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Chapter 6: Problem 46

Determine whether the linear transformation is invertible. If it is, find its inverse. $$T(x, y)=(x+2 y, x-2 y)$$

Short Answer

Expert verified
Yes, the given linear transformation is invertible. The inverse transformation is given by the matrix \[ \begin{array}{cc} 0.5 & 0.5 \ 0.25 & -0.25 \end{array} \]

Step by step solution

01

Express the given transformation as a matrix

Given transformation can be represented by a matrix. Let's denote \((x,y)\) as vector \(V_1\) and \((x+2y, x-2y)\) as vector \(V_2\), then \(T(V_1) = A \cdot V_1 = V_2\), where \(T\) is the transformation, \(A\) is the matrix representation of the transformation. The matrix \(A\) in this case would be: \[\[A\] = \(\[ \begin{array}{cc} 1 & 2 \ 1 & -2 \end{array} \]\)\]
02

Find the determinant of the matrix

In order for a matrix (and thus the corresponding linear transformation) to be invertible, its determinant must not be zero. The determinant of a 2x2 matrix \(\[ \begin{array}{cc} a & b \ c & d \end{array} \]\) is given by \(ad - bc\). Therefore, the determinant of our matrix is \((1 * -2) - (2 * 1) = -2 -2 = -4\). Because the determinant is not zero, the linear transformation is invertible.
03

Find the inverse of the matrix

The inverse of the matrix A exists since the determinant is non-zero. The inverse of a 2x2 matrix \(\[ \begin{array}{cc} a & b \ c & d \end{array} \]\), is given by \(\frac{1}{ad - bc}\[ \begin{array}{cc} d & -b \ -c & a \end{array} \]\). Applying this to our matrix, we get: \[\[A^{-1}\] = \(-0.25\) \(\[ \begin{array}{cc} -2 & -2 \ -1 & 1 \end{array} \]\) = \[ \begin{array}{cc} 0.5 & 0.5 \ 0.25 & -0.25 \end{array} \]\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Invertible Matrix
An invertible matrix is a crucial concept in linear algebra, especially when discussing linear transformations. If a matrix is invertible, it means there exists another matrix, called the inverse, that when multiplied with the original matrix, results in the identity matrix. The identity matrix is like the number 1 in multiplication; it doesn鈥檛 change the vector it multiplies. For a 2x2 matrix to be invertible, its determinant must not be zero.

Let's consider a transformation represented by a matrix \( A \). This matrix is invertible if there is another matrix \( A^{-1} \) such that:
  • \( A \cdot A^{-1} = I \)
  • \( A^{-1} \cdot A = I \)
Where \( I \) is the identity matrix. If such a matrix \( A^{-1} \) exists, the transformation is called invertible or non-singular.
Matrix Determinant
The determinant is a special number that can be calculated from a square matrix. For a 2x2 matrix \[\begin{bmatrix} a & b \ c & d \end{bmatrix}\]the determinant is calculated as \( ad - bc \).

Determinants have many applications:
  • They can tell us if a matrix is invertible. If the determinant is zero, the matrix isn't invertible or is singular.
  • They describe a property of the matrix that is linked to the volume scaling factor applied by the linear transformation.
      • In the given exercise, the matrix \[\begin{bmatrix} 1 & 2 \ 1 & -2 \end{bmatrix}\]has a determinant of \((1 \times -2) - (2 \times 1) = -4 \). Since \(-4 eq 0\), the matrix is invertible. This means the corresponding linear transformation can be reversed.
Inverse of a Matrix
An inverse of a matrix is like a reciprocal of a number. It reverses the effect of the original matrix. If you multiply a matrix by its inverse, you get the identity matrix.

For 2x2 matrices, if the determinant is non-zero, the inverse can be found using:\[A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]This formula only works when the determinant is not zero because you can't divide by zero.

In our exercise, given the matrix \[\begin{bmatrix} 1 & 2 \ 1 & -2 \end{bmatrix},\]the inverse is calculated as:1. Find determinant: \(-4\)2. Compute inverse using the formula: \[\begin{bmatrix} 0.5 & 0.5 \ 0.25 & -0.25 \end{bmatrix}.\]This inverse matrix will reverse the linear transformation, bringing us back to our original input vector.

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Most popular questions from this chapter

Let \(B=\\{(1,2),(-1,-1)\\}\) and \(B^{\prime}=\\{(-4,1),(0,2)\\}\) be bases for \(R^{2},\) and let \(A=\left[\begin{array}{rr}2 & 1 \\ 0 & -1\end{array}\right]\) be the matrix for \(T: R^{2} \rightarrow R^{2}\) relative to \(B\). (a) Find the transition matrix \(P\) from \(B^{\prime}\) to \(B\) (b) Use the matrices \(A\) and \(P\) to find \([\mathbf{v}]_{B}\) and \([T(\mathbf{v})]_{B},\) where \([\mathbf{v}]_{B^{\prime}}=\left[\begin{array}{r}-1 \\ 4\end{array}\right]\). (c) Find \(A^{\prime}\) (the matrix for \(T\) relative to \(B^{\prime}\) ) and \(P^{-1}\). (d) Find \([T(\mathbf{v})]_{B^{\prime}}\) in two ways: first as \(P^{-1}[T(\mathbf{v})]_{B}\) and then \(\operatorname{as} A^{\prime}[(\mathbf{v})]_{B^{\prime}}\)

Sketch the image of the rectangle with vertices at \((0,0),(0,2),(1,2),\) and (1,0) under the specified transformation. \(T\) is the shear represented by \(T(x, y)=(x, y+2 x)\)

(a) identify the transformation and (b) graphically represent the transformation for an arbitrary vector in the plane. $$T(x, y)=(x / 4, y)$$

Find the matrix that will produce the indicated rotation. \(60^{\circ}\) about the \(y\) -axis

Determine whether the linear transformation is invertible. If it is, find its inverse. $$T(x, y)=(2 x, 0)$$
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