Matrices - Infinite/No solutions

  • Thread starter planauts
  • Start date
  • #1
86
0

Homework Statement


Find the values of a and b such that the equations:
3x + ay = 2 and -6x + 4y = b
have i) an infinite set of solutions ii) no solutions

The Attempt at a Solution



[itex]\begin{pmatrix}
3 & a \\
-6 & 4
\end{pmatrix} * \begin{pmatrix}
x\\
y
\end{pmatrix}
= \begin{pmatrix}
2\\
b
\end{pmatrix}[/itex]

[itex]
\begin{pmatrix}
x\\
y
\end{pmatrix}
=
\begin{pmatrix}
3 & a \\
-6 & 4
\end{pmatrix}^{-1} *
\begin{pmatrix}
2\\
b
\end{pmatrix}
[/itex]


[itex]
\begin{pmatrix}
x\\
y
\end{pmatrix}
=
\tfrac{1}{12+6a} *

\begin{pmatrix}
4 & -a \\
6 & 3
\end{pmatrix} *
\begin{pmatrix}
2\\
b
\end{pmatrix}

[/itex]

I think that when the matrix is singular, then it does not have ONE solution (infinite OR no solution). So when a = -2 then it has infinite OR no solution. But what about b, how do I figure out the value for b?

Thanks
 

Answers and Replies

  • #2
34,678
6,387

Homework Statement


Find the values of a and b such that the equations:
3x + ay = 2 and -6x + 4y = b
have i) an infinite set of solutions ii) no solutions

The Attempt at a Solution



[itex]\begin{pmatrix}
3 & a \\
-6 & 4
\end{pmatrix} * \begin{pmatrix}
x\\
y
\end{pmatrix}
= \begin{pmatrix}
2\\
b
\end{pmatrix}[/itex]

[itex]
\begin{pmatrix}
x\\
y
\end{pmatrix}
=
\begin{pmatrix}
3 & a \\
-6 & 4
\end{pmatrix}^{-1} *
\begin{pmatrix}
2\\
b
\end{pmatrix}
[/itex]


[itex]
\begin{pmatrix}
x\\
y
\end{pmatrix}
=
\tfrac{1}{12+6a} *

\begin{pmatrix}
4 & -a \\
6 & 3
\end{pmatrix} *
\begin{pmatrix}
2\\
b
\end{pmatrix}

[/itex]

I think that when the matrix is singular, then it does not have ONE solution (infinite OR no solution). So when a = -2 then it has infinite OR no solution. But what about b, how do I figure out the value for b?

Thanks
Certainly, if a = -2, the matrix is singular (noninvertible), so solving for a vector <x, y> as you have done is invalid.

If a = -2, the two equations are:
3x - 2y = 2
-6x + 4y = b

In order for there to be an infinite number of solutions, the system above must actually represent a single line in the plane. For such a system, any point on one line is already on the other line, which is really the same line. Since there are an infinite number of points on the line, there are an infinite number of solutions to the system. What value of b gives us two equivalent equations, and hence one line?

If b is some other value, then there really are two distinct, parallel lines. No point on either line can also be on the other line, so the system is inconsistent, which means that the system has no solutions. What are the possible values of b so that the system above represents two parallel lines?
 
  • #3
86
0
Wow that makes sense. That's so easy. I could have simply guessed the a value by inspection :(

Thanks!
 
  • #4
34,678
6,387
It's helpful to not get too lost in matrix notation and all, and keep the geometry in mind, at least for simple systems of two or three equations. In a system of two equations in two unknowns, the system represents two lines. The lines can intersect at a single point, at every point, or not at all.

For a system of three equations in three unknowns, each equation represents a plane in space. The three planes can intersect at a single point, or in a line, in an entire plane, or not at all.
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
961
Personally, I would not have used matrices at all. To solve the equations 3x + ay = 2 and -6x + 4y = b, eliminate x by adding twice the first equation to the second: (6x+ 2ay)+ (-6x+ 4y)= (2a+ 4)y= 4+ b.

That equation has a unique solution as long as 2a+ 4 is not 0. If it is 0, then the left side of the equation is 0 for all y. There is no solution as long as 4+ b is not 0 but any y will be a solution if 4+ b= 0.
 

Related Threads on Matrices - Infinite/No solutions

  • Last Post
Replies
5
Views
7K
Replies
5
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
3K
Replies
1
Views
2K
Replies
2
Views
726
Replies
2
Views
2K
Replies
0
Views
2K
Top