Can systems of 3 linear equations with 3 unknowns have more than one solution?

Question

In each part,determine whether the given vector is a solution of the linear system

\begin{align} 2x-4y-z&=1\\ x-3y+z&=1\\ 3x-5y-3z&=1 \end{align}

(a) $(3,1,1)$ (b) $(3,-1,1)$ (c) $(13,5,2)$ (d) $(13/2,5/2,2)$ (e) $(17,7,5)$

It's easy to solve this question. Just plug in the given vector into 3 equations respectively and check that if the left and side = the right hand side. And it turns out that a.d.e meet the restrictions of all 3 equations.

But my question is: why is that possible? Since for a linear system with n equations and n unknowns, it has only 1 unique solution. Geometrically, this linear system is like 3 planes, and the solution is a point when these 3 planes coincide. So, I think there's only 1 point that can suit into this linear system.

Answer 1

As Doggyshakespeare said, your system is not full rank, which means there are redundant equations. In general, you can calculate the determinant of the matrix of your system to know if there is an infinite number of solutions for your system. If it is 0, it means that your system is not full rank and that you have an infinite number of solutions.

In your case, your system can be written as : $$ Ax=b $$ with $$ A = \left[ \begin{array}{ccc} 2 & -4 & -1 \\ 1 & -3 & 1 \\ 3 & -5 & -3 \end{array} \right] , x=\left[ \begin{array}{ccc} x \\ y \\ z \end{array} \right] , b=\left[ \begin{array}{ccc} 1 \\ 1 \\ 1 \end{array} \right] $$

And here you have $\det(A) = 0$ so that is the reason.

Answer 2

If you subtract the first equation from the third equation, you get the same thing as when you subtract the second from the first. That means that one of the equations is redundant, so you can have more than 1 answer. It's the same thing as all three equations parallel to and intersecting at one single line.

Answer 3

What about the system of equations: $$x + y =2$$ and $$2x+2y = 4$$ In fact, these two equations define the same line in the plane, so they have an infinite number of solutions. Any point of the form $(x, 2-x)$ will satisfy both equations.

What if I changed the system to be this: $$x + y =2$$ and $$2x+2y = 5$$ Now you will see that there are in fact no solutions to this system of equations. This system represents two parallel but distinct lines, so there are no common points of intersection.

This same idea extends as you say for $n$ equations with $n$ unknowns. There can either be $0$ solutions, exactly $1$ solution, or infinitely many solutions. To imagine a scenario where three planes can intersect in a line with infinitely many points of intersection, think about taking the x-y plane, the y-z plane and one that is between these two at say 45 degrees. They will all intersect along the y-axis. You could also have a situation where there are two parallel planes that never touch, which would give you no solutions.

By taking the determinant of the matrix corresponding to the system, you can tell immediately when there is a unique solution to the system precisely when the determinant is non-zero.

Answer 4

Yes, they can have more than one solution.

Each equation $\alpha \cdot x = \beta$ describes an affine plane.

• If 2 of your 3 planes are identical (and different from the 3rd), then the intersection might be a line, thus infinite many solutions, or is empty in case of parallel planes.

• If 3 of your 3 planes are identical, then the intersection is a plane and you have infinite many solutions.

Example: $$ 1 = x + y + z = (1,1,1) \cdot (x,y,z) \\ 2 = 2x + 2y + 2z = 2 \, (1,1,1) \cdot (x,y,z) \\ 4 = 3x + 3y + 3z = 3 \, (1,1,1) \cdot (x,y,z) $$ The second is just the first equation multiplied by 2 on each side of the equal sign.

Answer 5

the standard(?) way to do this problem is to row reduce the augmented matrix $[A|b].$ when i do that, this is what i get:

$$\left(\begin{array}{ccc|c}2&-4&-1&1\\1&-3&1&1\\3&-5&-3&1\end{array}\right) \to \left(\begin{array}{ccc|c}1&0&-3.5&-0.5\\0&1&-1,.5&-0.5\\0&0&0&0\end{array}\right)$$ so you have a free variable $z.$ now the equations/solutions are $$\pmatrix{x\\y} = \pmatrix{-0.5\\-0.5} + 0.5z\pmatrix{7\\1}, z \text{ arbitrary} $$