# 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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