Given the system:
$$\begin{align}
& x+3y-3z=4 \\
& y+2z=a \\
& 2x+5y+(a^2-9)z=9
\end{align}$$
For which values of a (if any) does the system have a unique solution, infinitely many solutions, and no solution?
So I am getting that it has:
infinitely many solutions at: (-1)
No solution at (1)
Unique solution at (-1,1)
AM I RIGHT??
Best Answer
Note that your system is equivalent to the matrix equation $$ \begin{bmatrix} 1 & 3 & -3\\ 0 & 1 & 2 \\ 2 & 5 & a^2-9 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 4\\ a\\9 \end{bmatrix} $$ Since $$ \det\begin{bmatrix} 1 & 3 & -3\\ 0 & 1 & 2 \\ 2 & 5 & a^2-9 \end{bmatrix}=a^2-1 $$ this system is guaranteed a unique solution for $a\neq\pm 1$ (do you know why?).
Now the augmented systems for $a=1$ is $$ \begin{bmatrix} 1&3&-3&4\\ 0&1&2&1\\ 2&5&-8&9 \end{bmatrix} $$ Row-reducing this matrix gives $$ \DeclareMathOperator{rref}{rref}\rref\begin{bmatrix} 1&3&-3&4\\ 0&1&2&1\\ 2&5&-8&9 \end{bmatrix} = \begin{bmatrix} 1&0&-9&0\\ 0&1&2&0\\ 0&0&0&1 \end{bmatrix} $$ This system is not consistent (why?) so the original system has no solution for $a=1$.
Can you repeat the process for $a=-1$?
Addendum. You mention in your question that you're having trouble taking determinants. To find the determinant computed above we can expand about the first column: \begin{align*} \det\begin{bmatrix} \color{blue}1 & \color{red}3 & \color{red}{-3}\\ \color{blue}0 & \color{green}1 & \color{green}2 \\ \color{blue}2 & \color{purple}5 & \color{purple}{a^2-9} \end{bmatrix} &= (\color{blue}{1})\cdot\det \begin{bmatrix}\color{green}1&\color{green}2\\\color{purple}5 & \color{purple}{a^2-9} \end{bmatrix}-(\color{blue}0)\det\begin{bmatrix}\color{red}3 & \color{red}{-3}\\ \color{purple}5 & \color{purple}{a^2-9}\end{bmatrix}+(\color{blue}{2})\det\begin{bmatrix}\color{red}3 & \color{red}{-3}\\ \color{green}1 & \color{green}2 \end{bmatrix} \\ &= \left(a^2-9-10\right)-(0)+2\,\left(6+3\right) \\ &= a^2-19+18 \\ &= a^2-1 \end{align*}