}, \mathbf{v}_3=\begin{bmatrix} 0 \\ 2 \\ b \end{bmatrix}.$

## Solution.

Let us consider the linear combination
$x_1\mathbf{v}_1+x_2\mathbf{v}_2+x_3\mathbf{v}_3=\mathbf{0}.$ We want to find conditions on $a, b$ so that there is $(x_1, x_2, x_3) \neq (0, 0, 0)$ satisfying this linear combination.
Since the linear combination can be written as
$A\mathbf{x}=\mathbf{0},$ where
$A=[\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3]=\begin{bmatrix} 1 & 1 & 0 \\ 3 &a &2 \\ 1 & 4 & b \end{bmatrix}, \text{ and } \mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix},$ our goal is the same as finding conditions on $a, b$ so that $A$ is a singular matrix.

We apply elementary row operations to the augmented matrix $[A\mid \mathbf{0}]$.
\begin{align*}
&[A\mid \mathbf{0}]= \left[\begin{array}{rrr|r}
1 & 1 & 0 & 0 \\
3 &a & 2 & 0 \\
1 & 4 & b & 0
\end{array} \right]\10pt] & \xrightarrow{\substack{R_2-3R_1\\R_3-R_1}} \left[\begin{array}{rrr|r} 1 & 1 & 0 & 0 \\ 0 &a-3 & 2 & 0 \\ 0 & 3 & b & 0 \end{array} \right] \xrightarrow{R_2\leftrightarrow R_3} \left[\begin{array}{rrr|r} 1 & 1 & 0 & 0 \\ 0 & 3 & b & 0 \\ 0 &a-3 & 2 & 0 \end{array} \right]\\[10pt] & \xrightarrow{\frac{1}{3}R_2} \left[\begin{array}{rrr|r} 1 & 1 & 0 & 0 \\ 0 & 1 & b/3 & 0 \\ 0 &a-3 & 2 & 0 \end{array} \right] \xrightarrow{R_3-(a-3)R_2} \left[\begin{array}{rrr|r} 1 & 1 & 0 & 0 \\ 0 & 1 & b/3 & 0 \\ 0 & 0 & 2-(a-3)b/3 & 0 \end{array} \right]. \end{align*} The last matrix is in echelon form. Let c=2-(a-3)b/3 be the number in (3,3)-entry. If c\neq 0, then it is easy to see that we can further reduce the matrix to \[ \left[\begin{array}{rrr|r} 1 & 0 & 0 & 0 \\ 0 &1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right] and the only solution is $x_1=x_2=x_3=0$.

On the other hand, if $c=0$, then the last row becomes a zero row and thus we see that $x_3$ is a free variable. Hence the system has a nonzero solution.

Therefore, the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent if and only if $c=0$.
Hence the condition we are looking for is
$c=2-(a-3)b/3=0,$ or equivalently,
$(a-3)b=6.$

## Related Question.

Problem.
Let
$\mathbf{v}_1=\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \mathbf{v}_2=\begin{bmatrix} 1 \\ a \\ 5 \end{bmatrix}, \mathbf{v}_3=\begin{bmatrix} 0 \\ 4 \\ b \end{bmatrix}$ be vectors in $\R^3$.

Determine a condition on the scalars $a, b$ so that the set of vectors $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is linearly dependent.

The solution is given in the post ↴
Determine a Condition on $a, b$ so that Vectors are Linearly Dependent

### 1 Response

1. 09/13/2017

[…] The solution is given in the post ↴ Determine Conditions on Scalars so that the Set of Vectors is Linearly Dependent […]

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