Let A be an n # n matrix, b be a nonzero vector, and x0 be a solution vector of the system Ax D b. Show that x is a solution of the nonhomogeneous system Ax D b if and only if y D x!x0 is a solution of the homogeneous system Ay D 0.

Complete Question

Let A be an n x n matrix, b be a nonzero vector, and x_0 be a solution vector of the system Ax = b. Show that x is a solution of the non-homogeneous system Ax = b if and only if y = x - x_0 is a solution of the homogeneous system Ay = 0.

Step-by-step explanation:

From the question we are told that

    A is an  n × n matrix

   b is a zero vector

  us the solution vector of  

Which implies that  

So first we show that

  if is the solution matrix of

  and   is the solution of  

Then

=>  

=>  

Secondly to show that

   if   is the solution of  

  then x is the solution of the non-homogeneous system

Now we know that is the solution of  

So

=>  

=>  

=>  

=>    

Thus this has been proved

the answer is a

step-by-step explanation:

because your just changing the position of where the shape is

4

step-by-step explanation:

6 (donuts) divided by 3 is 2.

so 2/3 means 2 times 2, which is 4/6.

you get two and your friend gets four.