For which sets of parameters will this system of equations have no solution?

Question

If $S$ is the set of distinct values of '$b$' for which the following system of linear equations

\begin{align} x + y + z = 1\\ x + ay + z = 1\\ ax + by + z = 0 \end{align} has no solution, then $S$ is:

1. a singleton
2. an empty set
3. an infinite set
4. a finite set containing two or more elements

There's no information given about value of $a$. You will be getting two cases:

• For $a=1, b$ has one value $b=1$
• For $a$ not equal to $1 , b$ has no value.

Answer 1

Suppose $a=1$. Then, the first two equations are the same.

The third equation is $x+by+z = 0$. We know that $x+y+z=1$, so subtracting, we get $y(1-b) = 1$, so that if $b\neq 1$, we have a solution for $y$, and trivially for $x,z$. Hence, $b=1$ is the only value that doesn't give solutions.

Suppose $a \neq 1$. Then, subtracting second equation from first, $(a-1)y = 0$, so that $y=0$.

Now the third equation changes to $ax + z =0$, and we know $x+z = 1$. These equations can be solved, with $x = \frac 1{1-a}$ and $z = 1-x$. These solutions exist regardless of the value of $b$, since they don't depend on $b$.

Hence, solutions will exist whenever $a \neq 1$.

This means that $b$ is a singleton (and you are correct, good job!)

Answer 2

$$\begin{bmatrix} 1 & 1 & 1 \\ 1 & a & 1 \\ a & b & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix}$$

$$\begin{bmatrix} 1 & 1 & 1 \\ 0 & a-1 & 0 \\ 0 & b-a & 1-a \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ -a\end{bmatrix}$$

The determinant of the matrix is $-(a-1)^2$, to have no solution, the determinant has to be $0$, hence $a=1$.

$$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & b-1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ -1\end{bmatrix}$$

To have no solution $b$ has to be $1$.

Hence $S=\{1 \}$.

Remark:

I interpret the question as $S=\{b: \exists a \text{ such that the system has no solution.} \}$

If we interpret $S=\{b: \forall a \text{ such that the system has no solution.} \}$, then $S = \emptyset$.

Answer 3

When $a=1$ and $b=1$ The first two equations are coincident planes ($x+y+z=1$)and the last one ($x+y+z=0$) is a plane parallel to them .

Therefore for $b=1$, these equations have no solutions at all.(Since parallel planes will intersect no where.)

Hence, your set $S=\{1 \}$