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

Question

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?

Answer 1

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.

Answer 2

$Ax=0$ 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 $a =-2$.