A Condition that a Linear System has Nontrivial Solutions
Problem 107
For what value(s) of $a$ does the system have nontrivial solutions?
\begin{align*}
&x_1+2x_2+x_3=0\\
&-x_1-x_2+x_3=0\\
& 3x_1+4x_2+ax_3=0.
\end{align*}
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Solution.
First note that the system is homogeneous and hence it is consistent. Thus if the system has a nontrivial solution, then it has infinitely many solutions.
This happens if and only if the system has at least one free variable. The number of free variables is $n-r$, where $n$ is the number of unknowns and $r$ is the rank of the augmented matrix.
To find the rank, we reduce the augmented matrix by elementary row operations.
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 2 & 1 & 0 \\
-1 &-1 & 1 & 0 \\
3 & 4 & a & 0
\end{array} \right]
\xrightarrow[R_3-3R_1]{R_2+R_1}
\left[\begin{array}{rrr|r}
1 & 2 & 1 & 0 \\
0 & 1 & 2 & 0 \\
0 & -2 & a-3 & 0
\end{array} \right]\\
\xrightarrow{R_3+2R_2}
\left[\begin{array}{rrr|r}
1 & 2 & 1 & 0 \\
0 & 1 & 2 & 0 \\
0 & 0 & a+1 & 0
\end{array} \right].
\end{align*}
The last matrix is in row echelon form.
Thus if $a+1=0$, then the third row is a zero row, hence the rank is $2$. In this case we have $n-r=3-2=1$ free variable. Thus there are infinitely many solutions. In particular, the system has nontrivial solutions.
On the other hand, if $a+1\neq 0$, then the rank is $3$ and there is no free variables since $n-r=3-3=0$.
In summary, the system has nontrivial solutions exactly when $a=-1$.
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