Question

Let T : R³→ R³ be defined by T (x1, x2, x3) = (x1 −x3, x2 −x3, x1). Is T invertible? If yes, find a rule for T–¹ like the one which defines T ?

Answer 1

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Answer on Question #80848 – Math – Linear Algebra

Question

Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be defined by $T(x1, x2, x3) = (x1 - x3, x2 - x3, x1)$. Is $T$ invertible? If yes, find a rule for $T^{-1}$ like the one which defines $T$?

Solution

We need to decide if the solution to

$$T(y_1, y_2, y_3) = (x_1, x_2, x_3)$$

exists and is unique for all $(x_1, x_2, x_3) \in \mathbb{R}^3$.

This means

$$\left\{ \begin{array}{l} y_1 - y_3 = x_1 \\ y_2 - y_3 = x_2 \\ y_1 = x_3 \end{array} \right.$$

Then

$$\left\{ \begin{array}{l} y_1 = x_3 \\ y_3 = x_3 - x_1 \\ y_2 = x_2 + y_3 = x_2 + x_3 - x_1 \end{array} \right.$$

The unique solution exists for any $x_1, x_2, x_3$. The rule is

$$T^{-1}(x_1, x_2, x_3) = (x_3, x_2 + x_3 - x_1, x_3 - x_1).$$

**Answer**: Yes, it is invertible. $T^{-1}(x_1, x_2, x_3) = (x_3, x_2 + x_3 - x_1, x_3 - x_1)$.