Answer to Question #279045 in Linear Algebra for Krstina

Question

Show that transformation 𝑇:𝑅

2 → 𝑅

2 defined by 

𝑇(𝑥, 𝑦) = (𝑎𝑥 + 𝑏𝑦, 𝑐𝑥 + 𝑑𝑦),

Answer 1

Given

$\begin{aligned} T: R^{2} \rightarrow R^{2} & \text { such map } \\ & T(x, y)=(a x+b y,(x+d y) .\end{aligned}$

$T(x, y)=(a x+b y, c x+d y) \cdot u, v \in R^{2}\ \text{such that}$

 $\begin{aligned} T(x, y) &=(u, v) \cdot \\ \text { i.e } \quad a x+b y &=u \\ c x+d y &=v \\ \left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right] &=\left[\begin{array}{l} u \\ v \end{array}\right] \\ \left[\begin{array}{l} x \\ y \end{array}\right] &=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]^{-1}\left[\begin{array}{l} u \\ v \end{array}\right] . \\ \left[\begin{array}{l} x \\ y \end{array}\right] &=\left[\begin{array}{ll} d & -b \\ -c & a \end{array}\right] \frac{1}{a d-b c}\left[\begin{array}{l} u \\ v \end{array}\right] \end{aligned}$

There exists (x, y) such that

 $T(x, y)=(u, v) \text { iff } a d-b c \neq 0 \text {. }$

a, b, c, d are scalars such that $a d-b c \neq 0.$