# Determine a Condition on $a, b$ so that Vectors are Linearly Dependent

## Problem 563

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.

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## Solution.

Consider the equation

\[x_1\mathbf{v}_1+x_2\mathbf{v}_2+x_3\mathbf{v}_3=\mathbf{0},\]
where $\mathbf{0}$ is the three-dimensional zero vector.

Our goal is to find a condition on $a, b$ so that the above equation has a nontrivial solution $x_1, x_2, x_3$.

This equation is equivalent to the $3\times 3$ homogeneous system of linear equations

\[\begin{bmatrix}

1 & 1 & 0 \\

2 &a &4 \\

0 & 5 & b

\end{bmatrix}\begin{bmatrix}

x_1 \\

x_2 \\

x_3

\end{bmatrix}=\mathbf{0}.\]

We solve the system by Gauss-Jordan elimination.

The augmented matrix of the system is

\[\left[\begin{array}{rrr|r}

1 & 1 & 0 & 0 \\

2 &a & 4 & 0 \\

0 & 5 & b & 0

\end{array} \right].\]

Applying elementary row operations, we have

\begin{align*}

\left[\begin{array}{rrr|r}

1 & 1 & 0 & 0 \\

2 &a & 4 & 0 \\

0 & 5 & b & 0

\end{array} \right]
\xrightarrow{R_2-2R_1}

\left[\begin{array}{rrr|r}

1 & 1 & 0 & 0 \\

0 &a-2 & 4 & 0 \\

0 & 5 & b & 0

\end{array} \right]
\xrightarrow{\frac{1}{5}R_3}\\[6pt]
\left[\begin{array}{rrr|r}

1 & 1 & 0 & 0 \\

0 &a-2 & 4 & 0 \\

0 & 1 & b/5 & 0

\end{array} \right]
\xrightarrow{R_2\leftrightarrow R_3}

\left[\begin{array}{rrr|r}

1 & 1 & 0 & 0 \\

0 & 1 & b/5 & 0 \\

0 &a-2 & 4 & 0 \\

\end{array} \right] \\[6pt]
\xrightarrow{R_3-(a-2)R_2}

\left[\begin{array}{rrr|r}

1 & 1 & 0 & 0 \\

0 & 1 & b/5 & 0 \\

0 &0 & 4-\frac{b(a-2)}{5} & 0 \\

\end{array} \right]
\end{align*}

If the $(3,3)$ entry $4-\frac{b(a-2)}{5}$ of the last matrix is $4-\frac{b(a-2)}{5}$ is zero, then we obtain a matrix in echelon form

\[ \left[\begin{array}{rrr|r}

1 & 1 & 0 & 0 \\

0 & 1 & b/5 & 0 \\

0 &0 & 0 & 0 \\

\end{array} \right].\]
This implies that $x_3$ is a free variable, hence the homogeneous system has a nonzero solution $x_1, x_2, x_3$. Hence in this case the set $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is linearly dependent.

On the other hand, if $4-\frac{b(a-2)}{5}\neq 0$, we divide the third row by this number and obtain

\[ \left[\begin{array}{rrr|r}

1 & 1 & 0 & 0 \\

0 & 1 & b/5 & 0 \\

0 &0 & 1 & 0 \\

\end{array} \right],\]
and from this we see that the solution is $x_1=x_2=x_3=0$.

Thus in this case the set $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is linearly independent.

In conclusion, the set $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is linearly dependent if and only if $4-\frac{b(a-2)}{5}= 0$.

Thus, the condition on $a, b$ is

## Related Question.

**Problem**.

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}.\]

The solution is given in the post ↴

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

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