# Check if vectors are the the fundamental set of solutions

$$\begin{cases} x_1+2x_2+x_3-x_4+x_5=0 \\ 2x_1+x_2-x_3+2x_4-x_5=0 \\ x_1+5x_2+4x_3-5x_4+4x_5=0 \\ 4x_1+5x_2+x_3+x_5=0 \end{cases}$$

$$\begin{cases} \vec{a}_1=(-5,4,3,3,-3) \\ \vec{a}_2=(-7,5,-3,6,6) \\ \vec{a}_3=(5,-4,6,-3,-6) \end{cases}$$

Solution:

$$\begin{bmatrix} 1&2&1&-1&1 \\ 2&1&-1&2&-1 \\ 1&5&4&-5&4 \\ 4&5&1&0&1 \end{bmatrix}$$ Reducing to the row echelon form gives us

$$\begin{bmatrix} 1&2&1&-1&1 \\ 0&3&-3&-4&-3 \end{bmatrix}$$ Let $$x_3 = c_1$$, $$x_4=c_2$$, $$x_5=c_3$$. $$\begin{cases} x_1+2x_2+c_1-c_2+c_3=0 \\ 0x_1+3x_2-3c_1-4c_2-3c_3=0\end{cases}$$ $$x_2=c_1+\frac{4}{3}c_2+c_3$$ $$x_1=-2c_1-\frac{8}{3}c_2-2c_3-2c_1+c_2-c_3=-4c_1-\frac{5}{3}c_2-3c_2$$ However, plugging the corresponding values of each vector into the resulting system does not work here. What did I miss?

• Looks to me like you’ve made a sign error in your row-reduced matrix. That aside, you could instead check that the three vectors satisfy the original system and that they’re linearly independent. From the row-reduction, you know that the nullity is 3, so you’re done.
– amd
Mar 21, 2019 at 23:42

$$\left( \begin{array}{rrrr} 1 & 0 & 0 &0 \\ -1 & 0 & 1 &0 \\ 3 &-1 &-1 &0 \\ -5 & 0 & 1& 1 \\ \end{array} \right) \left( \begin{array}{rrrrr} 1& 2 &1& -1 & 1\\ 2& 1 &-1 & 2 &-1\\ 1 &5 & 4 &-5 & 4\\ 4 &5& 1 & 0& 1\\ \end{array} \right) = \left( \begin{array}{rrrrr} 1& 2 &1& -1 & 1\\ 0& 3 &3 & -4 &3\\ 0 &0 & 0 &0 & 0\\ 0 &0& 0 & 0& 0\\ \end{array} \right)$$