# For which values $\alpha$ and $\beta$ the system has solution? Am I correct?

Consider this equations system of the form $$Ax=b$$ where

\begin{align*} A=\begin{pmatrix} 1 & 2 & 0 & 3\\ 2 & 1 & 0 & \alpha\\ 0 & 1 & 1 & 0\\ 0 & 0 & 1 & 3 \end{pmatrix} \ \ \ \ \ \ \ \ \ \ \ \ \ \ b=\begin{pmatrix} 0\\ \beta\\ 1\\ 0 \end{pmatrix} \end{align*}

1. For which values $$\alpha$$ and $$\beta$$ the system has solution?

2. For which values $$\alpha$$ and $$\beta$$ the system has an unique solution?

3. If the system doesn't have an unique solution, describe all the solutions

• What I did for 2 is to calculate:

\begin{align*} Det(A)=-15+\alpha \end{align*} So my answer is that the system of equations has an unique solution $$\iff$$ $$A^{-1}$$ exists $$\iff$$ $$\alpha\neq15$$

i.e., for $$\alpha\neq15$$ and $$\forall \beta$$ the system of equations has an unique solution. Am I correct?

• For 3 what I did is this:

\begin{align*} x_1+2x_2+3x_4&=0\\ 2x_1+x_2+\alpha x_4&=\beta\\ x_2+x_3&=1\\ x_3+x_4&=0 \end{align*}

From that, I got that the solutions has this form:

\begin{align*} \begin{pmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{pmatrix}=\begin{pmatrix} -2(2x_4+1)\\ 1+x_4\\ -x_4\\ x_4 \end{pmatrix} \ \ \ \ \ \ \ \ \ \text{with } x_4=\frac{3+\beta}{\alpha-9} \end{align*}

• My doubts:

a) Am I correct in 2?

b) In 3, I got $$x_4=\frac{3+\beta}{\alpha-9}$$, but what happens if $$\alpha=9$$? Does that means that with $$\alpha=9$$ then the system hasn't solution?

c) For 1, how can I determine if the system has solution?

I would really appreciate your help!

## 1 Answer

For Cramer's theorem the system has one and only one solution if $$\det(A)\ne 0$$ that is for $$a\ne 15$$.

If $$\alpha=15$$ the matrix $$A$$ has rank 3, while the augmented matrix $$A|b$$ has rank 4, if $$\beta\ne -3$$, therefore there are no solutions for Rouché–Capelli theorem.

If $$\alpha=15;\;\beta=-3$$ the rank of $$A|b$$ is 3 and the system has infinite solutions depending on $$4-3=1$$ parameter.

$$\left( \begin{array}{ccc} 2 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right).(x,y,z)=(-3-15 t,1,-3 t)$$ give the solutions $$x= -2 - 9 t, y = 1 + 3 t, z = -3t;\quad\forall t\in\mathbb{R}$$

• Thanks you very very much! Just one little question, how did you get the condition $\beta \neq -3$? @Raffaele Commented Oct 19, 2020 at 22:13