$3\times3$ linear system organization

Question

How to organize the system below? Especially the 2nd row of the system.

$$\left\{\begin{eqnarray} 4x-3y+2z+4&=&0\\ x-\frac y3+\frac z2&=&-\frac16\\ 5x+2z&=&3y-3\\ \end{eqnarray}\right.$$

Answer 1

The second row is simply $$x - \frac 13 y + \frac 12 z = -\frac 16$$

Tip for greater ease in computation: the above equation is equivalent to $$6\left(x - \frac 13 y + \frac 12 z\right) = 6\left(-\frac 16\right) \iff6x - 2y + 3z = -1$$

In the first row, subtract $4$ from both sides.

In the third row, subtract $3y$ from both sides of the equation.

This will give you three rows of the form $ax + by + cz = d$, with real constants $a, b, c, d$.

Now, simply set up an augmented coefficient matrix to solve, and/or calculate the determinant of the $3\times 3$ coefficient matrix.

$$\begin{pmatrix} 4 & -3 & 2&|& -4\\ 6 & -2 & 3 &|&-1\\ 5&-3&2&|&-3\end{pmatrix}$$

Answer 2

The systems are more easily read when variables are grouped by colums. For instance, every term in $x$ should be on the same columns as others from other lines. And the constant terms are usually placed to the right of the equal signs.

As for your question concerning the second line, it is no different from the other lines if you realize that $y/3$ is the same as $\frac{1}{3} y$.

Eventually, you get: $$ \left\{ \begin{array}{llll} 4x &-3y &+ 2z &= -4\\ x &-\frac{1}{3} y &+ \frac{1}{2} z &= -\frac{1}{6}\\ 5x &-3y &+2z &= -3\\ \end{array} \right. $$

edit: fixed alignment of terms

Answer 3

Usually for a linear system, you put variables on the same side, and align columns like this:

$$\left\{\begin{eqnarray} 4x &-& 3y &+& 2z &=& -4\\ x &-& \frac 13y &+& \frac 12z&=&-\frac16\\ 5x&-&3y&+&2z&=&-3\\ \end{eqnarray}\right.$$

And you may multiply some rows by a constant to eliminate fractions:

$$\left\{\begin{eqnarray} 4x &-& 3y &+& 2z &=& -4\\ 6x &-& 2y &+& 3z&=&-1\\ 5x&-&3y&+&2z&=&-3\\ \end{eqnarray}\right.$$

Now, one thing that would be obvious, is that, subtracting third and first rows, you get $x=1$.

Then, computing $3R_1-2R_2$ gives $-5y=-10$ or $y=2$.

You finish with $2z=-4-4x+3y=-2$ so $z=-1$.