Define linear transformation with an example.

Let A = \(\begin{bmatrix}1 & -3\\ 3 & 5\\ -1 & 7\end{bmatrix}\), v = \(\begin{bmatrix}2\\ -1\end{bmatrix}\), b = \(\begin{bmatrix}3\\ 2\\ 4\end{bmatrix}\), x = \(\begin{bmatrix}x_1\\ x_2\end{bmatrix}\)

and define a transformation T:R2 → R2 by T(x) = Ax then

  1. find T(v)
  2. find x ∈ R2 whose image under T is b

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Let T:V → W be a tranformation (mapping or function) such that,

  1. T(c, v) = c . T(v)
  2. T(u + v) = T(u) + T(v)  ∀ c ∈ K and u, v ∈ V

Example: Let A = (aij)m × n be an m × n matrix.

Let T:Rn → Rm

T: multiplication be the matrix A

i.e. T(x) = Ax is a linear transformation

Solution:

Given Matrices are

A = \(\begin{bmatrix}1 & -3\\ 3 & 5\\ -1 & 7\end{bmatrix}\)

v = \(\begin{bmatrix}2\\ -1\end{bmatrix}\)

b = \(\begin{bmatrix}3\\ 2\\ 4\end{bmatrix}\)

x = \(\begin{bmatrix}x_1\\ x_2\end{bmatrix}\)

Given  that transformation T:R2 → R2 by T(x) = Ax

a) First Part:

T(u) = Au

= \(\begin{bmatrix}1 & -3\\ 3 & 5\\ -1 & 7\end{bmatrix}\) \(\begin{bmatrix}2\\ -1\end{bmatrix}\)

= \(\begin{bmatrix}2+3\\ 6-5 \\-2-7\end{bmatrix}\)

= \(\begin{bmatrix}5\\ 1 \\-9\end{bmatrix}\)

b) Second Part:

Let x = \(\begin{bmatrix}x_1\\ x_2\end{bmatrix}\)

Suppose x in R2 where image under T is b then

T(x) = b

Ax = b

\(\begin{bmatrix}1 & -3\\ 3 & 5\\ -1 & 7\end{bmatrix}\)\(\begin{bmatrix}x_1\\ x_2\end{bmatrix}\) = \(\begin{bmatrix}3\\ 2\\ 4\end{bmatrix}\)

The augmented matrix of Ax = b is

\(\begin{bmatrix}1 & -3 & : & 3\\ 3 & 5 & : & 2\\ -1 & 7 & : & 4\end{bmatrix}\)

R2 → R2 – 3R1,   R3 → R3 + R1

\(\begin{bmatrix}1 & -3 & : & 3\\ 0 & 14 & : & -7\\ 0 & 4 & : & 7\end{bmatrix}\)

R2 → R2 / 14

\(\begin{bmatrix}1 & -3 & : & 3\\ 0 & 1 & : & \frac{-1}{2}\\ 0 & 4 & : & 7\end{bmatrix}\)

R3 → R3 – 4R2,   R1 → R1 + 3R2

\(\begin{bmatrix}1 & 0 & : & \frac{3}{2}\\ 0 & 1 & : & \frac{-1}{2}\\ 0 & 0 & : & 9\end{bmatrix}\)

This implies \(x_1 = \frac{3}{2}\) and \(x_2 = \frac{-1}{2}\)

Thus, x = \(\begin{bmatrix}\frac{3}{2}\\ \frac{-1}{2}\end{bmatrix}\) in R2 whose image under T is b.

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