Question

Let T be the linear operator on R

3 defined by

T(x1, x2, x3) = (3x1, x1 − x2 , 2x1 + x2 + x3 )

Is T invertible? If so, find a rule for T −1 like the one which defines T.

Answer 1

Given that ,

$T:\R^3 \rightarrow \R^3$ is a linear transformation defined by

$T(x_1,x_2,x_3)=(3x_1,x_1-x_2,2x_1+x_2+x_3)$

$\therefore$ We can define , $T=\begin{pmatrix} 3 &0&0 \\ 1&-1&0\\ 2&1&1 \end{pmatrix}$ with respect to usual basis .

$\therefore Det(T)=\begin{vmatrix} 3&0&0 \\ 1&-1&0\\ 2&1&1 \end{vmatrix}$ $=-3$

$\therefore T$ is invertible .

Now , we have to find $T^{-1}$ .

For $T^{-1}$ , first we have to find adj($T$ ).

$adj(T)=\begin{pmatrix} -1&-1 &3 \\ 0&3&-3 \\ 0&0&-3 \end{pmatrix}$ .

$\therefore T^{-1}=\frac{(adjT)'}{Det(T)}$ $=\frac{1}{-3} \begin{pmatrix} -1&0&0\\ -1&3&0 \\ 3&-3&-3 \end{pmatrix}$

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

$\therefore T^{-1}:\R^3\rightarrow \R^3$ defined by

$T^{-1}(x_1,x_2,x_3)=(\frac{x_1}{3},\frac{x_1}{3}-x_2,-x_1+x_2+x_3)$