Does the equation Axequalsb have a solution for each b in set of real numbers RSuperscript 4? A. No, because A has a pivot position in every row. B. Yes, because the columns of A span set of real numbers RSuperscript 4. C. Yes, because A does not have a pivot position in every row. D. No, because the columns of A do not span set of real numbers R

C. Yes, because A does not have a pivot position in every row.

Step-by-step explanation:

The pivot position in the matrix is determined by entries in non zero rows. The pivot position may be in the row or a column. By Invertible Matrix Theorem the equation Axequalsb has non trivial solution. A has fewer pivot positions therefore A is not invertible. Ax will map RSuperscript into real numbers for n times. A has pivot position if left parenthesis bold x right parenthesis.

for the points

step-by-step explanation: