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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

Prove Theorem \(2.2(\) iii) and (iv): (iii) \((B+C) A=B A+C A\) (iv) \(k(A B)=(k A) B=A(k B)\)

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 \(\mathrm{T}\left(a_{1}, a_{2}, \ldots, a_{n}\right)=a_{1}+a_{n}\).

Label the following statements as true or false. (a) Suppose that \(\beta=\left\\{x_{1}, x_{2}, \ldots, x_{n}\right\\}\) and $\beta^{\prime}=\left\\{x_{1}^{\prime}, x_{2}^{\prime}, \ldots, x_{n}^{\prime}\right\\}\( are ordered bases for a vector space and \)Q$ is the change of coordinate matrix that changes \(\beta^{\prime}\)-coordinates into \(\beta\)-coordinates. Then the \(j\) th column of \(Q\) is \(\left[x_{j}\right]_{\beta^{\prime}}\). (b) Every change of coordinate matrix is invertible. (c) Let \(\mathrm{T}\) be a linear operator on a finite-dimensional vector space \(\mathrm{V}\), let \(\beta\) and \(\beta^{\prime}\) be ordered bases for \(\mathrm{V}\), and let \(Q\) be the change of coordinate matrix that changes \(\beta^{\prime}\)-coordinates into \(\beta\)-coordinates. Then \([\mathrm{T}]_{\beta}=Q[\mathrm{~T}]_{\beta^{\prime}} Q^{-1}\). (d) The matrices \(A, B \in \mathrm{M}_{n \times n}(F)\) are called similar if \(B=Q^{t} A Q\) for some \(Q \in \mathrm{M}_{n \times n}(F)\). (e) Let \(\mathrm{T}\) be a linear operator on a finite-dimensional vector space \(\mathrm{V}\). Then for any ordered bases \(\beta\) and \(\gamma\) for \(\mathrm{V},[\mathrm{T}]_{\beta}\) is similar to \([\mathrm{T}]_{\gamma}\).

Let \(V\) and \(W\) be vector spaces with subspaces \(V_{1}\) and \(W_{1}\), respectively. If \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{W}\) is linear, prove that \(\mathrm{T}\left(\mathrm{V}_{1}\right)\) is a subspace of \(\mathrm{W}\) and that $\left\\{x \in \mathrm{V}: \mathrm{T}(x) \in \mathrm{W}_{1}\right\\}\( is a subspace of \)\mathrm{V}$.

Define \(\mathrm{T}: \mathrm{M}_{2 \times 2}(R) \rightarrow \mathrm{P}_{2}(R) \quad\) by $\quad \mathrm{\top}\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)=(a+b)+(2 d) x+b x^{2} .$ Let $$ \beta=\left\\{\left(\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right),\left(\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right),\left(\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right),\left(\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right)\right\\} \quad \text { and } \quad \gamma=\left\\{1, x, x^{2}\right\\} $$ Compute \([T]_{\beta}^{\gamma}\).
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