Question

2. Define a linear transformation T:R^2 --> R^2 HRas follows: T(x1, x2) = (x1 - 12,21 + x2). Show that T is invertible and find T-1:

Answer 1

Answer:

$T^{-1}(x_1,x_2)=(\frac{x_1+x_2}{2},\frac{x_2-x_1}{2})$

Step-by-step explanation:

Given:

Linear transformation,

$T:R^2\rightarrow R^2$   defined as

$T(x_1,x_2)=(x_1-x_2,x_1+x_2)$

To Show: T is invertible

To find: $T^{-1}$

We know that Standard Basis of R² is $\{e_1=(1,0)\:,\:e_2=(0,1)\}$

$T(e_1)=T(1,0)=(1,1)=1e_1+1e_2$

$T(e_2)=T(0,1)=(-1,1)=-1e_1+1e_2$

So, The matrix representation of T is $\begin{bmatrix}1&1\\-1&1\end{bmatrix}^T=\begin{bmatrix}1&-1\\1&1\end{bmatrix}$

Now, Determinant of T = 1 - (-1) = 1 + 1 = 2 ≠ 0

⇒ Matrix Representation of T is Invertible matrix.

⇒ T is invertible Linear Transformation.

Hence Proved.

let,

$x_1-x_2=u.........................(1)$

$x_1+x_2=v.........................(2)$

Add (1) and (2),

$2x_1=u+v$

$x_1=\frac{u+v}{2}$

Putting this value in (1),

$\frac{u+v}{2}-x_2=u$

$x_2=\frac{u+v}{2}-u$

$x_2=\frac{u+v-2u}{2}$

$x_2=\frac{v-u}{2}$

Now,

$T(x_1,x_2)=(x_1-x_2,x_1+x_2)=(u,v)$

$\implies(x_1,x_2)=T^{-1}(u,v)$

$\implies T^{-1}(u,v)=(\frac{u+v}{2},\frac{v-u}{2})$

$\implies T^{-1}(x_1,x_2)=(\frac{x_1+x_2}{2},\frac{x_2-x_1}{2})$

Therefore, $T^{-1}(x_1,x_2)=(\frac{x_1+x_2}{2},\frac{x_2-x_1}{2})$

Answer 2

Answer with explanation:

T is a Linear transformation such that

 $T:R^2\rightarrow R^2\\\\T(x_{1},x_{2})=(x_{1}-12,21+x_{2})$

To show that ,T is invertible  that is inverse of a matrix exist ,we need to show that ,T is non singular.

⇒ |T|≠0

→→For a Homogeneous system

  $(x_{1}-12,21+x_{2})=(0,0)\\\\x_{1}=12,x_{2}=-21\\\\|\text{Matrix}|=\left[\begin{array}{cc}12&0\\0&-21\end{array}\right]\\\\ |\text{Matrix}|\neq 0$

so it is invertible.We have considered equation is of the form

   $\rightarrow ax_{1}+bx_{2}+c=0$

→→For a Non Homogeneous system

  $(x_{1}-12,21+x_{2})=(s,v)\\\\x_{1}=12+s,x_{2}=-21+v\\\\|\text{Matrix}|=\left[\begin{array}{cc}12+s&0\\0&v-21\end{array}\right]\\\\ |\text{Matrix}|=(s+12)\times (v-21)\neq 0$

so T, is invertible.

$T^{-1}=\frac{Adj.T}{|T|}\\\\T=\left[\begin{array}{cc}s+12&0\\0&v-21\end{array}\right]\\\\Adj.T=\left[\begin{array}{cc}v-21&0\\0&s+12\end{array}\right]\\\\T^{-1}=\frac{\left[\begin{array}{cc}v-21&0\\0&s+12\end{array}\right]}{(s+12)\times (v-21)}\\\\T^{-1}={\left[\begin{array}{cc}\frac{1}{s+12}&0\\0&\frac{1}{v-21}\end{array}\right]}\\\\\text{Replacing s by} x_{1}\\\\ \text{and v by} x_{2}, \text{we get}\\\\T^{-1}={\left[\begin{array}{cc}\frac{1}{x_{1}+12}&0\\0&\frac{1}{x_{2}-21}\end{array}\right]}$