Find the Inverse Linear Transformation if the Linear Transformation is an Isomorphism

Linear Transformation problems and solutions

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

\[T^{-1}\left(\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \,\right)=\begin{bmatrix}
3x_1-x_2-6x_3 \\
-2x_1+x_2+4x_3 \\
-2x_1+x_2+3x_3
\end{bmatrix}.\]

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2 Responses

  1. Mahendra Reddy says:

    Formula for inverse of T is wrong. Infact, inverse matrix if T is wrong.

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