# The standard ordered basis of ℝ2 is {e1, e2}. Let T : ℝ2 → ℝ2 be the linear transformation such that T reflects the points through the line x1 = -x2.

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The standard ordered basis of ℝ2 is {e1, e2}. Let T : ℝ2 → ℝ2 be the linear transformation such that T reflects the points through the line x1 = -x2. The standard matrix of T is:
1. $\left( {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right)$
2. $\left( {\begin{array}{*{20}{c}} 0&-1\\ -1&0 \end{array}} \right)$
3. $\left( {\begin{array}{*{20}{c}} -1&0\\ 0&-1 \end{array}} \right)$
4. $\left( {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right)$

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Correct Answer - Option 2 : $\left( {\begin{array}{*{20}{c}} 0&-1\\ -1&0 \end{array}} \right)$

Concept:

If x1 and x2 are the components of a vector X with respect to a standard basis, this means

X = [x1, x2]T = x1 e1 + x2 e2 where {e1, e2} is standard ordered basis.

Reflection theorem:

Let T : R2 → R2 be a linear transformation given by reflecting vectors over the line x2 = m x1. Then the matrix of T is given by

$\frac {1}{1+{m^2}} \left[ {\begin{array}{*{20}{c}} {1-m^2}&2m\\ 2m&m^2-1 \end{array}} \right]$

Calculation:

Given T is the linear transformation that reflects the points through the line x1 = -x2

⇒ m = -1

The line x1 = -x2 is the bisector of the second and fourth quadrant and the reflection through this line is represented by a matrix

$T = \left( {\begin{array}{*{20}{c}} 0&-1\\ -1&0 \end{array}} \right)$