Answer to Question #278119 in Linear Algebra for Mayuri

Question

Investigate the linear dependence and independence of the vector X1=(1,2,3), X2=(2,-1,3), X3=(0,1,2), X4=(-3,7,2)

Answer 1

We are tasked with finding a nonzero solution $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ to the equation:

 $a_{1} x_{1}+a_{2} x_{2}+a_{3} x_{3}+a_{4} x_{4}=\mathbf{0}$

This is the same as finding a nonzero solution $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ to the system

 $\left\{\begin{array}{l} 1 a_{1}-3 a_{2}+1 a_{3}+2 a_{2}=0 \\ -1 a_{1}+2 a_{2}+2 a_{3}+3 a_{4}=0 \\ 2 a_{1}+1 a_{2}-3 a_{3}+1 a_{4}=0 \end{array}\right.$

This is the same as finding a solution $\left[a_{1}, a_{2}, a_{3}, a_{4}\right]^{T}$ to the matrix equation

$\left[\begin{array}{cccc} 1 & -3 & 1 & 2 \\ -1 & 2 & 2 & 3 \\ 2 & 1 & -3 & 1 \end{array}\right]\left[\begin{array}{l} a_{1} \\ a_{2} \\ a_{3} \\ a_{4} \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]$  

In other words, we are trying to find the kernel of the matrix.

Row reducing

 $\operatorname{rref}\left(\left[\begin{array}{cccc} 1 & -3 & 1 & 2 \\ -1 & 2 & 2 & 3 \\ 2 & 1 & -3 & 1 \end{array}\right]\right)=\left[\begin{array}{llll} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{array}\right]$

 Remembering that row-reducing a system of equations retains the solution-set of the system and reinterpreting this as a system of equations this is the system:

 $\left\{\begin{array}{rrrrr} 1 a_{1} & & & +3 a_{4} & =0 \\ & 1 a_{2} & & +1 a_{4} & =0 \\ & & 1 a_{3} & +2 a_{4} & =0 \end{array}\right.$

This tells us that for $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ to be a solution that $a_{1}=-3 a_{4}, a_{2}=-a_{4}$ and $a_{3}=-2 a_{4}$

Picking any number (other than zero) for $a_{4}$ , we can then write a linear combination of $x_{1}, \ldots, x_{4}$ resulting in zero.