Further Exercises: Linear systems

1
For which values of $a$ is the following linear system solvable? What are the solutions in that case?
$\begin{array}{cccccccc} x_{1} & + & 2 x_{2} + & x_{3} & = & 2 \\ x_{1} & + & 4 x_{2} + & 3 x_{3} & = & 4 \\ - 2 x_{1} & - & 3 x_{2} - & x_{3} & = & a \end{array}$

For this task you need the following basic knowledge: System of linear equations
The linear system can be written down as a matrix and simplified by the Gauss-Jordan algorithm.
$(\begin{array}{rrr} 1 & 2 & 1 \\ 1 & 4 & 3 \\ - 2 & - 3 & - 1 \end{array} | \begin{array}{r} 2 \\ 4 \\ a \end{array})$
$(\begin{array}{rrr} 1 & 2 & 1 \\ 0 & 2 & 2 \\ 0 & 1 & 1 \end{array} | \begin{array}{r} 2 \\ 2 \\ a + 4 \end{array})$
$(\begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array} | \begin{array}{r} 2 \\ a + 4 \\ 2 - 2 (a + 4) \end{array})$
Since the last line contains no more coefficients, $0 = 2 - 2 (a + 4)$ or $a = - 3$ . The system of equations is thus "under-defined" and we can choose $x_{3}$ arbitrarily, e.g. $x_{3} = z$ . We add this equation to the system as a new line and set $a = - 3$ .
$(\begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} | \begin{array}{r} 2 \\ - 3 + 4 \\ z \end{array})$
$(\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} | \begin{array}{r} z \\ 1 - z \\ z \end{array})$
That is, $a = - 3$ and $x_{1} = z$ , $x_{2} = 1 - z$ , and $x_{3} = z$ with $z \in ℝ$ .
2
Conseder the linear system of equations:
$\begin{array}{cccccccc} (I) & - x_{1} & + & 2 x_{2} & = & 2 \\ (I I) & 2 x_{1} & - & x_{2} & = & 2 \end{array}$
First solve the system graphically.

For this task you need the following basic knowledge: System of linear equations
Solve equation $(I)$ for $x_{2}$ :
$x_{2} = \frac{1}{2} \cdot x_{1} + 1$
The line has slope $m = \frac{1}{2} $ and $x_{2}$ -axis intercept $1$ .
Draw this line into a coordinate system. Below, this is the $purple line$ .
Solve equation $(I I)$ for $x_{2}$ :
$x_{2} = 2 \cdot x_{1} - 2$
The line has slope $m = 2$ and $x_{2}$ -axis intercept $- 2$ .
Draw this line into the same coordinate system. Below, this is the $red line$ .
The two lines intersect at the point $S (2 | 2)$ .

Now, determine the precise solution to the linear system of equations using a method of your choice.

