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.

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.

Answer:

Step-by-step explanation:

From the question we are told that

A is an n × n matrix

b is a zero vector

[tex]x_o[/tex] us the solution vector of [tex]Ax = b[/tex]

Which implies that

[tex]Ax_o = b[/tex]

So first we show that

if [tex]x[/tex] is the solution matrix of [tex]Ax = b[/tex]

and [tex]y= x-x_o[/tex] is the solution of [tex]Ay = 0[/tex]

Then

[tex]A(x-x_o) = 0[/tex]

=> [tex]Ax -Ax_o = 0[/tex]

=> [tex]b-b = 0[/tex]

Secondly to show that

if [tex]y= x-x_o[/tex] is the solution of [tex]Ay =0[/tex]

then x is the solution of the non-homogeneous system

[tex]Ax = b[/tex]

Now we know that [tex]y = x-x_o[/tex] is the solution of [tex]Ay =0[/tex]

So

[tex]Ay = 0[/tex]

=> [tex]A(x- x_o) = 0[/tex]

=> [tex]Ax - Ax_o = 0[/tex]

=> [tex]Ax - b = 0[/tex]

=> [tex]Ax = b[/tex]

Thus this has been proved