Question

Suppose \(A\) is a \(3 \times 3\) matrix and \(y\) is a vector in \({\mathbb{R}^3}\) such that the equation \(Ax = y\) does not have a solution. Does there exist a vector \(z\) in \({\mathbb{R}^3}\) such that the equation \(Ax = z\) has a unique solution? Discuss.

Short answer

The equation \(Ax = z\) contains at most two basic variables and at least one free variable. Therefore, the solution set for the equation \(Ax = z\) is either empty or contains infinitely many elements.

Step-by-step solution

Determine the pivot position of matrix \(A\)

If \(Ax = y\) does not have a solution, then \(A\) cannot have a pivot in every row. \(A\) has at most two pivot positions since it is a \(3 \times 3\) matrix.

Determine the basic variable and free variable for the vector

The variable corresponding to pivot columns in the matrix is called the basic variable.The other variable is called the free variable.

For any value of \(z\), the equation \(Ax = z\) contains at most two basic variables and at least one free variable.

Determine whether the equation \(Ax = z\) has a unique solution

The equation \(Ax = z\) contains at most two basic variables and at least one free variable. It does not have a unique solution. Therefore, the solution set for the equation \(Ax = z\) is either empty or contains infinitely many elements.