# Linear System without solution

Find all real numbers $\alpha$ and $\beta$ for which the linear system: \begin{cases} X_1 + X_3 = 0 \\[4px] \alpha X_1 + X_2 + 2X_3 = 0 \\[4px] 3X_1 + 4X_2 + \beta X_3 = 2 \end{cases} does not have a solution.

I can't use Gaussian Elimination as I don't know how to put $\alpha X_1 + X_2 + 2X_3 = 0$ in reduced row-echelon form. I'm aware that $0X_3$ should equal to $c$ where $c$ is not $0$ for the system to not have a solution but I don't know how to get there.

Why can't you use Gaussian elimination? \begin{align} \left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ \alpha & 1 & 2 & 0 \\ 3 & 4 & \beta & 2 \end{array}\right] &\to \left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 2-\alpha & 0 \\ 0 & 4 & \beta-3 & 2 \end{array}\right] && \begin{aligned}R_2&\gets R_2-\alpha R_1,\\R_3&\gets R_3-3R_1\end{aligned} \\&\to \left[\begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 2-\alpha & 0 \\ 0 & 0 & 4\alpha+\beta-11 & 2 \end{array}\right] && R_3\gets R_3-4R_2 \end{align}

Now it's quite easy, isn't it?

• Thanks! so that means 4α+β−11 has to be zero for the system to be inconsistent but can I use the second equation to find α and β? Jan 18, 2017 at 23:24
• @user408176 That's as much as you can get: there are infinitely many values of $\alpha$ and $\beta$ so that the system is inconsistent. Jan 18, 2017 at 23:31

Using SymPy:

>>> from sympy import *
>>> a, b = symbols('a b')
>>> M = Matrix([[1,0,1,0],[a,1,2,0],[3,4,b,2]])
>>> M
Matrix([
[1, 0, 1, 0],
[a, 1, 2, 0],
[3, 4, b, 2]])
>>> M.rref()
(Matrix([
[1, 0, 0,          -2/(4*a + b - 11)],
[0, 1, 0, -2*(-a + 2)/(4*a + b - 11)],
[0, 0, 1,           2/(4*a + b - 11)]]), [0, 1, 2])


Can you take it from here?

Since $X_3=-X_1$,

we have

$$(\alpha-2) X_1 + X_2 = 0$$

$$(3- \beta) X_1 + 4X_2 = 2$$

Since $X_2=(2-\alpha)X_1$,

we have

$$(3-\beta+8-4\alpha) X_1 = 2$$

Can you take it from here?

An alternative way (and easier method) is to use determinants.

If the determinant of the linear system is zero, either the system has an infinite number of solutions, or none.

Calculate the following determinant using Laplace expansion and set it equal to zero:

$$\begin{vmatrix} 1 & 0 & 1 \\ \alpha & 1 & 2 \\ 3 & 4 & \beta \end{vmatrix}=0$$

This should give you the appropriate values of $\alpha$ and $\beta$.