Suppose A is a 3 times 33*3 matrix and y is a vector in set of real numbers R cubedℝ3 such that the equation Upper A Bold x equals Bold yAx=y does not have a solution. Does there exist a vector z in set of real numbers R cubedℝ3 such that the equation Upper A Bold x equals Bold zAx=z has a unique solution?

Question

Mathematics

Ana Odom

4 January, 11:57

Suppose A is a 3 times 33*3 matrix and y is a vector in set of real numbers R cubedℝ3 such that the equation Upper A Bold x equals Bold yAx=y does not have a solution. Does there exist a vector z in set of real numbers R cubedℝ3 such that the equation Upper A Bold x equals Bold zAx=z has a unique solution?

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Answers (1)

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Payton Stevens · Answer 1

No

Step-by-step explanation:

If A were reversible, then such y coudnt not exist, because A⁻¹y would be a solution of the equation Ax=y. This means that A in not reversible, thus, there exist a non zero vector w in R³ such that A*w = 0. If x' were a solution of Ax = z, then x'' = x'+w is a different solution of the equation: Ax'' = A (x'+w) = Ax'+Aw = z+0 = z. This means that a vector z such that there exist a unique solution for the equation Ax=z cant exist.