Decide if the vector $(1,1,1)$ is in the row space of the matrix
$$ \begin{bmatrix} 1& 1& 3\\-1&0&1\\-1&2&7 \end{bmatrix}$$
Yes. To see if there are $c_1$ and $c_2$ such that
$c_1 \cdot (1,1,3) + c_2 \cdot ( -1,0,1) + c_3\cdot(-1,2,7) = (1,1,1)$ has a solution
I solve :
first row $c_1 – c_2 – c_3 = 1$
second row $c_1 + 2c_3 = 1$
third row $3c_1 + c_2+ 7c_3 = 1$
coefficient matrix
$$\begin{array}{|ccc|c|} 1&-1&-1&1\\
1&0&2&1\\
3&1&7&1\end{array}$$
I use Gauss method $-3p_1+p_3$,$-p_1+p_2$, and $-3p_2+p_3$ from the above matrix to get the below matrix.
$$ \begin{array}{|ccc|c|}0&4&10&-2\\
0&1&3&0\\
0&1&1&-2\end{array}$$
I use $-4p_2+p_1$ and $-p_2+p_3$ to get the below matrix
$$\begin{array}{|ccc|c|}0&0&-2&-2\\
0&1&3&0\\
0&0&-2&-2\end{array}$$
From the final matrix, $-2c_3 = -2$ which means $c3 =1$.
Plugging $c_3$ to obtain $c_2$:
$$\begin{align}c_2 + 3c_3 &= 0\\
c_2 + 3(1) &= 0\\
c_2 &= -3\end{align}$$
I know it has a solutions because I got plug in $c_3$ and $c_2$ to the original linear system to obtain $c_1$.
$$\begin{align} 3c_1 + c_2+7c_3 &=1\\
3c_1 -3+7 &= 1\\
3c_1 +4&=1\\
3c_1 &= -3\\
c_1 &= -1\end{align}$$
plugging $c_1,c_2$ and $c_3$ to the original linear system, it has a solution.
Edit: I just want verification from users.
Best Answer
Note that, you have been asked "Decide if the vector $(1,1,1)$ is in the row space of the matrix ?". So, you do not have to find $c_1,c_2,c_3$, instead, it is enough to prove that the system has a unique solution by proving the determinant of the coefficient matrix $|A|\neq 0$.