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?
 A: The determinant of $A_a$ is $\det(A_a)=-4a^2+9a-5$ so:
$$
\det(A_a)=0 \iff a=1 \;\lor \; a=\frac{5}{4}
$$
and, since the matrix is invertible only if its determinant is not null, for $a\ne 1$ and $a \ne \frac{5}{4}$ the system have only one solution.
Now test these values, for which the determinant is null and the matrix is not invertible, for $A_a(x,y,z)^T=(1,0,1)^T$ and see when the system is impossible or have infinitely many solutions.
A: 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
$$
\begin{bmatrix}2\\2\\0\end{bmatrix}+
t\begin{bmatrix}1\\-2\\1\end{bmatrix}
$$
A: Writing $A_a$ short as $A$, and the inhomogenous system $A x = b$ as augmented matrix $[A|b]$ we get these transformations:
$$
\left[
\begin{array}{rrr|r}
1 & a & 1 & 1 \\ 
a & 2 & 3 & 0 \\
2 & 3 & 4 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1   & a    & 1 & 1 \\ 
a-3 & 2-3a & 0 & -3 \\
-2  & 3-4a & 0 & -3
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1   & a    & 1 & 1 \\ 
a-1 & a-1 & 0 & 0 \\
-2  & 3-4a & 0 & -3
\end{array}
\right]
$$
The second row would benefit from division by $a-1$, so it is important if $a-1$ is zero or not.
Case 1:
If $a = 1$ this augmented matrix turns into
$$
\left[
\begin{array}{rrr|r}
1   & 1    & 1 & 1 \\ 
0 & 0 & 0 & 0 \\
-2  & 3 & 0 & -3
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1   & 1    & 1 & 1 \\ 
-2  & 3 & 0 & -3 \\
0 & 0 & 0 & 0 \\
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1   & 1    & 1 & 1 \\ 
0  & 5 & 2 & -1 \\
0 & 0 & 0 & 0 \\
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 0 & 3/5 & 6/5 \\ 
0 & 1 & 2/5 & -1/5 \\
0 & 0 & 0 & 0 \\
\end{array}
\right]
$$
Only two equations for three variables, one variable stays free and in this case we have infinite many solutions
$$
(x, y, z) = (6/5-3/5 z, -1/5 - 2/5z, z) \quad (z \in \mathbb{R})
$$
where we used $z$ as free variable.
Other Cases:
If $a \ne 1$ we are led to
$$
\left[
\begin{array}{rrr|r}
1   & a    & 1 & 1 \\ 
1 & 1 & 0 & 0 \\
-2  & 3-4a & 0 & -3
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1-a   & 0    & 1 & 1 \\ 
1 & 1 & 0 & 0 \\
-2+4a  & 3 & 0 & -3
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1-a   & 0    & 1 & 1 \\ 
1 & 1 & 0 & 0 \\
-5+4a  & 0 & 0 & -3
\end{array}
\right]
$$
Here it makes a difference if $-5 + 4a$ is zero or not.
Case 2:
If $a = 5/4$ we have
$$
\left[
\begin{array}{rrr|r}
-1/4 & 0 & 1 & 1 \\ 
1 & 1 & 0 & 0 \\
0 & 0 & 0 & -3
\end{array}
\right]
$$
and the equation $0 x + 0 y + 0 z = -3$ corresponding to the last row can not be fulfilled, this means no solution in this case.
Case 3:
If $a \ne 1$ and $a \ne 5/4$ we have:
$$
\left[
\begin{array}{rrr|r}
1-a   & 0    & 1 & 1 \\ 
1 & 1 & 0 & 0 \\
1  & 0 & 0 & -3/(-5+4a)
\end{array}
\right]
$$
and the unique solution
$$
(x, y, z) = \left( 
-\frac{3}{-5+4a}, 
\frac{3}{-5+4a}, 
1+\frac{3}{(1-a)(-5+4a)}
\right)
$$
