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!
Best 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 $$x= -2 - 9 t, y = 1 + 3 t, z = -3t;\quad\forall t\in\mathbb{R}$$