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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

Let \(\beta\) and \(\gamma\) be the standard ordered bases for \(\mathrm{R}^{n}\) and \(\mathrm{R}^{m}\), respectively. For each linear transformation \(\mathrm{T}: \mathrm{R}^{n} \rightarrow \mathrm{R}^{m}\), compute \([\mathrm{T}]_{\beta}^{\gamma}\). (a) \(\mathrm{T}: \mathrm{R}^{2} \rightarrow \mathrm{R}^{3}\) defined by $\mathrm{T}\left(a_{1}, a_{2}\right)=\left(2 a_{1}-a_{2}, 3 a_{1}+4 a_{2}, a_{1}\right)$. (b) \(\mathrm{T}: \mathrm{R}^{3} \rightarrow \mathrm{R}^{2}\) defined by $\mathrm{T}\left(a_{1}, a_{2}, a_{3}\right)=\left(2 a_{1}+3 a_{2}-a_{3}, a_{1}+a_{3}\right)$. (c) \(\mathrm{T}: \mathrm{R}^{3} \rightarrow R\) defined by \(\mathrm{T}\left(a_{1}, a_{2}, a_{3}\right)=2 a_{1}+a_{2}-3 a_{3}\). (d) \(\mathrm{T}: \mathrm{R}^{3} \rightarrow \mathrm{R}^{3}\) defined by $$ \mathrm{T}\left(a_{1}, a_{2}, a_{3}\right)=\left(2 a_{2}+a_{3},-a_{1}+4 a_{2}+5 a_{3}, a_{1}+a_{3}\right) . $$ (e) \(\mathrm{T}: \mathrm{R}^{n} \rightarrow \mathrm{R}^{n}\) defined by $\mathrm{T}\left(a_{1}, a_{2}, \ldots, a_{n}\right)=\left(a_{1}, a_{1}, \ldots, a_{1}\right)$. (f) \(\mathrm{T}: \mathrm{R}^{n} \rightarrow \mathrm{R}^{n}\) defined by $\mathrm{T}\left(a_{1}, a_{2}, \ldots, a_{n}\right)=\left(a_{n}, a_{n-1}, \ldots, a_{1}\right)$. (g) \(\mathrm{T}: \mathrm{R}^{n} \rightarrow R\) defined by \(\mathrm{T}\left(a_{1}, a_{2}, \ldots, a_{n}\right)=a_{1}+a_{n}\).

Compute \(A B\) using block multiplication, where $$A=\left[\begin{array}{ccc} 1 & 2 & 1 \\ 3 & 4 & 0 \\ 0 & 0 & 2 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{cccc} 1 & 2 & 3 & 1 \\ 4 & 5 & 6 & 1 \\ 0 & 0 & 0 & 1 \end{array}\right].$$

Find the diagonal and trace of each matrix: (a) \(A=\left[\begin{array}{rrr}1 & 3 & 6 \\ 2 & -5 & 8 \\ 4 & -2 & 9\end{array}\right]\) (b) \(B=\left[\begin{array}{rrr}2 & 4 & 8 \\ 3 & -7 & 9 \\ -5 & 0 & 2\end{array}\right]\) (c) \(\quad C=\left[\begin{array}{rrr}1 & 2 & -3 \\ 4 & -5 & 6\end{array}\right]\).

Let \(V\) and \(W\) be finite-dimensional vector spaces with ordered bases \(\beta=\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) and \(\gamma=\left\\{w_{1}, w_{2}, \ldots, w_{m}\right\\}\), respectively. By Theorem \(2.6\) (p. 73), there exist linear transformations $\mathrm{T}_{i j}: \mathrm{V} \rightarrow \mathrm{W}$ such that $$ \mathrm{T}_{i j}\left(v_{k}\right)= \begin{cases}w_{i} & \text { if } k=j \\\ 0 & \text { if } k \neq j\end{cases} $$ First prove that $\left\\{\mathbf{T}_{i j}: 1 \leq i \leq m, 1 \leq j \leq n\right\\}\( is a basis for \)\mathcal{L}(\mathbf{V}, \mathbf{W})$. Then let \(M^{i j}\) be the \(m \times n\) matrix with 1 in the \(i\) th row and \(j\) th column and 0 elsewhere, and prove that $\left[\mathrm{T}_{i j}\right]_{\beta}^{\gamma}=M^{i j}$. Again by Theorem 2.6, there exists a linear transformation $\Phi_{\beta}^{\gamma}: \mathcal{L}(\mathrm{V}, \mathrm{W}) \rightarrow \mathrm{M}_{m \times n}(F)$ such that \(\Phi_{\beta}^{\gamma}\left(T_{i j}\right)=M^{i j}\). Prove that \(\Phi_{\beta}^{\gamma}\) is an isomorphism.

For each matrix \(A\) and ordered basis \(\beta\), find \(\left[\mathrm{L}_{A}\right]_{\beta}\). Also, find an invertible matrix \(Q\) such that \(\left[\mathrm{L}_{A}\right]_{\beta}=Q^{-1} A Q\). (a) \(A=\left(\begin{array}{ll}1 & 3 \\ 1 & 1\end{array}\right)\) and $\beta=\left\\{\left(\begin{array}{l}1 \\\ 1\end{array}\right),\left(\begin{array}{l}1 \\ 2\end{array}\right)\right\\}$ (b) \(A=\left(\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right)\) and $\beta=\left\\{\left(\begin{array}{l}1 \\\ 1\end{array}\right),\left(\begin{array}{r}1 \\ -1\end{array}\right)\right\\}$ (c) $A=\left(\begin{array}{rrr}1 & 1 & -1 \\ 2 & 0 & 1 \\ 1 & 1 & 0\end{array}\right) \quad\( and \)\quad \beta=\left\\{\left(\begin{array}{l}1 \\\ 1 \\ 1\end{array}\right),\left(\begin{array}{l}1 \\ 0 \\\ 1\end{array}\right),\left(\begin{array}{l}1 \\ 1 \\\ 2\end{array}\right)\right\\}$ (d) $A=\left(\begin{array}{rrr}13 & 1 & 4 \\ 1 & 13 & 4 \\ 4 & 4 & 10\end{array}\right)\( and \)\beta=\left\\{\left(\begin{array}{r}1 \\ 1 \\\ -2\end{array}\right),\left(\begin{array}{r}1 \\ -1 \\\ 0\end{array}\right),\left(\begin{array}{l}1 \\ 1 \\\ 1\end{array}\right)\right\\}$
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