<s>[INST] Say for what values of $a \in \mathbb {R} $ this matrix system has solutions 
Let $a \in \mathbb{R}$ and
  $$
 A_a =
 \begin{pmatrix}
  1 & a & 1 \\
  a & 2 & 3 \\
  2 & 3 & 4
 \end{pmatrix}.
$$
  
  
*
  
*Say for each values of $a \in \mathbb{R}$ the system:
  $$
 A_a
 \begin{pmatrix}
  x \\ y \\ z
 \end{pmatrix}
 =
 \begin{pmatrix}
  1 \\ 0 \\ 1
 \end{pmatrix}
$$
  accepts one solution only and for what values have no solutions.
  
*Resolve the system for $a = 1$.
  

I got this problem from this image.
I think I know how to resolve system of equations using inverse matrices $3 \times 3$, $2 \times 2$ and more by finding the determinant. Then, I can reflect the matrix about the diagonal and multiply it with the inverse to find the solution. However, here, I get a little lost, because it seems like I should work backwards. What should I do?
 [/INST] You can do the exercise in one swoop by using Gaussian elimination:
\begin{align}
\left[\begin{array}{ccc|c}
  1 & a & 1 & 1\\
  a & 2 & 3 & 0\\
  2 & 3 & 4 & 1
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
  1 & a & 1 & 1\\
  0 & 2-a^2 & 3-a & -a\\
  0 & 3-2a & 2 & -1
\end{array}\right] && \begin{matrix}R_2\gets R_2-aR_1\\R_3\gets R_3-2R_1\end{matrix}
\\[6px]&\to
\left[\begin{array}{ccc|c}
  1 & a & 1 & 1 \\
  0 & 3-2a & 2 & -1 \\
  0 & 2-a^2 & 3-a & -a
\end{array}\right] && R_2\leftrightarrow R_3
\\[6px]\color{red}{(3-2a\ne0)}\quad&\to
\left[\begin{array}{ccc|c}
  1 & a & 1 & 1\\
  0 & 1 & 2/(3-2a) & 1/(2a-3)\\
  0 & 2-a^2 & 3-a & -a
\end{array}\right] && R_2\gets (3-2a)^{-1}R_2
\\[6px]&\to
\left[\begin{array}{ccc|c}
  1 & a & 1 & 1 \\
  0 & 1 & 2/(3-2a) & 1/(2a-3) \\
  0 & 0 & \frac{(4a-5)(a-1)}{3-2a} & \frac{(a-2)(a-1)}{3-2a}
\end{array}\right] && R_3\gets R_3-(2-a^2)R_2
\end{align}
If $a=1$, the last row is zero, which means the system has infinitely many solutions; the matrix $A_a$ has rank $2$.
If $a=5/4$, the matrix $A_a$ has rank $2$, and the system has no solution.
If $a\notin\{1,5/4,3/2\}$, the matrix $A_a$ has rank $3$ and the system has a single solution.
It remains to see the case when $a=3/2$.
\begin{align}
\left[\begin{array}{ccc|c}
  1 & a & 1 & 1\\
  0 & 2-a^2 & 3-a & -a\\
  0 & 3-2a & 2 & -1
\end{array}\right]
&=
\left[\begin{array}{ccc|c}
  1 & 3/2 & 1 & 1\\
  0 & -1/4 & 3/2 & -3/2\\
  0 & 0 & 2 & -1
\end{array}\right]
\end{align}
It's clear that the matrix $A_a$ has rank $3$ and the system has a single solution.
The solution for $a=1$ is readily available, because the last matrix in the elimination process is
$$
\left[\begin{array}{ccc|c}
  1 & 1 & 1 & 1 \\
  0 & 1 & 2 & -1 \\
  0 & 0 & 0 & 0
\end{array}\right]
$$
Do the transformation $R_1\gets R_1-R_2$ to get
$$
\left[\begin{array}{ccc|c}
  1 & 0 & -1 & 2 \\
  0 & 1 & 2 & -1 \\
  0 & 0 & 0 & 0
\end{array}\right]
$$
so the system can be written
$$
\begin{cases}
x=2+z\\[4px]
y=-2-2z
\end{cases}
$$
so the solutions are of the form
$$
\begin{bmatrix}2\\2\\0\end{bmatrix}+
t\begin{bmatrix}1\\-2\\1\end{bmatrix}
$$
</s>