Suppose A is a 3X3 matrix and b is a vector in R3 such that the equation Ax = b does not have a solution. Does there exist a vector y in R3 such that the equation Ax = y has a unique solution?

Respuesta :

A 3 × 3 matrix has three rows and three columns, while a zero row is a row

that has all the elements as zero

The correct response option is as follows;

No. Since Ax = y has no solution, then A cannot have a pivot in every row.

Since A is a 3 × 3, it has at most two pivot positions. So the equation Ax = z

has at most two basic variables and at least one free variable for any z.

Thus there is no solution for Ax = z

Reason:

There are no free variables if Ax = z has a solution that is unique, given that

all the rows would have a pivot. However, there exist a zero row '0 0 0' in

the row reduced Echelon form of the coefficient matrix A, given that Ax = y

has no solution. and the matrix A cannot have a pivot at every row [0 0 0 1]

Also there would be no free columns in the row reduced matrix A where Ax

= z has a solution that is unique, because it will result in [Az] having free

variables, which is contradicting the fact that matrix A has a free column

Therefore, a vector y ∈ R³ such that Ax = y has a unique solution does not exist

Therefore;

No. Since Ax = y has no solution, then A cannot have a pivot in every row.

Since A is a 3 × 3, it has at most two pivot positions. So the equation Ax = z

has at most two basic variables and at least one free variable for any z.

Thus there is no solution for Ax = z

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