Determine Linearly Independent or Linearly Dependent. Express as a Linear Combination
Problem 277
Determine whether the following set of vectors is linearly independent or linearly dependent. If the set is linearly dependent, express one vector in the set as a linear combination of the others.
\[\left\{\, \begin{bmatrix}
1 \\
0 \\
-1 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
2 \\
3 \\
4
\end{bmatrix}, \begin{bmatrix}
-1 \\
-2 \\
0 \\
1
\end{bmatrix},
\begin{bmatrix}
-2 \\
-2 \\
7 \\
11
\end{bmatrix}\, \right\}.\]
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Solution.
Consider the linear combination
\[x_1\begin{bmatrix}
1 \\
0 \\
-1 \\
0
\end{bmatrix}+x_2 \begin{bmatrix}
1 \\
2 \\
3 \\
4
\end{bmatrix}+x_3\begin{bmatrix}
-1 \\
-2 \\
0 \\
1
\end{bmatrix}+x_4
\begin{bmatrix}
-2 \\
-2 \\
7 \\
11
\end{bmatrix}=\mathbf{0} \tag{*}\]
with variables $x_1, x_2, x_3, x_4$.
We determine whether there is $(x_1, x_2, x_3, x_4)\neq (0,0,0,0)$ satisfying the linear combination (*).
The linear combination (*) is written as the matrix equation
\[\begin{bmatrix}
1 & 1 & -1 & -2 \\
0 &2 & -2 & -2 \\
-1 & 3 & 0 & 7 \\
0 & 4 & 1 & 11
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}=\mathbf{0}.\]
To find the solutions of the equation, we apply the Gauss-Jordan elimination.
The augmented matrix $[A\mid \mathbf{0}]$ can be reduced by elementary row operations as follows.
\begin{align*}
[A\mid \mathbf{0}]= \left[\begin{array}{rrrr|r}
1 & 1 & -1 & -2 & 0\\
0 &2 & -2 & -2 & 0 \\
-1 & 3 & 0 & 7& 0 \\
0 & 4 & 1 & 11& 0 \\
\end{array} \right]
\xrightarrow{\substack{R_3+R_1 \\ \frac{1}{2}R_2}}
\left[\begin{array}{rrrr|r}
1 & 1 & -1 & -2 & 0\\
0 & 1 & -1 & -1 & 0 \\
0 & 4 & -1 & 5& 0 \\
0 & 4 & 1 & 11& 0 \\
\end{array} \right]\\[10pt]
\xrightarrow{\substack{R_1-R_2\\ R_3-4R_2\\ R_4-4R_2}}
\left[\begin{array}{rrrr|r}
1 & 0 & 0 & -1 & 0\\
0 & 1 & -1 & -1 & 0 \\
0 & 0 & 3 & 9& 0 \\
0 & 0 & 5 & 15 & 0 \\
\end{array} \right]
\xrightarrow{\substack{\frac{1}{3}R_3\\ \frac{1}{5}R_4}}
\left[\begin{array}{rrrr|r}
1 & 0 & 0 & -1 & 0\\
0 & 1 & -1 & -1 & 0 \\
0 & 0 & 1 & 3 & 0 \\
0 & 0 & 1 & 3 & 0 \\
\end{array} \right]\\[10pt]
\xrightarrow{\substack{R_2+R_3\\R_4-R_3}}
\left[\begin{array}{rrrr|r}
1 & 0 & 0 & -1 & 0\\
0 & 1 & 0 & 2 & 0 \\
0 & 0 & 1 & 3 & 0 \\
0 & 0 & 0 & 0 & 0 \\
\end{array} \right].
\end{align*}
Thus, the general solution is given by
\begin{align*}
x_1&=x_4\\
x_2&=-2x_4\\
x_3&=-3x_4,
\end{align*}
where $x_4$ is a free variable.
If we let $x_4=1$, then we have a nonzero solution $x_1=1, x_2=-2, x_3=-3, x_4=1$.
Thus the set is linearly dependent.
Substituting these values into (*), we have
\[\begin{bmatrix}
1 \\
0 \\
-1 \\
0
\end{bmatrix}-2 \begin{bmatrix}
1 \\
2 \\
3 \\
4
\end{bmatrix}-3\begin{bmatrix}
-1 \\
-2 \\
0 \\
1
\end{bmatrix}+
\begin{bmatrix}
-2 \\
-2 \\
7 \\
11
\end{bmatrix}=\mathbf{0}.\]
Solving this for the last vector we obtain the linear combination
\[ \begin{bmatrix}
-2 \\
-2 \\
7 \\
11
\end{bmatrix}
=
-\begin{bmatrix}
1 \\
0 \\
-1 \\
0
\end{bmatrix}+2 \begin{bmatrix}
1 \\
2 \\
3 \\
4
\end{bmatrix}+3\begin{bmatrix}
-1 \\
-2 \\
0 \\
1
\end{bmatrix}.\]
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