# Solutions to a linear system of equations.

I want to find the solution of this system when the parameter $a \in R$ varies.

\begin{cases} (a+2)x_2 + x_4 = 1 \\ -x_1 +x_3 = a+1 \\ (a+1)x_1 + 2x_2 -x_3 = 0 \\ x_1 -2x_2 -(a+1)x_3 = -2 \end{cases}

I notice that the matrix associated with the homogeneous system has rank $= 3$ for $a \ne 0,-2$ and never has $rank = 4$.

Now taking the matrix \begin{bmatrix} 0 & a+2 & 0 & 1 \\ -1 & 0 & 1 & a+1 \\ a+2 & 2 & -1 & 0 \\ 0 & -2 & -a-1 & -2 \end{bmatrix}

this has rank $= 4 \quad \forall a \ne 3,2$

and taking \begin{bmatrix} 1 & a+2 & 0 & 1 \\ 0 & 0 & 1 & a+1 \\ 0 & 2 & -1 & 0 \\ 0 & -2 & -a-1 & -2 \end{bmatrix} I get that this has rank $= 4 \quad \forall a \ne \frac{-2 + 2 \sqrt{11}}{4}, \frac{-2 - 2 \sqrt{11}}{4}$

So the original system has no solutions, correct?

• It is not a homogeneous system, and the augmented matrix should contain 5 columns (including both sides of the equal signs in the equations).
– Dave
Jun 13 '16 at 1:05
• You're missing a column for $\ x_4 \$, aren't you? And the last row is not correct. Jun 13 '16 at 1:21
• I found that $a=-3$ and $a=0$ yields a consistent system.
– Dave
Jun 13 '16 at 1:39
• @Dave Thank you for your answer, I was comparing the rank of the matrix associated with the homogeneous system with the one of the augmented matrix, if they are different I do not have solutions. When calculating the rank of the augmented matrix I choose a 4 by 4 submatrix, this is the reason of the missing column. How did you see that $a = -3$ and $a = 0$ yields a consistent system? same method I was using? Jun 13 '16 at 13:37

Transforming the system of equations into an augmented matrix, we have: $$\begin{bmatrix} 0 & (a+2) & 0 & 1 &|& 1 \\ -1 & 0 & 1 & 0 &|& (a+1) \\ (a+1) & 2 & -1 & 0 &|& 0 \\ 1 & -2 & -(a+1) & 0 &|& -2\end{bmatrix}$$ (note that the vertical lines between columns 4 and 5 seperate the sides of the equations). Row reduction (not all the way to the RREF form) yields: $$\begin{bmatrix} 1 & 0 & -1 & 0 &|& -(a+1) \\ 0 & 1 & \frac{1}{2}a & 0 &|& -\frac{1}{2}(a-1) \\ 0 & 0 & a(a+2) & -2 &|& -a(a+1) \\ 0 & 0 & 0 & 0 &|& a(a+3) \end{bmatrix}$$ We can now see that the last row must have $0$'s on both sides of the vertical line (otherwise the system is inconsistent), which means $a(a+3)=0$ which gives $a=-3$ or $a=0$. The solution set for when $a$ equals either of those values can be easily determined by plugging-in the values for $a$ and row reducing further.

Hopefully this helps.