Let \(T\) be the linear operator on \(\RR^3\) defined by \[ T(x_1, x_2, x_3) = (3x_1,\; x_1 - x_2,\; 2x_1 + x_2 + x_3). \] Is \(T\) invertible? If so, find a rule for \(T^{-1}\) like the one which defines \(T\).
Solution
Write \[ T(x_1,x_2,x_3)=(y_1,y_2,y_3). \] Then \[ y_1=3x_1,\qquad y_2=x_1-x_2,\qquad y_3=2x_1+x_2+x_3. \] These equations can be solved successively: \[ x_1=\frac{y_1}{3},\qquad x_2=\frac{y_1}{3}-y_2,\qquad x_3=-y_1+y_2+y_3. \] Thus every \(Y\) has a unique preimage, so \(T\) is invertible, with \[ T^{-1}(y_1,y_2,y_3)= \left(\frac{y_1}{3},\frac{y_1}{3}-y_2,-y_1+y_2+y_3\right). \] A direct substitution verifies both \(T^{-1}T=I\) and \(TT^{-1}=I\).