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QuestionSuppose T:R2→R3T: R^{2} \rightarrow \mathbb{R}^{3} is a linear transformation. Three vectors u1,u2\mathbf{u}_{1}, \mathbf{u}_{2} and u3\mathbf{u}_{3} are given below together with their images by TT. Find T(w)T(\mathbf{w}) for the given vector w\mathbf{w}. u1=[22]u2=[−6−6]u3=[23]w=[−10−13]T(u1)=[208−16]T(u2)=[−60−2448]T(u3)=[2511−21]T(w)=[000]\begin{array}{l} \mathbf{u}_{1}=\left[\begin{array}{l} 2 \\ 2 \end{array}\right] \mathbf{u}_{2}=\left[\begin{array}{l} -6 \\ -6 \end{array}\right] \mathbf{u}_{3}=\left[\begin{array}{l} 2 \\ 3 \end{array}\right] \mathbf{w}=\left[\begin{array}{l} -10 \\ -13 \end{array}\right] \quad T\left(\mathbf{u}_{1}\right)=\left[\begin{array}{c} 20 \\ 8 \\ -16 \end{array}\right] \quad T\left(\mathbf{u}_{2}\right)=\left[\begin{array}{c} -60 \\ -24 \\ 48 \end{array}\right] \quad T\left(\mathbf{u}_{3}\right)=\left[\begin{array}{c} 25 \\ 11 \\ -21 \end{array}\right] \\ T(\mathbf{w})=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \end{array} SUBMIT AND MARK

Studdy Solution

STEP 1

1. T:R2→R3 T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} is a linear transformation.
2. The vectors u1,u2,u3\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3} and their images under T T are given.
3. We need to find T(w) T(\mathbf{w}) for the vector w\mathbf{w}.

STEP 2

1. Express w\mathbf{w} as a linear combination of u1,u2,u3\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}.
2. Use the linearity of T T to find T(w) T(\mathbf{w}) .

STEP 3

Express w\mathbf{w} as a linear combination of u1,u2,u3\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}.
Assume w=au1+bu2+cu3\mathbf{w} = a\mathbf{u}_{1} + b\mathbf{u}_{2} + c\mathbf{u}_{3}.
This gives the equation:
[−10−13]=a[22]+b[−6−6]+c[23]\begin{bmatrix} -10 \\ -13 \end{bmatrix} = a \begin{bmatrix} 2 \\ 2 \end{bmatrix} + b \begin{bmatrix} -6 \\ -6 \end{bmatrix} + c \begin{bmatrix} 2 \\ 3 \end{bmatrix}

STEP 4

Set up the system of equations from the above vector equation:
1. 2a−6b+2c=−10 2a - 6b + 2c = -10
2. 2a−6b+3c=−13 2a - 6b + 3c = -13
Solve this system of equations to find a,b, a, b, and c c .

STEP 5

Subtract equation 1 from equation 2:
(2a−6b+3c)−(2a−6b+2c)=−13−(−10)(2a - 6b + 3c) - (2a - 6b + 2c) = -13 - (-10)
Simplify:
c=−3c = -3
Substitute c=−3 c = -3 back into equation 1:
2a−6b+2(−3)=−102a - 6b + 2(-3) = -10
Simplify:
2a−6b−6=−102a - 6b - 6 = -10
2a−6b=−42a - 6b = -4
Divide the entire equation by 2:
a−3b=−2a - 3b = -2

STEP 6

Solve for a a in terms of b b :
a=3b−2a = 3b - 2
Since we have two variables and one equation, we can choose a convenient value for b b . Let b=0 b = 0 .
Then:
a=3(0)−2=−2a = 3(0) - 2 = -2
Thus, a=−2,b=0,c=−3 a = -2, b = 0, c = -3 .

STEP 7

Use the linearity of T T :
T(w)=T(au1+bu2+cu3)T(\mathbf{w}) = T(a\mathbf{u}_{1} + b\mathbf{u}_{2} + c\mathbf{u}_{3})
=aT(u1)+bT(u2)+cT(u3)= aT(\mathbf{u}_{1}) + bT(\mathbf{u}_{2}) + cT(\mathbf{u}_{3})
Substitute the values of a,b,c a, b, c :
=−2T(u1)+0T(u2)−3T(u3)= -2T(\mathbf{u}_{1}) + 0T(\mathbf{u}_{2}) - 3T(\mathbf{u}_{3})
Calculate:
=−2[208−16]−3[2511−21]= -2 \begin{bmatrix} 20 \\ 8 \\ -16 \end{bmatrix} - 3 \begin{bmatrix} 25 \\ 11 \\ -21 \end{bmatrix}

STEP 8

Perform the scalar multiplications and addition:
=[−40−1632]+[−75−3363]= \begin{bmatrix} -40 \\ -16 \\ 32 \end{bmatrix} + \begin{bmatrix} -75 \\ -33 \\ 63 \end{bmatrix}
Add the vectors:
=[−115−4995]= \begin{bmatrix} -115 \\ -49 \\ 95 \end{bmatrix}
Thus, T(w)=[−115−4995] T(\mathbf{w}) = \begin{bmatrix} -115 \\ -49 \\ 95 \end{bmatrix} .
The solution is:
T(w)=[−115−4995] T(\mathbf{w}) = \begin{bmatrix} -115 \\ -49 \\ 95 \end{bmatrix}

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