Question

A is a 3 times 3 matrix with three pivot positions.

(a) Does the equation Ax = 0 have a nontrivial solution?
(b) Does the equation Ax = b have at least one solution for every possible b?
(a) Does the equation Ax = 0 have a nontrivial solution? No Yes
(b) Does the equation Ax = b have at least one solution for every possible b? No Yes

Answer 1

Matrix A with three pivot positions means Ax = 0 only has the trivial solution, and Ax = b will have at least one solution for any given b because A is invertible and its columns span the entire three-dimensional vector space.

Step-by-step explanation:

The question pertains to linear algebra and the solutions to systems of linear equations represented in matrix form. Specifically, we are dealing with a 3x3 matrix A which has three pivot positions.

(a) If matrix A has three pivot positions, it means that A is of full rank, and thus, the equation Ax = 0 only has the trivial solution, x=0. There are no free variables, and as a result, there are no nontrivial solutions to this homogeneous equation.

(b) Since A is a 3x3 matrix with three pivot positions, it is invertible, which implies that the equation Ax = b has exactly one solution for every possible b. This is because the columns of A span ℝ^3, the entire space of 3-dimensional vectors, ensuring that every b in ℝ^3 can be expressed as a combination of these columns.

Answer 2

The following are the answer:

In option a "No".

In option b "Yes".

Step-by-step explanation:

In choice a:

Ax = 0 has no nontrivial solution. A would be the three-pivot matrix, it may assume, that the function has no free variable, and only if the function has had at least one free factor are their nontrivial formulas for the equations of the form Ax=0.

It implies that since A is a 3x3 matrix, has no free variables so that it has no non-trivial choices, and Ax = 0.

In choice b:

we assume that every potential has at least one solution that is Ax=b . If A does have a three-pivot matrix, It will be a pivot element for each row and column, and for each possible b∈ R³, Ax = b has at least one solution.