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Chapter 6: Problem 34

(a) find the standard matrix \(A\) for the linear transformation \(T,(\mathrm{b})\) use \(A\) to find the image of the vector \(\mathbf{v},\) and \((\mathrm{c})\) use a graphing utility or computer software program and \(A\) to verify your result from part (b). $$\begin{array}{l} T(x, y, z)=(2 x+3 y-z, 3 x-2 z, 2 x-y+z) \\ \mathbf{v}=(1,2,-1) \end{array}$$

Short Answer

Expert verified
The standard matrix is \[\begin{bmatrix}2 & 3 & -1 \3 & 0 & -2 \2 & -1 & 1 \end{bmatrix}\]The image of the vector \(\mathbf{v}=(1,2,-1)\) under the transformation is \(T(\mathbf{v}) = (7, 5, 1)\).

Step by step solution

01

Finding the Standard Matrix

The linear transformation \(T(x, y, z)=(2x+3y-z, 3x-2z, 2x-y+z)\) can be represented by the standard matrix \(A\), where the coefficients of \(x, y, z\) in \(T(x, y, z)\) are the elements of the matrix. Here, the standard matrix \(A\) is \[\begin{bmatrix}2 & 3 & -1 \3 & 0 & -2 \2 & -1 & 1 \end{bmatrix}\]
02

Finding the Image of the Vector

To find the image of \(\mathbf{v}=(1,2,-1)\), multiply the matrix \(A\) with the vector \(\mathbf{v}\). This can be done by taking the dot product of the rows of \(A\) with \(\mathbf{v}\).\[\mathbf{Av} = \begin{bmatrix} 2 & 3 & -1 \ 3 & 0 & -2 \ 2 & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 \ 2 \ -1 \end{bmatrix} = \begin{bmatrix} 2*1 + 3*2 - (-1)*1 \ 3*1 + 0*2 - 2*(-1) \ 2*1 - 2*1 + 1*(-1) \ \end{bmatrix} = \begin{bmatrix} 7 \ 5 \ 1 \end{bmatrix}\]So, the image of \(\mathbf{v}\) under the transformation \(T\) is \(T(\mathbf{v}) = (7, 5, 1)\).
03

Verification using a Graphing Utility

This step entails that you use software with the capability to perform matrix computations and visualize the result. Compare the result with the manually obtained outcome for a confirmation.

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Most popular questions from this chapter

Let \(T_{1}: V \rightarrow V\) and \(T_{2}: V \rightarrow V\) be one-to-one linear transformations. Prove that the composition \(T=T_{2} \circ T_{1}\) is one-to- one and that \(T^{-1}\) exists and is equal to \(T_{1}^{-1} \circ T_{2}^{-1}\). Getting Started: To show that \(T\) is one-to-one, you can use the definition of a one-to-one transformation and show that \(T(\mathbf{u})=T(\mathbf{v})\) implies \(\mathbf{u}=\mathbf{v} .\) For the second statement, you first need to use Theorems 6.8 and 6.12 to show that \(T\) is invertible, and then show that \(T \circ\left(T_{1}^{-1} \circ T_{2}^{-1}\right)\) and \(\left(T_{1}^{-1} \circ T_{2}^{-1}\right) \cdot T\) are identity transformations. (i) Let \(T(\mathbf{u})=T(\mathbf{v}) .\) Recall that \(\left(T_{2} \circ T_{1}\right)(\mathbf{v})=T_{2}\left(T_{1}(\mathbf{v})\right)\) for all vectors \(\mathbf{v} .\) Now use the fact that \(T_{2}\) and \(T_{1}\) are one-to-one to conclude that \(\mathbf{u}=\mathbf{v}\) (ii) Use Theorems 6.8 and 6.12 to show that \(T_{1}, T_{2},\) and \(T\) are all invertible transformations. So \(T_{1}^{-1}\) and \(T_{2}^{-1}\) exist. (iii) Form the composition \(T^{\prime}=T_{1}^{-1} \circ T_{2}^{-1} .\) It is a linear transformation from \(V\) to \(V .\) To show that it is the inverse of \(T,\) you need to determine whether the composition of \(T\) with \(T^{\prime}\) on both sides gives an identity transformation.

Determine whether the linear transformation is invertible. If it is, find its inverse. $$T\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=\left(x_{1}-2 x_{2}, x_{2}, x_{3}+x_{4}, x_{3}\right)$$

(a) find the matrix \(A^{\prime}\) for \(T\) relative to the basis \(B^{\prime}\) and \((b)\) show that \(A^{\prime}\) is similar to \(A,\) the standard matrix for \(T\). $$\begin{array}{l} T: R^{3} \rightarrow R^{3}, T(x, y, z)=(0,0,0) \\ B^{\prime}=\\{(1,1,0),(1,0,1),(0,1,1)\\} \end{array}$$

(a) find the matrix \(A^{\prime}\) for \(T\) relative to the basis \(B^{\prime}\) and \((b)\) show that \(A^{\prime}\) is similar to \(A,\) the standard matrix for \(T\). $$\begin{aligned} &T: R^{3} \rightarrow R^{3}, T(x, y, z)=(x, x+2 y, x+y+3 z)\\\ &B^{\prime}=\\{(1,-1,0),(0,0,1),(0,1,-1)\\} \end{aligned}$$

(a) find the matrix \(A^{\prime}\) for \(T\) relative to the basis \(B^{\prime}\) and \((b)\) show that \(A^{\prime}\) is similar to \(A,\) the standard matrix for \(T\). $$T: R^{2} \rightarrow R^{2}, T(x, y)=(2 x+y, x-2 y), B^{\prime}=\\{(1,2),(0,4)\\}$$
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