Linear System without solution

Question

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.

Answer 1

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?

Answer 2

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?

Answer 3

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?

Answer 4

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 $\beta$.