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Math  /  Algebra

QuestionSuppose that T:R3→R3T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3} is such that its action on a vector [xyz]\left[\begin{array}{l}x \\ y \\ z\end{array}\right] is given below: T[xyz]=[x−y2x−3y−2z2x−y+z]T\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} x-y \\ 2 x-3 y-2 z \\ 2 x-y+z \end{array}\right]
Find the inverse transformation T−1T^{-1} and give its action on a general vector [xyz]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]. T−1[xyz]=[000]T^{-1}\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]

Studdy Solution

STEP 1

1. The transformation T:R3→R3 T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3} is linear.
2. The transformation is represented by a matrix A A such that T(v)=Av T(\mathbf{v}) = A\mathbf{v} .
3. We need to find the inverse transformation T−1 T^{-1} .

STEP 2

1. Write the transformation T T as a matrix equation.
2. Construct the matrix A A representing the transformation.
3. Calculate the inverse of the matrix A A .
4. Verify the inverse by checking T(T−1(v))=v T(T^{-1}(\mathbf{v})) = \mathbf{v} .
5. Express the action of T−1 T^{-1} on a general vector.

STEP 3

Write the transformation T T as a matrix equation:
Given: T[xyz]=[x−y2x−3y−2z2x−y+z] T\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} x-y \\ 2x-3y-2z \\ 2x-y+z \end{array}\right]
This can be expressed as: T(v)=Av T(\mathbf{v}) = A \mathbf{v}

STEP 4

Construct the matrix A A representing the transformation:
From the transformation equations, we can deduce: A=[1−102−3−22−11] A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & -3 & -2 \\ 2 & -1 & 1 \end{bmatrix}

STEP 5

Calculate the inverse of the matrix A A :
To find A−1 A^{-1} , we use the formula for the inverse of a 3x3 matrix. We need to find the determinant of A A and the adjugate matrix.
Calculate the determinant of A A : det(A)=1(−3⋅1−(−2)(−1))−(−1)(2⋅1−(−2)(2))+0(2⋅(−1)−(−3)(2)) \text{det}(A) = 1(-3 \cdot 1 - (-2)(-1)) - (-1)(2 \cdot 1 - (-2)(2)) + 0(2 \cdot (-1) - (-3)(2)) det(A)=1(−3−2)−(−1)(2−(−4)) \text{det}(A) = 1(-3 - 2) - (-1)(2 - (-4)) det(A)=1(−5)+(−1)(6) \text{det}(A) = 1(-5) + (-1)(6) det(A)=−5−6=−11 \text{det}(A) = -5 - 6 = -11
Since the determinant is non-zero, the matrix is invertible.
Calculate the adjugate matrix and divide by the determinant to find A−1 A^{-1} .

STEP 6

Calculate the adjugate matrix:
The adjugate matrix is the transpose of the cofactor matrix. Calculate each cofactor and transpose.
adj(A)=[1−102−3−22−11] \text{adj}(A) = \begin{bmatrix} 1 & -1 & 0 \\ 2 & -3 & -2 \\ 2 & -1 & 1 \end{bmatrix}
After calculating the cofactors and transposing, we find: adj(A)=[122−1−3−10−21] \text{adj}(A) = \begin{bmatrix} 1 & 2 & 2 \\ -1 & -3 & -1 \\ 0 & -2 & 1 \end{bmatrix}
Calculate A−1 A^{-1} : A−1=1−11[122−1−3−10−21] A^{-1} = \frac{1}{-11} \begin{bmatrix} 1 & 2 & 2 \\ -1 & -3 & -1 \\ 0 & -2 & 1 \end{bmatrix}

STEP 7

Verify the inverse by checking T(T−1(v))=v T(T^{-1}(\mathbf{v})) = \mathbf{v} :
Multiply A A by A−1 A^{-1} and check if it results in the identity matrix.

STEP 8

Express the action of T−1 T^{-1} on a general vector:
T−1[xyz]=1−11[122−1−3−10−21][xyz] T^{-1}\left[\begin{array}{l} x \\ y \\ z \end{array}\right] = \frac{1}{-11} \begin{bmatrix} 1 & 2 & 2 \\ -1 & -3 & -1 \\ 0 & -2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}
Perform the matrix multiplication to express the inverse transformation.
The inverse transformation T−1 T^{-1} is:
T−1[xyz]=[1−11(x+2y+2z)1−11(−x−3y−z)1−11(−2y+z)] T^{-1}\left[\begin{array}{l} x \\ y \\ z \end{array}\right] = \begin{bmatrix} \frac{1}{-11}(x + 2y + 2z) \\ \frac{1}{-11}(-x - 3y - z) \\ \frac{1}{-11}(-2y + z) \end{bmatrix}

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