# 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.

## 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

$b(a-2)=20.$

## 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

### 1 Response

1. 09/13/2017

[…] The solution is given in the post ↴ Determine a Condition on $a$ so that Vectors are Linearly Dependent […]

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