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?
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.
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.
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
