Linear transformations - Find x such that T(x) = b

Question

I need help with a problem in Linear Algebra:

Given a matrix $$A = \begin{bmatrix} 1 & 2 & 1\\ 1 & -1 & 1\\ 2 & 1 & 1 \end{bmatrix}$$

and a vector

$$b = \begin{bmatrix} 4\\ 1\\ 4 \end{bmatrix}$$

Find $x$ such that $T(x) = b$

What I have tried:

I know that $T(x) = Ax = b$, so plugging in yields Ax = b. Rewriting as an augmented matrix and simplifying, we get the reduced row echelon form

$$ \begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1\\ \end{bmatrix}$$

However, I do not know how to proceed.

Answer 1

You get the reduced row echelon form $$ \begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1\\ \end{bmatrix},$$ which gives the solution
$$ x=\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}.$$ Indeed, $$Ax = \begin{bmatrix} 1 & 2 & 1\\ 1 & -1 & 1\\ 2 & 1 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}=\begin{bmatrix} 4\\ 1\\ 4 \end{bmatrix}.$$

Answer 2

Find $A^{-1}$

Then $a= A^{-1}.b$

This could be a method to find $a$ if $A$ is non singular matrix, because

$a= A^{-1}.b$

$ \implies b=A.a$