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I'm studying, in this class we are given practice problems (but no answer sheet, go figure!). So my question is:

Is my answer correct? Please point out anything I might have gotten wrong.

Determine a basis for the solution set of the linear system:

$$\left\{\begin{array}{l} 3x + 4y - 2z =0\\ 2x + 3y + z =0\\ -2x -2y + 6z =0\\ \end{array}\right.$$

My solution:

Then the augmented matrix is: \begin{align*} \left[ \begin{array}{ccc|c} 3 & 4 & -2 &0\\ 2 & 3 & 1 &0\\ -2 & -2 & 6 &0\\ \end{array} \right] \end{align*}

Reduced Row Echelon Form \begin{align*} \left[ \begin{array}{ccc|c} 1 & 0 & -10 &0\\ 0 & 1 & 7 &0\\ 0 & 0 & 0 &0\\ \end{array} \right] \end{align*}

This means that any vector $\vec x$ can be represented as \begin{align*} \vec x = \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} = \begin{bmatrix} 10z\\ -7z\\ z\end{bmatrix}=z \begin{bmatrix} 10\\ -7\\ 1 \end{bmatrix} \end{align*}

Where $z$ is any scalar.

This means that the linear system represents a one-dimensional subspace and its basis is:

\begin{align*}\left(\begin{bmatrix} 10\\ -7\\ 1 \end{bmatrix}\right) \end{align*}

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  • $\begingroup$ You can perform at least a partial check of your answer yourself by seeing if the vector that you computed satisfies the original equations. $\endgroup$ – amd Nov 22 '17 at 1:36
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Your computations are correct, and so is your result.

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