01 - Linear System - Example 1 - Solving Linear Systems (Unique Solution)

Determine the solutions (if any) to the following equation

$$3x+y-2z=-2,x+y+z=2;2x+4y+z=0$$

Solution 1 (Substitution)

Substituting $z=2-x-y$ (using second equation) into the first and third give

$$\begin{array}{rl}3x+y-2(2-x-y)=-2& ⟹5x+3y=2;\\ 2x+4y+(2-x-y)=0& ⟹x+3y=-2;\end{array}$$

Hence we find $4x=4$ so we get the solutions

$$x=1,y=\frac{-2-x}{3}=-1,z=2-x-y=2$$

Solution 2 (Row Reduction - EROs)

Write the linear system in Augmented Matrix Form

$$\left(\begin{array}{cccc}3& 1& -2& -2\\ 1& 1& 1& 2\\ 2& 4& 1& 0\end{array}\right)$$

Aim is to try use EROs in order to get it in a augmented matrix form $(I\mid \mathbf{x})$ where $\mathbf{x}$ is then your unique solution
(Note that we only care about what happens with the 3x3 part of the matrix)

Getting a $1$ in the top left to make it easier to get $0$s for each row underneath the $1$

$$\left(\begin{array}{cccc}3& 1& -2& -2\\ 1& 1& 1& 2\\ 2& 4& 1& 0\end{array}\right)\overrightarrow{S_{12}}\left(\begin{array}{cccc}1& 1& 1& 2\\ 3& 1& -2& -2\\ 2& 4& 1& 0\end{array}\right)$$

Eliminating the numbers beneath the top-left $1$ (ensuring the first column is $(1,0,0{)}^T$)

$$\left(\begin{array}{cccc}1& 1& 1& 2\\ 3& 1& -2& -2\\ 2& 4& 1& 0\end{array}\right)\overrightarrow{A_{12}(-3)}\left(\begin{array}{cccc}1& 1& 1& 2\\ 0& -2& -5& -8\\ 2& 4& 1& 0\end{array}\right)\overrightarrow{A_{13}(-2)}\left(\begin{array}{cccc}1& 1& 1& 2\\ 0& -2& -5& -8\\ 0& 2& -1& -4\end{array}\right)$$

Getting a $1$ in the middle-middle part of the matrix

$$\left(\begin{array}{cccc}1& 1& 1& 2\\ 0& -2& -5& -8\\ 0& 2& -1& -4\end{array}\right)\overrightarrow{M_2\left(-\frac{1}{2}\right)}\left(\begin{array}{cccc}1& 1& 1& 2\\ 0& 1& \frac{5}{2}& 4\\ 0& 2& -1& -4\end{array}\right)$$

Making the second column $(0,1,0{)}^T$

$$\left(\begin{array}{cccc}1& 1& 1& 2\\ 0& 1& \frac{5}{2}& 4\\ 0& 2& -1& -4\end{array}\right)\overrightarrow{A_{21}(-1)}\left(\begin{array}{cccc}1& 0& -\frac{3}{2}& -2\\ 0& 1& \frac{5}{2}& 4\\ 0& 2& -1& -4\end{array}\right)\overrightarrow{A_{23}(-2)}\left(\begin{array}{cccc}1& 0& -\frac{3}{2}& -2\\ 0& 1& \frac{5}{2}& 4\\ 0& 0& -6& -12\end{array}\right)$$

Getting a $1$ in the bottom-right of the matrix

$$\left(\begin{array}{cccc}1& 0& -\frac{3}{2}& -2\\ 0& 1& \frac{5}{2}& 4\\ 0& 0& -6& -12\end{array}\right)\overrightarrow{M_3\left(-\frac{1}{6}\right)}\left(\begin{array}{cccc}1& 0& -\frac{3}{2}& -2\\ 0& 1& \frac{5}{2}& 4\\ 0& 0& 1& 2\end{array}\right)$$

Making the third column $(0,0,1{)}^T$

$$\left(\begin{array}{cccc}1& 0& -\frac{3}{2}& -2\\ 0& 1& \frac{5}{2}& 4\\ 0& 0& 1& 2\end{array}\right)\overrightarrow{A_{31}\left(\frac{3}{2}\right)}\left(\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& \frac{5}{2}& 4\\ 0& 0& 1& 2\end{array}\right)\overrightarrow{A_{32}\left(-\frac{5}{2}\right)}\left(\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 0& -1\\ 0& 0& 1& 2\end{array}\right)$$

Converting the Augmented matrix back to a Linear System gives

$$x=1,y=-2,z=2$$

The solution is the same but the sequence of EROs is not necessarily unique so you could have made the top left number $1$ without swapping the row $1$ and row $2$

Non-unique cases will be covered later on