Find the Inverse Linear Transformation if the Linear Transformation is an Isomorphism
Problem 553
Let $T:\R^3 \to \R^3$ be the linear transformation defined by the formula
\[T\left(\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \,\right)=\begin{bmatrix}
x_1+3x_2-2x_3 \\
2x_1+3x_2 \\
x_2-x_3
\end{bmatrix}.\]
Determine whether $T$ is an isomorphism and if so find the formula for the inverse linear transformation $T^{-1}$.
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Solution.
Let $B=\{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\}$ be the standard basis of $\R^3$, where
\[\mathbf{e}_1=\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}, \mathbf{e}_2=\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}, \mathbf{e}_3=\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}.\]
We determine the matrix representation $[T]_B$ of $T$ with respect to the basis $B$.
Since we have
\begin{align*}
T(\mathbf{e}_1)=\begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}, T(\mathbf{e}_2)=\begin{bmatrix}
3 \\
3 \\
1
\end{bmatrix}, T(\mathbf{e}_3=\begin{bmatrix}
-2 \\
0 \\
-1
\end{bmatrix},
\end{align*}
we have
\[[T]_B=\begin{bmatrix}
T(\mathbf{e}_1) & T(\mathbf{e}_2) & T(\mathbf{e}_3)
\end{bmatrix}=\begin{bmatrix}
1 & 3 & -2 \\
2 &3 &0 \\
0 & 1 & -1
\end{bmatrix}\\.\]
This matrix is invertible and the inverse matrix is given by
\[[T]_B^{-1}=\begin{bmatrix}
3 & -1 & -6 \\
-2 &1 &4 \\
-2 & 1 & 3
\end{bmatrix}.\]
(See the post Find the Inverse Matrices if Matrices are Invertible by Elementary Row Operations for details of how to find the inverse matrix of this matrix.)
This implies that the matrix $T$ is an isomorphism.
Observe that we have $[T]_B^{-1}=[T^{-1}]_B$.
Thus, we obtain
\begin{align*}
T^{-1}\left(\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \,\right)&=[T^{-1}]_B\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}
\\[6pt]
&=\begin{bmatrix}
3 & -1 & -6 \\
-2 &1 &4 \\
-2 & 1 & 3
\end{bmatrix}\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}\\[6pt]
&=\begin{bmatrix}
3x_1-x_2-6x_3 \\
-2x_1+x_2+4x_3 \\
-2x_1+x_2+3x_3
\end{bmatrix}.
\end{align*}
In summary, the formula for the inverse linear transformation $T^{-1}$ is given by
x_1 \\
x_2 \\
x_3
\end{bmatrix} \,\right)=\begin{bmatrix}
3x_1-x_2-6x_3 \\
-2x_1+x_2+4x_3 \\
-2x_1+x_2+3x_3
\end{bmatrix}.\]
Add to solve later
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Formula for inverse of T is wrong. Infact, inverse matrix if T is wrong.
Dear Mahendra Reddy,
Thank you for catching the error. I modified the problem and now it should be fine.