homogeneous linear systems, finding scalar I'm struggling with finding a solution to this problem.
For what values of b is the solution set of this linear system:
$x_{1} + x_{2} + bx_{3} = 0$
$x_{1} + bx_{2} + x_{3} = 0$
$bx_{1} + x_{2} + x_{3} = 0$
equal to the origin only, a line through the origin, a plane through the origin, or all of $R^3$
I understand that for the solution set of a linear system Ax=0 to be a single point through the origin only, Ax=0 has only the trivial solution.
Also that a line through the origin means Ax=0 has one free variable.
And a plane through the origin means Ax=0 has 2 free variables.
I can't figure out how to accurately describe b to reflect each of these, I think I'm confusing myself. Would someone mind walking through this step by step with me? Thanks.
 A: If we perform row reduction,
$
\begin{bmatrix}
1&1&b\\
1&b&1\\
b&1&1
\end{bmatrix}
$
$\xrightarrow[R3-R1]{R2-R1}$
$
\begin{bmatrix}
1&1&b\\
0&b-1&1-b\\
b-1&0&1-b
\end{bmatrix}
$
We see that if $b=1$, the second and third row are zero, and the system is reduced to the plane $x_1+x_2+x-3=0$. 
If $b\ne1$, we can continue
$
\begin{bmatrix}
1&1&b\\
0&b-1&1-b\\
b-1&0&1-b
\end{bmatrix}
$
$\xrightarrow{(b-1)R1}$
$
\begin{bmatrix}
b-1&b-1&b(b-1)\\
0&b-1&1-b\\
b-1&0&1-b
\end{bmatrix}
$
$
\xrightarrow{R3-R1}
$
$
\begin{bmatrix}
b-1&b-1&b(b-1)\\
0&b-1&1-b\\
0&1-b&(1-b)^2
\end{bmatrix}
$
$
\xrightarrow{R3+R2}
$
$
\begin{bmatrix}
b-1&b-1&b(b-1)\\
0&b-1&1-b\\
0&0&(1-b)(2-b)
\end{bmatrix}
$
So if the $3,3$ entry is nonzero the matrix is in reduced echelon form and so the system has unique solution. 
The conclusion is: $b=1$ gives $2$ free variables, $b=2$ gives one free variable, any other value of $b$ gives unique solution. 
A: HINT:
Start by writing your set of linear equations:
$x_{1} + x_{2} + bx_{3} = 0$
$x_{1} + bx_{2} + x_{3} = 0$
$bx_{1} + x_{2} + x_{3} = 0$
in matrix determinant form:
$$\begin{vmatrix}
1 & 1 & b \\
1 & b & 1 \\
b & 1 & 1 \\
\end{vmatrix}
$$
A: The rank of the coefficient matrix determines everything. The rank of the matrix may be $3$ (which means non-zero determinant) or $< 3$ (which means zero determinant).
Consider the determinant of the matrix and equate it to $0$
$ 
\begin{vmatrix}
1 & 1 & b \\ 
1 & b & 1 \\ 
b & 1 & 1 \notag
\end{vmatrix} = 0$
Try $R_1 \leftarrow R_1 - R_2$ and have 
$ 
\begin{vmatrix}
0 & 1-b & b-1 \\ 
1-b & b-1 & 0 \\ 
b & 1 & 1 \notag
\end{vmatrix} = 0$
Next try $C_3 \leftarrow C_3+C_2$ followed by  $C_2 \leftarrow C_2 + C_1$ to have 
$ 
\begin{vmatrix}
0 & 1-b & 0 \\ 
1-b & 0 & b-1 \\ 
b & b+1 & 2 \notag
\end{vmatrix} = 0$
Next take the factor $(1-b)$ out from both $R_1$ and $R_2$ to have 
$ (b-1)^2
\begin{vmatrix}
0 & 1 & 0 \\ 
1 & 0 & -1 \\ 
b & b+1 & 2 \notag
\end{vmatrix} = 0$
$\implies (b-1)^2(b+2) = 0$
Note that the rank of the matrix is 
$1$ for $b = 1$ ($\implies$ Plane) and 
$2$ for $b = -2$ ($\implies$ Straight line) and
$3$ ($\implies$ a point, i.e., the origin) otherwise.
