Using matrices for real values of t either solve the following simultaneous equations or show they have no solution:
x + y - z = -1
2x + 3y + tz = 3
x + ty + 3z = 2
x + y - z = -1
2x + 3y + tz = 3
x + ty + 3z = 2
1*3*3 + 2*t*(-1) + 1*1*t - 1*3*(-1) - 1*t*t - 3*2*1 =
9 - 2t + t + 3 - t^2 -6 = -t^2 -t +6
t^2 + t - 6 = 0 has one Real, Positive Solution and one Real Negative Solution. This is sufficient to state the system has a solution if you avoid those two places.
Your system is $\displaystyle \left(\begin{array}{ccc}1&1&\!\!\!\!-1\\2&3&t\\1&t&3\end{array}\right) \left(\begin{array}{c}x\\y\\z\end{array}\right)$ $\displaystyle =\left(\begin{array}{c}\!\!\!\!-1\\3\\2\end{array}\right)$ . Now, a way to show that a non-homogeneous square system has a solution is to show the coefficients matrix's determinant is not zero...but ever when it's zero there can be solutions, though this time you have to bring to echelon form the augmented matrix, once with each value of t that makes the determinant vanish (there are two such values), and check whether you get a row of zeroes or else the last entry (the rightmost one) is non-zero while all the others are zero...
Tonio
Tonio