## Linear Algebra and Its Applications, Review Exercise 2.25

Review exercise 2.25. Suppose that $T$ is a linear transformation from $\mathbb{R}^3$ to itself, and $T$ transforms the point $(u, v, w)$ to the point $(u+v+w, u+v, u)$. What does the inverse transformation $T^{-1}$ do to the point $(x, y, z)$?

Answer: The effect of $T^{-1}$ is to reverse the effect of $T$. Since $T$ takes the first entry of a vector and makes it the last entry of the resulting vector, $T^{-1}$ must take the last entry of a vector and make it the first. So applying $T^{-1}$ to the point $(x, y, z)$ results in a point whose first entry is $z$.

Next, $T$ takes the second entry of a vector, adds to it the first entry, and makes the  sum the second entry of the resulting vector. In reversing this $T^{-1}$ must take the second entry and subtract from it the original first entry. So applying $T^{-1}$ to the point $(x, y, z)$ results in a point whose second entry is $y-z$ (since $z$ was the original first entry, as discussed in the previous paragraph).

Finally, $T$ takes the third entry of a vector, adds to it the first and second entries, and makes the sum the first entry of the resulting vector . In reversing this $T^{-1}$ must take the first entry and subtract from it the original first entry and second entry. So applying $T^{-1}$ to the point $(x, y, z)$ results in a point whose third entry is $x - z - (y-z) = x - y$. (Recall that $z$ was the original first entry, and $y-z$ the original second entry.)

The inverse transformation $T^{-1}$ thus transforms the point $(x, y, z)$ into the point $(z, y-z, x-y)$. To confirm this, we apply the transformation $T$ to $(z, y-z, x-y)$ resulting in the point $(z +y-z + x-y, z + y-z, z) = (x, y, z)$

The transformation $T^{-1}$ is thus indeed the inverse of the transformation $T$.

Note that another way to compute $T^{-1}$ is to take the matrix $[T]$ corresponding to the transformation $T$ and compute its inverse $[T]^{-1}$.

The linear transformation $T$ corresponds to the matrix $[T] = \begin{bmatrix} 1&1&1 \\ 1&1&0 \\ 1&0&0 \end{bmatrix}$

so that applying $T$ to $(u, v, w)$ gives $\begin{bmatrix} 1&1&1 \\ 1&1&0 \\ 1&0&0 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} = \begin{bmatrix} u+v+w \\ u+v \\ w \end{bmatrix}$

We can compute the inverse of $[T]$ using Gauss-Jordan elimination. Start with $\begin{bmatrix} 1&1&1&\vline&1&0&0 \\ 1&1&0&\vline&0&1&0 \\ 1&0&0&\vline&0&0&1 \end{bmatrix}$

Subtract 1 times the first row from the second row: $\Rightarrow \begin{bmatrix} 1&1&1&\vline&1&0&0 \\ 0&0&-1&\vline&-1&1&0 \\ 1&0&0&\vline&0&0&1 \end{bmatrix}$

Subtract 1 times the first row from the third row: $\Rightarrow \begin{bmatrix} 1&1&1&\vline&1&0&0 \\ 0&0&-1&\vline&-1&1&0 \\ 0&-1&-1&\vline&-1&0&1 \end{bmatrix}$

Exchange the second and third rows: $\Rightarrow \begin{bmatrix} 1&1&1&\vline&1&0&0 \\ 0&-1&-1&\vline&-1&0&1 \\ 0&0&-1&\vline&-1&1&0 \end{bmatrix}$

Subtract 1 times the third row from the second row: $\Rightarrow \begin{bmatrix} 1&1&1&\vline&1&0&0 \\ 0&-1&0&\vline&0&-1&1 \\ 0&0&-1&\vline&-1&1&0 \end{bmatrix}$

Subtract -1 times the third row from the first row: $\Rightarrow \begin{bmatrix} 1&1&0&\vline&0&1&0 \\ 0&-1&0&\vline&0&-1&1 \\ 0&0&-1&\vline&-1&1&0 \end{bmatrix}$

Subtract -1 times the second row from the first row: $\Rightarrow \begin{bmatrix} 1&0&0&\vline&0&0&1 \\ 0&-1&0&\vline&0&-1&1 \\ 0&0&-1&\vline&-1&1&0 \end{bmatrix}$

Multiply both the second row and the third row by -1: $\Rightarrow \begin{bmatrix} 1&0&0&\vline&0&0&1 \\ 0&1&0&\vline&0&1&-1 \\ 0&0&1&\vline&1&-1&0 \end{bmatrix}$

We thus have $[T]^{-1} = \begin{bmatrix} 0&0&1 \\ 0&1&-1 \\ 1&-1&0 \end{bmatrix}$

so that $\begin{bmatrix} 0&0&1 \\ 0&1&-1 \\ 1&-1&0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} z \\ y-z \\ x-y \end{bmatrix}$

and $[T][T]^{-1} = \begin{bmatrix} 1&1&1 \\ 1&1&0 \\ 1&0&0 \end{bmatrix} \begin{bmatrix} 0&0&1 \\ 0&1&-1 \\ 1&-1&0 \end{bmatrix}$ $= \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} = I$

So the linear transformation $T^{-1}$ as defined by $[T]^{-1}$ is indeed the inverse of the original transformation $T$.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

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