# Matrix vector form. Is this in the correct form?

I have this question:

Write the linear system $$\begin{array}{rcr}-2x_1+x_2-4x_3 & = & 1 \\ x_1-2x_2 & = & -3 \\ x_1+x_2-4x_3 & = & 0 \end{array}$$ in the matrix-vector form $A\mathbf{x}=\mathbf{b}$.

Is this what they want?

$$x_1* \begin{bmatrix} -2 \\ 1 \\ 1 \end{bmatrix} + x_2* \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} + x_3* \begin{bmatrix} -4 \\ 0 \\ -4 \end{bmatrix} = \begin{bmatrix} 1 \\ -3 \\ 0 \end{bmatrix}$$

• For general MathJax tips, you can take a look here : math.meta.stackexchange.com/questions/5020/… (See in particular the top-voted answer for matrices). – Arnaud D. Aug 1 '18 at 18:48
• Is there anything I could improve which prevents you right now from accepting one of the given answers? – mrtaurho May 19 at 18:25

I guess the matrix-vector form here refers to the matrix A and the vector b. I would suggest to rewrite the equation in the following way

$$\begin{pmatrix}-2&1&-4\\1&-2&0\\1&1&-4\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}~=~\begin{pmatrix}1\\-3\\0\end{pmatrix}$$

To verify the L.H.S. you can just multiply the vector by the matrix and then your will get first guess.

• What does pmatrix mean? – Jwan622 Aug 1 '18 at 18:45
• Why is the column of x's called a vector? – Jwan622 Aug 1 '18 at 18:46
• The pmatrix-command gives you the $()$ braces. For other kinds of matrices just search for the LaTeX commands. How would you call a single column instead? – mrtaurho Aug 1 '18 at 18:47
• You can interpret this system of equation as the point of intersection within $\mathbb{R}^3$ of the three given functions. For that every single variable represents on direction of the $3$-dimensional space you can just conclude setting the vector x as $\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$ where $x_1,x_2,x_3$ are the three directions towards the the $\mathbb{R}^3$-space. – mrtaurho Aug 1 '18 at 18:54

basically!

$A = \begin{pmatrix} -2 & 1 & -4 \\ 1 & -2 & 0 \\ 1 & 1 & -4 \end{pmatrix}$ and $b = \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix}$ yielding $Ax = b$.