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Chapter 2: Problem 11

Prove that there exists a linear transformation \(T: R^{2} \rightarrow R^{3}\) such that \(\mathrm{T}(1,1)=(1,0,2)\) and \(\mathrm{T}(2,3)=(1,-1,4)\). What is \(\mathrm{T}(8,11)\) ?

Short Answer

Expert verified
We can represent the given linear transformation as a matrix product, T\(\begin{pmatrix} x \\ y \end{pmatrix}\) = A\(\begin{pmatrix} x \\ y \end{pmatrix}\). By analyzing the given conditions, we obtain the system of equations for the components of matrix A. Solving these equations, we find the matrix A = \(\begin{pmatrix} 5 & -4 \\ 3 & -3 \\ 3 & -1 \end{pmatrix}\). Using this matrix, we can compute the transformation T(8,11) = \(\begin{pmatrix} 11 \\ 3 \\ 8 \end{pmatrix}\). Hence, \(\mathrm{T}(8,11) = (11, 3, 8)\).

Step by step solution

01

Linear transformation representation

Given the transformation T(1,1) = (1,0,2) and T(2,3) = (1,-1,4), we can represent the transformation as the following matrix product: T\(\begin{pmatrix} x \\ y \end{pmatrix}\) = A\(\begin{pmatrix} x \\ y \end{pmatrix}\) Where A is a 3x2 matrix, and x and y are the components of the input vector. We can represent the conditions from the problem statement as: A\(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\) = \(\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}\), A\(\begin{pmatrix} 2 \\ 3 \end{pmatrix}\) = \(\begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}\).
02

Find the matrix A

We'll find the matrix A by analyzing the two given conditions: 1. A\(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\) = \(\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}\), Consider \(\begin{pmatrix} a & b \\ c & d \\ e & f \end{pmatrix}\) as the matrix A, where a, b, c, d, e, and f are the components of the matrix. This condition can be expressed as: \(a+b = 1\), \(c+d = 0\), \(e+f = 2\). 2. A\(\begin{pmatrix} 2 \\ 3 \end{pmatrix}\) = \(\begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}\), This condition can be expressed as: \(2a+3b = 1\), \(2c+3d = -1\), \(2e+3f = 4\). Now, we'll solve these equations to find the matrix A.
03

Solve the equations

Solving the system of equations obtained in Step 2: From \(a+b = 1\) and \(2a+3b = 1\): \(a = 5\), \(b = -4\). From \(c+d = 0\) and \(2c+3d = -1\): \(c = 3\), \(d = -3\). From \(e+f = 2\) and \(2e+3f = 4\): \(e = 3\), \(f = -1\). Thus, the matrix A we obtain is: A = \(\begin{pmatrix} 5 & -4 \\ 3 & -3 \\ 3 & -1 \end{pmatrix}\)
04

Compute T(8,11)

Now, we can use the matrix A to compute T(8,11): T\(\begin{pmatrix} 8 \\ 11 \end{pmatrix}\) = A\(\begin{pmatrix} 8 \\ 11 \end{pmatrix}\) T\(\begin{pmatrix} 8 \\ 11 \end{pmatrix}\) = \(\begin{pmatrix} 5 & -4 \\ 3 & -3 \\ 3 & -1 \end{pmatrix}\begin{pmatrix} 8 \\ 11 \end{pmatrix}\) T\(\begin{pmatrix} 8 \\ 11 \end{pmatrix}\) = \(\begin{pmatrix} 11 \\ 3 \\ 8 \end{pmatrix}\) Hence, \(\mathrm{T}(8,11) = (11, 3, 8)\).

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Most popular questions from this chapter

Find real numbers \(x, y, z\) such that \(A\) is Hermitian, where \(A=\left[\begin{array}{ccc}3 & x+2 i & y i \\ 3-2 i & 0 & 1+z i \\ y i & 1-x i & -1\end{array}\right]\)

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