MHB Question for null space of a matrix

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For a 4×3 matrix A, if the null space N(A) is {0}, the linear system Ax=c has a unique solution, indicating that A is full rank. Conversely, if N(A) is not {0}, the system Ax=c will have either no solutions or infinitely many solutions, depending on the relationship between c and the column space of A. The discussion emphasizes the importance of understanding the null space in determining the nature of solutions to linear systems. Participants are encouraged to share their progress to facilitate more effective assistance. Overall, the nature of the null space directly impacts the solution set of the matrix equation.
shiecldk
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Let A be a 4×3 matrix and let
c=2a1+a2+a3

(a) If N(A) = {0}, what can you conclude about the solutions to the linear system Ax=c?

(b) If N(A) ≠ {0}, how many solutions will the system Ax=c have? Explain.
 
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shiecldk said:
Let A be a 4×3 matrix and let
c=2a1+a2+a3

(a) If N(A) = {0}, what can you conclude about the solutions to the linear system Ax=c?

(b) If N(A) ≠ {0}, how many solutions will the system Ax=c have? Explain.

Hello and welcome to MHB! :D

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Can you post what you have done so far?
 
Thread 'How to define a vector field?'

Thread 'How to define a vector field?'

Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...
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