What Value of 'a' Yields One, None, or Infinite Solutions?

[TTaJTa4]

(a-2)x+2y+3z=-1
y+z-w=0
2(1+a)z+(1+a)w=3

what a will give us:
1. One Solution
2. No solution at all.
3. Infinite Solutions

Thank you!

 

[MarkFL]

Hello and welcome to MHB! :D

Can you post what you have tried so far so our helpers can see where you are having difficulty and offer the best help?

 

[TTaJTa4]

Hello and welcome to MHB! :D

Can you post what you have tried so far so our helpers can see where you are having difficulty and offer the best help?

I understand that if a=-1 than there is no solution. I tried to row the matrix but I couldn't finish it.

 

[Evgeny.Makarov]

The augmented matrix of this system:
\[
\left(\begin{array}{@{}{cccc}|c@{}}
a-2 & 2 & 3 & 0 & -1\\
0 & 1 & 1 & -1 & 0\\
0 & 0 & 2(1+a) & 1+a & 3
\end{array}\right)
\]
is already in row echelon form. You are correct that if $a=-1$, then the last equation becomes $0z+0w=3$, and it has no solutions. If $a\ne-1$, then we can solve that equation for $z$:
\[
z=\frac{3-(1+a)w}{2(1+a)}
\]
so every value of $w$ gives a unique value of $z$ that satisfies the last equation. From the second equation, given $w$ and $z$, we can find a unique $y$ that satisfies the second equation. Finally, if $a\ne2$, then given $w$, $z$ and $y$ we can find a unique $x$ that satisfies the first equation. All this happens for every value of $w$, so in this case there are infinitely many solutions.

What happens when $a=2$? Consider the system without $x$ (whose augmented matrix lacks the first column) and determine how many solutions it has. But the coefficient of $x$ is 0 in all equations, so $x$ can be arbitrary.

 

[blue_raver22]

The augmented matrix for this problem would be:

| a-2 2 3 -1 |
| 0 1 1 0 |
| 2(1+a) 0 1+a 3 |

To solve this matrix, we can use Gaussian elimination or other matrix solving methods. The value of "a" will determine the type of solution we will get.

1. If a is any real number, we will get one unique solution for x, y, and z. This is because all the variables are present in the first two equations, and the third equation is a multiple of the second equation, so it does not add any new information.

2. If a = 2, the third equation becomes 4z + 4w = 3, which is a contradiction to the second equation. This means that there is no solution that satisfies all three equations. This can also happen if the first two equations are inconsistent (e.g. if a = 0, the first equation becomes -2x + 2y + 3z = -1, and the second equation becomes y + z - w = 0, which has no solution).

3. If a = -1, the third equation becomes 0 = 3, which is also a contradiction to the second equation. However, in this case, the first two equations are consistent and the third equation is redundant. This means that there are infinitely many solutions that satisfy the first two equations, but the third equation does not add any new information.

In summary, the value of "a" can give us different types of solutions for this augmented matrix problem. It is important to carefully examine the equations and their relationships to determine the type of solution.