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Question: 3 unknowns, 1 matrix, and 1 variable

Question:3 unknowns, 1 matrix, and 1 variable

The function 110 Maple 2019
linear-algebra
October 03 2022
1

Dear everyone,

Ive been looking at this problem ive got, but i cant find out what would make it work. I can put numbers in a matrix, i can solve 3 unknowns with 3 equations. But now there is a variable, and the variable causes the solution to give either: 1 solution, no solution, or more than one solution. 

I can "brush the matrix" as they say it in Dutch, but im learning Maple so it can go a lot faster than manual labor. So is there a way to do it with Maple? 

Figure 4.1 gives from left to right: 1 solution, no solution, and an indefinite amount of solutions.

Questions: For what value of p, a. gives 1 solution? b. gives no solution? c. gives more than 1 solution?

Now i got Maple to tell me how much p x1, x2, and x3 was, but its not bringing me anywhere to where i can get to a solution. 

A manual method is given in this example. Manual is way way to slow for what i need to achieve, ive got some large big books with mechanics, dynamics, statics, fluid mechanics, concrete calcutions, chemistry etc. waiting for me. I can never achieve that manually in the same time as i could do it with maple. :S 

> 

#example of how to do a solve of a matrix with 3 variables:
> 

stelsel := {x1+x2+x3 = 4, 3*x1+x2-5*x3 = -10, 3*x1+2*x2-x3 = 1}

{x1+x2+x3 = 4, 3*x1+x2-5*x3 = -10, 3*x1+2*x2-x3 = 1}

 (1)
> 

solve(stelsel, {x1, x2, x3})

{x1 = -7+3*x3, x2 = 11-4*x3, x3 = x3}

 (2)
> 

#now the example number 15
> 

stelsel := {(2-p)*x2+x3 = 2, -x1+2*x2+(2-p)*x3 = 0, (1-p)*x1+2*x2+2*x3 = p+3}

{(2-p)*x2+x3 = 2, -x1+2*x2+(2-p)*x3 = 0, (1-p)*x1+2*x2+2*x3 = p+3}

 (3)
> 

solve(stelsel, {x1, x2, x3})

{x1 = -(p^2-6)/(p^2-4*p+4), x2 = -(2*p-5)/(p^2-4*p+4), x3 = 1/(-2+p)}

 (4)
> 

``

Thank you!

Greetings,

The Function

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