For this task you need the following basic knowledge: System of linear equations
Solution by equating coefficients
You may equivalently write equations $(I)$ and $(I I)$ as $x_{2} = \frac{1}{2} \cdot x_{1} + 1$ and $x_{2} = 2 \cdot x_{1} - 2$ . Setting both equations equal renders:
$\frac{1}{2} \cdot x_{1} + 1$ $=$ $2 \cdot x_{1} - 2$ $+ 2$
↓
Solve for $x_{1}$ .
$\frac{1}{2} \cdot x_{1} + 3$ $=$ $2 \cdot x_{1}$ $- \frac{1}{2} \cdot x_{1}$
$3$ $=$ $\frac{3}{2} \cdot x_{1}$ $\cdot \frac{2}{3}$
$2$ $=$ $x_{1}$
Plug $x_{1} = 2$ into one of the two equations and calculate $x_{2}$ .
$x_{2} = \frac{1}{2} \cdot x_{1} + 1 \Rightarrow x_{2} = \frac{1}{2} \cdot 2 + 1 = 2$
Result: The solution of the linear system is the point $(2 | 2)$ .
Solution by substitution method
You may equivalently write equation $(I)$ as $x_{2} = \frac{1}{2} \cdot x_{1} + 1$ . Now, plug this expression for $x_{2}$ into equation $(I I)$ :
$2 \cdot x_{1} - x_{2}$ $=$ $2$
↓
Plug in $x_{2} = \frac{1}{2} \cdot x_{1} + 1$ .
$2 \cdot x_{1} - (\frac{1}{2} \cdot x_{1} + 1)$ $=$ $2$
↓
Multiply out the bracket.
$2 \cdot x_{1} - \frac{1}{2} \cdot x_{1} - 1$ $=$ $2$ $+ 1$
↓
Solve for $x_{1}$ .
$\frac{3}{2} \cdot x_{1}$ $=$ $3$ $\cdot \frac{2}{3}$
$x_{1}$ $=$ $2$
Plug $x_{1} = 2$ into one of the two equations and calculate $x_{2}$ .
$x_{2} = \frac{1}{2} \cdot x_{1} + 1 \Rightarrow x_{2} = \frac{1}{2} \cdot 2 + 1 = 2$
Result: The solution of the linear system is the point $(2 | 2)$ .
Solution by addition method
$\begin{array}{cccccccc} (I) & - x_{1} & + & 2 x_{2} & = & 2 \\ (I I) & 2 x_{1} & - & x_{2} & = & 2 \end{array}$
Eliminate e.g. the variable $x_{1}$ by calculating: $2 \cdot (I) + (I I)$
$\Rightarrow 4 x_{2} - x_{2} = 4 + 2 \Rightarrow 3 x_{2} = 6 \Rightarrow x_{2} = 2$
Plug the calculated value $x_{2} = 2$ into equation $(I I)$ :
$\Rightarrow 2 x_{1} - 2 = 2 \Rightarrow 2 x_{1} = 4 \Rightarrow x_{1} = 2$
Result: The solution of the linear system is the point $(2 | 2)$ .

$\frac{1}{2} \cdot x_{1} + 1$	$=$	$2 \cdot x_{1} - 2$	$+ 2$
	↓	Solve for $x_{1}$ .
$\frac{1}{2} \cdot x_{1} + 3$	$=$	$2 \cdot x_{1}$	$- \frac{1}{2} \cdot x_{1}$
$3$	$=$	$\frac{3}{2} \cdot x_{1}$	$\cdot \frac{2}{3}$
$2$	$=$	$x_{1}$

$2 \cdot x_{1} - x_{2}$	$=$	$2$
	↓	Plug in $x_{2} = \frac{1}{2} \cdot x_{1} + 1$ .
$2 \cdot x_{1} - (\frac{1}{2} \cdot x_{1} + 1)$	$=$	$2$
	↓	Multiply out the bracket.
$2 \cdot x_{1} - \frac{1}{2} \cdot x_{1} - 1$	$=$	$2$	$+ 1$
	↓	Solve for $x_{1}$ .
$\frac{3}{2} \cdot x_{1}$	$=$	$3$	$\cdot \frac{2}{3}$
$x_{1}$	$=$	$2$

Further Exercises: Linear systems

Solution by equating coefficients

You may equivalently write equations (I) and (II) as x2=12⋅x1+1 and x2=2⋅x1−2 . Setting both equations equal renders:

Solution by substitution method

You may equivalently write equation (I) as x2=12⋅x1+1. Now, plug this expression for x2 into equation (II):

Solution by addition method

You may equivalently write equations $(I)$ and $(I I)$ as $x_{2} = \frac{1}{2} \cdot x_{1} + 1$ and $x_{2} = 2 \cdot x_{1} - 2$ . Setting both equations equal renders:

You may equivalently write equation $(I)$ as $x_{2} = \frac{1}{2} \cdot x_{1} + 1$ . Now, plug this expression for $x_{2}$ into equation $(I I)$ :