# Determine value of '$a$' for which the system is inconsistent and has infinitely many solutions.

Consider the matrix $$A$$, to be equal to: $$\begin{bmatrix}1&2&1\\-1&4&3\\2&-2&a\end{bmatrix}$$

Then we can rewrite this as: $$\begin{bmatrix}1&2&1\\0&6&4\\2&-2&a\end{bmatrix}$$ $$\begin{bmatrix}1&2&1\\0&1&2/3\\2&-2&a\end{bmatrix}$$ $$\begin{bmatrix}1&2&1\\0&1&2/3\\0&-6&a-2\end{bmatrix}$$ $$\begin{bmatrix}1&2&1\\0&1&2/3\\0&0&a+2\end{bmatrix}$$

Now consider the system $$Ax=0$$. If $$\mathbf{a = -2}$$, then $$x_3$$ is a free variable, because a+2 turns to zero. The solution becomes: $$x_1 = (1/3)x_3$$ $$x_2 = -(2/3)x_3$$ $$x_3 = free$$ If $$\mathbf{a}$$ does not equal 2 , then the system seems to have trivial solutions every time, because you divide $$a+2$$ by itself, which becomes $$1$$, regardless of the value of $$a$$. The solutions become: $$x_1 = 0$$ $$x_2 = 0$$ $$x_3 = 0$$

But then how can you determine when the system $$Ax=0$$ is inconsistent or has infinitely many solutions? What mistake have I made? Could someone provide me your own method if my method seems to be wrong?

We are talking of the homogeneous system of equations, right?

The solution for $$a=-2$$ is correct. As you have said, for $$a\neq -2$$, the only solution is $$x_1=x_2=x_3=0$$.

The dimension of the solution space is thus $$0$$ if $$a=-2$$ and $$1$$ otherwise.

• OP is talking about infinitely many solutions, not infinitely many independent solutions – Kabo Murphy Mar 24 at 11:59
• @Stallmp There is always at least one solution for homogeneous systems of equations (the trivial solution $0$); there are infinitely many solutions for $a=-2$. – st.math Mar 24 at 12:02
• @Stallmp Yes, there is always a solution. – st.math Mar 24 at 12:04
• @Stallmp Your calculations are correct. – st.math Mar 24 at 12:06
• @Stallmp: When $a=-2$, as you wrote, you have as solutions $S = \{ (v/3, -2v/3, v): v \in \mathbb{R} \}$. – Ertxiem Mar 24 at 12:07

$$Ax=0$$ has a unique solution if $$\det(A)\neq 0$$ and it has infintely many solutions if $$\det(A)= 0$$ (any multiple of a solution is also a solution). In this case $$\det(A)=0$$ iff $$a =-2$$.

• Alright I see, thank you. Would this imply that inconsistency is not possible with systems Ax = 0 ? – Stallmp Mar 24 at 12:07
• Yes, the zero vector is always a solution, so there is no question of inconsistency. – Kabo Murphy Mar 24 at 12:09
• Alright thanks a lot! – Stallmp Mar 24 at 12:09