# 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}.\]

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## 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

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