Determine the values of $a$ for which the following linear system has (i) no
solutions, (ii) infinitely many solutions, (iii) exactly one solution:
\begin{cases}
x + 2y − 3z = 4 \\[4px]
3x − y + 5z = 2 \\[4px]
4x + y + (a^2 − 14)z = a + 2
\end{cases}
Best Answer
Without knowing what tools you have available, it is difficult to give an answer as expected by your instructor. However, a standard procedure is Gaussian elimination on the system's matrix. \begin{align} \left[\begin{array}{ccc|c} 1 & 2 & -3 & 4 \\ 3 & -1 & 5 & 2 \\ 4 & 1 & a^2-14 & a+2 \end{array}\right] &\to \left[\begin{array}{ccc|c} 1 & 2 & -3 & 4 \\ 0 & -7 & 14 & -10 \\ 0 & -7 & a^2-2 & a-10 \end{array}\right] &&\begin{aligned}R_2&\gets R_2-3R_1\\ R_3&\gets R_3-4R_1\end{aligned} \\ &\to \left[\begin{array}{ccc|c} 1 & 2 & -3 & 4 \\ 0 & -7 & 14 & -10 \\ 0 & 0 & a^2-16 & a \end{array}\right] &&R_3\gets R_3-R_2 \end{align} You can clearly see that the transformed system has a unique solution if and only if $a^2-16\ne0$.
For $a^2-16=0$, the third row has a nonzero entry in the last column, which says the system has no solution.