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drek231 [11]
3 years ago
12

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 fi

nd T-1:
Mathematics
2 answers:
Semenov [28]3 years ago
8 0

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})

yulyashka [42]3 years ago
4 0

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]}

 

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Answer:

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m<b = 180 - m<a

m<c = 180 - m<d

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Reasons:

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