homogeneous linear systems, finding scalar

Question

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.

Answer 1

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: $$\begin{vmatrix} 1 & 1 & b \\ 1 & b & 1 \\ b & 1 & 1 \\ \end{vmatrix} $$

Answer 2

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.

Answer 3

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.