Answer
$B=\begin{bmatrix}
12 & 10 & -3 & 10\\
-7 & -6 & 7 & 5\\
9 & 9 & -5 & -1\\
-4 & -3 & 6 & 9\\
8 & 7 & -9 & -8\\
\end{bmatrix}$
Work Step by Step
We need to calculate vector B such that it doesnt have any nontrivial solutions.
By finding the reduced row echelon form of matrix A, we get
$A=\begin{bmatrix}
1 & 0 & 2 & 0 & 2 & 0\\
0 & 1 & -3 & 0 & -2 & 0\\
0 & 0 & 0 & 1 & -1 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
\end{bmatrix}$
This results shows us that column 3 and 5 don't have pivot positions and therefore $x_3$ and $x_5$ are free variables. Therefore, we can delete columns 3 and 5.
$B=\begin{bmatrix}
12 & 10 & -3 & 10\\
-7 & -6 & 7 & 5\\
9 & 9 & -5 & -1\\
-4 & -3 & 6 & 9\\
8 & 7 & -9 & -8\\
\end{bmatrix}$
The reduced row echelon form is
$B=\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0\\
\end{bmatrix}$
Because each column is a pivot column, the equation
$Bx=0$ has only the trivial solution.
