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Question: Maple gives trivial solution to linear system with variable parameters, but there are non-trivial solutions for certain parameter values.

Question:Maple gives trivial solution to linear system with variable parameters, but there are non-trivial solutions for certain parameter values.

zenterix 350 Maple 2022
linear-algebra parametric
October 06 2023
1

Consider the following. 

We have a matrix A with five parameters in it.

If we tell Maple to solve Ax=0, it gives us the trivial solution.

Then, if I make a matrix B where I choose specific values for those five parameters and I tell Maple to solve Bx=0 I get a non-trivial solution.

Why doesn't Maple give me a more informative result in the Ax=0 case? 

A := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = a__1, (2, 1) = 0, (2, 2) = 1, (2, 3) = 0, (2, 4) = 0, (2, 5) = b__1, (3, 1) = -216, (3, 2) = -18, (3, 3) = 0, (3, 4) = 0, (3, 5) = c__1, (4, 1) = 0, (4, 2) = 0, (4, 3) = 1, (4, 4) = 0, (4, 5) = d__1, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = 1, (5, 5) = e__1}) = Matrix(%id = 36893488151959194068)NULL

with(LinearAlgebra)

GaussianElimination(A)

Matrix(%id = 36893488151959184316)

 (1)

LinearSolve(A, `<,>`(0, 0, 0, 0, 0))

Vector[column](%id = 36893488151959183100)

 (2)

B := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 0, (2, 2) = 1, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (3, 1) = -216, (3, 2) = -18, (3, 3) = 0, (3, 4) = 0, (3, 5) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 1, (4, 4) = 0, (4, 5) = 1, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = 1, (5, 5) = 4}) = Matrix([[1, 0, 0, 0, 0], [0, 1, 0, 0, 0], [-216, -18, 0, 0, 0], [0, 0, 1, 0, 1], [0, 0, 0, 1, 4]])NULL

GaussianElimination(B) = Matrix(%id = 36893488151959174788)NULL

LinearSolve(B, `<,>`(0, 0, 0, 0, 0))

Vector[column](%id = 36893488151959164916)

 (3)

``

NULL

An even easier way to show a non-trivial solution is to just make the entire last column zero.

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