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!
 
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