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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.

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    $\begingroup$ $ is your friend. Anyway, looks good for first glance. What is your question? $\endgroup$ – Berci Dec 17 '13 at 1:18
  • $\begingroup$ I just wanted verification. It does looks fine to me too. $\endgroup$ – user983246 Dec 17 '13 at 1:20
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    $\begingroup$ Hint: \pmatrix{1&1&3\\-1&0&1\\-1&2&7} enclosed in $ signs. $\endgroup$ – Berci Dec 17 '13 at 1:20
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    $\begingroup$ For future reference, how to type math nicely (i.e. readably). $\endgroup$ – Omnomnomnom Dec 17 '13 at 1:21
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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 $|A|\neq 0$.

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    $\begingroup$ neat suggestion. $\endgroup$ – user983246 Dec 17 '13 at 2:21
  • $\begingroup$ @user983246: You are very welcome. $\endgroup$ – Mhenni Benghorbal Dec 17 '13 at 2:23

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