Math  /  Algebra

QuestionIf T:R2R3T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} is a linear transformation and the action of TT on the special vectors u\mathbf{u} and v\mathbf{v} is as given, find a formula for T(x)T(\mathbf{x}), where x\mathbf{x} is any vector in R2\mathbb{R}^{2}. u=[34]v=[45]T(u)=[222]T(v)=[323]T[xy]=[000]\begin{array}{l} \mathbf{u}=\left[\begin{array}{l} -3 \\ -4 \end{array}\right] \quad \mathbf{v}=\left[\begin{array}{l} 4 \\ 5 \end{array}\right] \quad T(\mathbf{u})=\left[\begin{array}{c} -2 \\ -2 \\ 2 \end{array}\right] \quad T(\mathbf{v})=\left[\begin{array}{c} 3 \\ 2 \\ -3 \end{array}\right] \\ T\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \end{array}

Studdy Solution

STEP 1

1. T:R2R3 T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} is a linear transformation.
2. The vectors u\mathbf{u} and v\mathbf{v} are given as u=[34]\mathbf{u} = \begin{bmatrix} -3 \\ -4 \end{bmatrix} and v=[45]\mathbf{v} = \begin{bmatrix} 4 \\ 5 \end{bmatrix}.
3. The transformation T T acts on u\mathbf{u} and v\mathbf{v} as follows: T(u)=[222] T(\mathbf{u}) = \begin{bmatrix} -2 \\ -2 \\ 2 \end{bmatrix} and T(v)=[323] T(\mathbf{v}) = \begin{bmatrix} 3 \\ 2 \\ -3 \end{bmatrix} .
4. We need to find a formula for T(x) T(\mathbf{x}) where x=[xy]\mathbf{x} = \begin{bmatrix} x \\ y \end{bmatrix}.

STEP 2

1. Use the properties of linear transformations.
2. Express x\mathbf{x} as a linear combination of u\mathbf{u} and v\mathbf{v}.
3. Apply the linear transformation T T to x\mathbf{x}.
4. Simplify the expression to find the formula for T(x) T(\mathbf{x}) .

STEP 3

Since T T is a linear transformation, it satisfies the properties: T(au+bv)=aT(u)+bT(v) T(a\mathbf{u} + b\mathbf{v}) = aT(\mathbf{u}) + bT(\mathbf{v}) for any scalars a a and b b .

STEP 4

Express x=[xy]\mathbf{x} = \begin{bmatrix} x \\ y \end{bmatrix} as a linear combination of u\mathbf{u} and v\mathbf{v}.
We need to find scalars a a and b b such that: a[34]+b[45]=[xy] a\begin{bmatrix} -3 \\ -4 \end{bmatrix} + b\begin{bmatrix} 4 \\ 5 \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix}
This gives us the system of equations: 3a+4b=x -3a + 4b = x 4a+5b=y -4a + 5b = y

STEP 5

Solve the system of equations for a a and b b .
Using the method of substitution or elimination, solve for a a and b b .

STEP 6

Let's solve the system using elimination:
Multiply the first equation by 4 and the second by 3: 12a+16b=4x -12a + 16b = 4x 12a+15b=3y -12a + 15b = 3y
Subtract the second equation from the first: b=4x3y b = 4x - 3y
Substitute b=4x3y b = 4x - 3y into the first equation: 3a+4(4x3y)=x -3a + 4(4x - 3y) = x 3a+16x12y=x -3a + 16x - 12y = x 3a=x16x+12y -3a = x - 16x + 12y 3a=15x+12y -3a = -15x + 12y a=5x4y a = 5x - 4y

STEP 7

Apply the linear transformation T T to x\mathbf{x} using the linear combination: T(x)=T(au+bv)=aT(u)+bT(v) T(\mathbf{x}) = T(a\mathbf{u} + b\mathbf{v}) = aT(\mathbf{u}) + bT(\mathbf{v})
Substitute a=5x4y a = 5x - 4y and b=4x3y b = 4x - 3y : T(x)=(5x4y)T(u)+(4x3y)T(v) T(\mathbf{x}) = (5x - 4y)T(\mathbf{u}) + (4x - 3y)T(\mathbf{v})

STEP 8

Substitute the values of T(u) T(\mathbf{u}) and T(v) T(\mathbf{v}) : T(x)=(5x4y)[222]+(4x3y)[323] T(\mathbf{x}) = (5x - 4y)\begin{bmatrix} -2 \\ -2 \\ 2 \end{bmatrix} + (4x - 3y)\begin{bmatrix} 3 \\ 2 \\ -3 \end{bmatrix}
Calculate each component: T(x)=[(5x4y)(2)+(4x3y)(3)(5x4y)(2)+(4x3y)(2)(5x4y)(2)+(4x3y)(3)] T(\mathbf{x}) = \begin{bmatrix} (5x - 4y)(-2) + (4x - 3y)(3) \\ (5x - 4y)(-2) + (4x - 3y)(2) \\ (5x - 4y)(2) + (4x - 3y)(-3) \end{bmatrix}

STEP 9

Simplify each component: T(x)=[10x+8y+12x9y10x+8y+8x6y10x8y12x+9y] T(\mathbf{x}) = \begin{bmatrix} -10x + 8y + 12x - 9y \\ -10x + 8y + 8x - 6y \\ 10x - 8y - 12x + 9y \end{bmatrix}
Combine like terms: T(x)=[2xy2x+2y2x+y] T(\mathbf{x}) = \begin{bmatrix} 2x - y \\ -2x + 2y \\ -2x + y \end{bmatrix}
The formula for T(x) T(\mathbf{x}) is: T(x)=[2xy2x+2y2x+y] T(\mathbf{x}) = \begin{bmatrix} 2x - y \\ -2x + 2y \\ -2x + y \end{bmatrix}

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