1
$\begingroup$

How can I find $T^{-1}(x,y,z)$ (inverted matrix) of a linear operator $T:V_3 \to V_3$, which matrix relative to the basis: $A=\{ (1,0,0), (1,1,0), (1,1,1)\}$ is:

$$T_A= \begin{bmatrix} 2 &0 &1\\ 1 &-1 &1\\ -1 &2 &-1\\ \end{bmatrix} $$

$\endgroup$

1 Answer 1

1
$\begingroup$

Hint:

Find the matrix of $T$ wrt standard ordered basis,then find its inverse

Solution:

let $a=(1,0,0),b=(1,1,0),c=(1,1,1)$

Hence $T(a)=2a+b-c=(2,0,-1),T(b)=-b+2c=(1,1,2),T(c)=a+b-c=(1,0,-1)$ [This is concluded from the given matrix]

Thus $(0,1,0)=b-a\Rightarrow T(0,1,0)=T(b)-T(a)=(-1,1,3)$

And $(0,0,1)=c-b\Rightarrow T(0,0,1)=T(c)-T(b)=(0,-1,-3)$

So we have that matrix of $T$ wrt standard ordered basis is

$$\begin{pmatrix} 2 & -1 & 0\\ 0 & 1 & -1 \\ -1 & 3 & -3\end{pmatrix}$$

Now find its inverse.

(Please let me know if there is any calculative mistake)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .