Answer
See solution
Work Step by Step
$\begin{bmatrix}
11&14&19\\
-5&-8&-13\\
10&13&18\\
7&10&15
\end{bmatrix}$~$\begin{bmatrix}
11&14&19\\
0&3&8\\
0&0&0\\
0&0&0
\end{bmatrix}$
Because the augmented matrix is consistent, x is in H.
$\begin{bmatrix}
19\\
-13\\
18\\
15
\end{bmatrix}=-\frac{5}{3}\begin{bmatrix}
11\\
-5\\
10\\
7
\end{bmatrix}+\frac{8}{3}\begin{bmatrix}
14\\
-8\\
13\\
10
\end{bmatrix}$
$[x]_B=\begin{bmatrix}
-\frac{5}{3}\\
\frac{8}{3}\\
\end{bmatrix}$
