Gaussian elimination - number of solutions

Question

How do I know how many solutions does a system of linear equations have?

I have such system of linear equations:

$\begin{cases} (1+b)x+y+z=1\\ x+(1+b)y+z=b\\ x+y+(1+b)z=b^2 \end{cases}$

And I would like to know if it has zero, one, or more solutions depanding on $b$ ($b\in\mathbb{R}$).

How do I do that?

Answer 1

Hint: write down the augumented matrix and apply Gaussian elimination to it.

Added: Here is what you should get

$$ \left[ \begin {array}{cccc} 1+b&1&1&1\\0&{\frac {b \left( 2+b \right) }{1+b}}&{\frac {b}{1+b}}&{\frac {b+{b}^{2}-1}{1+b} }\\0&0&{\frac { \left( b+3 \right) b}{2+b}}&{\frac { {b}^{3}+2\,{b}^{2}-b-1}{2+b}}\end {array} \right] .$$

Now, you should analyse the last row. See here for another approach.

Answer 2

suppose that $b=0$ then you get

$x+y+z=1$

$x+y+z=0$

$x+y+z=0$

clearly it has not solution,of course you can choose different $b$ and test it,or try to reudce it to row echelon form and determine how given system is consistent or has solution

Answer 3

Your matrix is $$bI_{3 \times 3} + \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix}\begin{bmatrix} 1& 1& 1\end{bmatrix}$$ The inverse of this matrix by Sherman Morrison Woodbury formula is $$\dfrac{I_{3 \times 3}}b - \dfrac1{b(b+3)}\begin{bmatrix} 1\\ 1\\ 1\end{bmatrix}\begin{bmatrix} 1& 1& 1\end{bmatrix}$$ Hence, $b$ should not be equal to $0$ or $-3$ for the system to have a solution.

In the $n \times n$ case, $b$ should not be equal to $0$ or $-n$ for the system to have a solution.