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

Problems and solutions in Linear Algebra

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

\[b(a-2)=20.\]

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


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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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