Question

Vectors$\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right],\mathit{}\left[\begin{array}{c}5\\ 6\\ 7\\ 8\end{array}\right],\mathit{}\left[\begin{array}{c}9\\ 8\\ 7\\ 6\end{array}\right],\mathit{}\left[\begin{array}{c}5\\ 4\\ 3\\ 2\end{array}\right]\mathit{}\mathit{a}\mathit{n}\mathit{d}\mathit{}\left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right]$are linearly independent.

Short answer

The above statement is false.

The given vectors$\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right],\mathit{}\left[\begin{array}{c}5\\ 6\\ 7\\ 8\end{array}\right],\mathit{}\left[\begin{array}{c}9\\ 8\\ 7\\ 6\end{array}\right],\mathit{}\left[\begin{array}{c}5\\ 4\\ 3\\ 2\end{array}\right]\mathit{}\mathit{a}\mathit{n}\mathit{d}\mathit{}\left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right]$ are linearly dependent, because, the system, in echelon form, has two equations in four unknowns; hence the system has a non-zero solution. Thus the given vectors are linearly dependent.

Step-by-step solution

Definition of linearly dependent set

Let V be a vector space over field K. The vectors ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,...,{\overrightarrow{v}}_n$in V are said to be linearly dependent over field K if there exist scalars$a_1,a_2,...,a_n\in K$ , not all of them 0, such that

$a_1{\overrightarrow{v}}_1+a_2{\overrightarrow{v}}_2+...+a_n{\overrightarrow{v}}_n=0$

Finding the scalars a1,a2,...,an

Let$a_1,a_2,a_3,a_4\text{and}{a}_5$ are scalars and

${\overrightarrow{v}}_1\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}5\\ 6\\ 7\\ 8\end{array}\right],{\overrightarrow{v}}_3=\left[\begin{array}{c}9\\ 8\\ 7\\ 6\end{array}\right],{\overrightarrow{v}}_4=\left[\begin{array}{c}5\\ 4\\ 3\\ 2\end{array}\right]\text{and}{\overrightarrow{v}}_5=\left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right]$

are the given vectors. Then we have

$a_1\left(1,2,3,4\right)+a_2\left(5,6,7,5\right)+a_3\left(9,8,7,6\right)+a_4\left(5.4,3,2\right)+a_5\left(1,0,-1,2\right)=\left(0,0,0,0\right)$

Checking linear dependency/independency of a given system of vectors

The coefficient matrix of the above equation is-

Let$A=\left[\begin{array}{ccccc}1& 5& 9& 5& 1\\ 2& 6& 8& 4& 0\\ 3& 7& 7& 3& -1\\ 4& 8& 6& 2& -2\end{array}\right]$

Applying row operations:

$$\begin{gathered} R_2\to 2R_1-R_2 \\ {R}_3\to 3R_1-R_3 \\ {R}_4\to 4R_1-R_4 \end{gathered}$$

We get,

$A=\left[\begin{array}{ccccc}1& 5& 9& 5& 1\\ 0& 4& 10& 6& 2\\ 0& 8& 20& 12& 4\\ 0& 12& 30& 18& 6\end{array}\right]$

Applying row operations:

$$\begin{gathered} R_3\to 2R_2-R_3 \\ {R}_4\to 3R_2-R_4 \end{gathered}$$

We get,

$A=\left[\begin{array}{ccccc}1& 5& 9& 5& 1\\ 0& 4& 10& 6& 2\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\end{array}\right]$

Number of non-zero solutions

As there are two linearly independent rows in A, therefore rank(A) = 2 and the number of unknowns is 5.

As, rank(A) < number of unknowns, therefore, the system of equations will have 4 – 2 = 2 non-zero solutions, and hence the set of vectors is linearly dependent.

Final answer

The given vectors $\left[\begin{array}{c}1\\ 2\\ 3\\ 4\end{array}\right],\left[\begin{array}{c}5\\ 6\\ 7\\ 8\end{array}\right],\left[\begin{array}{c}9\\ 8\\ 7\\ 6\end{array}\right],\left[\begin{array}{c}5\\ 4\\ 3\\ 2\end{array}\right]\text{and}\left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right]$are linearly dependent because, the system, in echelon form, has two equations in four unknowns. Hence, the system has non-zero solutions, and therefore, is linearly dependent.