MHB Creating a 4D System with a Specific Solution: How Can It Be Done?

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To create a 4D system with the solution (-2, 5, -6, 1), the simplest approach is to directly assign each variable to the corresponding value. A basic system can be structured as x1 = -2, x2 = 5, x3 = -6, and x4 = 1. If a matrix format is required, using the identity matrix with these values as the right-hand side is also valid. The choice of format may depend on the instructor's preferences regarding what constitutes a "system." Ultimately, both methods yield the same solution.
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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?
 
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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?

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.
 
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.

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?
 
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?

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.
 
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.
I think this is exactly what he wanted, thank you!
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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