The discussion focuses on creating a four-dimensional system with a specified solution of (-2, 5, -6, 1). The simplest approach is to directly assign the values to the variables: x1 = -2, x2 = 5, x3 = -6, and x4 = 1. Alternatively, for a matrix representation, an identity matrix can be used alongside the specified solution vector. This method is effective and aligns with standard mathematical practices for representing systems of equations.
PREREQUISITESStudents studying linear algebra, educators teaching systems of equations, and anyone interested in mathematical problem-solving techniques involving multi-dimensional systems.
sawdee said:I know how to solve systems, but the question asks to work backwards.
"Create a 4-d system that has a solution (-2,5,-6,1)."
How would i do this?
I like Serena said:Hi sawdee! Welcome to MHB! (Smile)
Without any other conditions, we can go for the simplest system that satisfies your requirement.
We can pick the system:
\begin{cases}
x_1=-2 \\
x_2 = 5 \\
x_3 = -6 \\
x_4 = 1
\end{cases}
No point in making it more complicated unless there are more restrictions.
We're talking about math - the simplest solution possible should be considered the best one.
sawdee said:Thank you! :) Happy to be here.
I picked that solution as well, but in terms of format I'm not sure how to set up the answer. Should I use a matrix with those 4 as my x values, and leave everything else as 0?
I think this is exactly what he wanted, thank you!I like Serena said:It depends a bit on what your teacher considers to be a "system".
If it's supposed to involve a matrix, we can pick the identity matrix with right hand values that correspond to the solution.
That is:
$$\begin{bmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1\\
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2\\
x_3 \\
x_4
\end{bmatrix} =
\begin{bmatrix}
-2 \\
5 \\
-6 \\
1
\end{bmatrix}$$
It's really just the same thing as the "system" I suggested though.
The only difference being what we pick as the notation.