Question

Consider the basis S = {v₁, v₂} for R², where v₁ = (− 2, 1) and v₂ = (1, 3), and let T:R² → R³ be the linear transformation such that

T(v₁) = (− 1, 4, 0) and T(v₂) = (0, − 5, 9)

Find a formula for T(x₁, x₂), and use that formula to find T(4, − 5).

Give exact answers in the form of a fraction.

Answer 1

$Solution~:~Let~ us ~consider~ basis~ for ~ R^2 ~ given ~by: \\S= \{ \begin{bmatrix}-2 \\1 \end{bmatrix},\begin{bmatrix}1 \\3 \end{bmatrix} \} ~and ~let ~ us ~consider ~ linear ~ operator~ T:R^2 \rightarrow R^3~ for ~which : \\T(v_1)= \begin{bmatrix}-1 \\4 \\0 \end{bmatrix}~ and~ T(v_2)=\begin{bmatrix}0 \\-5 \\9 \end{bmatrix} \\Let ~ us ~ first ~ find ~ formula ~ for ~ T(x_1,x_2): \\(x_1,x_2)=av_1+bv_2 \\~~~~~~~~~~~~~~=a\begin{bmatrix}-2 \\1 \end{bmatrix}+b\begin{bmatrix}1 \\3 \end{bmatrix} \\\begin{bmatrix}x_1 \\x_2 \end{bmatrix}=\begin{bmatrix}-2(a) \\1(a) \end{bmatrix}+\begin{bmatrix}1(b) \\3(b) \end{bmatrix} \\\begin{bmatrix}x_1 \\x_2 \end{bmatrix}=\begin{bmatrix}-2a \\a \end{bmatrix}+\begin{bmatrix}b \\3b \end{bmatrix} \\\begin{bmatrix}x_1 \\x_2 \end{bmatrix}=\begin{bmatrix}-2a+b \\a+3b \end{bmatrix} \\\Rightarrow~x_1=-2a+b~ and ~x_2 =a+3b, Solving ~these ~ equations ~ for ~a ~and ~b, we ~get \\a=\frac{1}{7}(-3x_1+x_2) ~and ~b=\frac{1}{7}(x_1+2x_2)~ which ~gives ~us: \\(x_1,x_2)=av_1+bv_2=\frac{1}{7}(-3x_1+x_2)v_1+\frac{1}{7}(x_1+2x_2)v_2 \\Therefore, we ~have: \\T(x_1,x_2)=T(av_1+bv_2)=aT(v_1)+bT(v_2) \\~~~~~~~~~~~~~~~~~=\frac{1}{7}(-3x_1+x_2)T(v_1)+\frac{1}{7}(x_1+2x_2)T(v_2) \\~~~~~~~~~~~~~~~~~=\frac{1}{7}(-3x_1+x_2)\begin{bmatrix}-1 \\4 \\0 \end{bmatrix}+\frac{1}{7}(x_1+2x_2)\begin{bmatrix}0 \\-5 \\9 \end{bmatrix} \\~~~~~~~~~~~~~~~~~=\frac{1}{7}~\{(-3x_1+x_2)\begin{bmatrix}-1 \\4 \\0 \end{bmatrix}+(x_1+2x_2)\begin{bmatrix}0 \\-5 \\9 \end{bmatrix}\} \\~~~~~~~~~~~~~~~~~=\frac{1}{7}\{\begin{bmatrix} 3x_1-x_2 \\-12x_1+4x_2 \\0 \end{bmatrix}+\begin{bmatrix}0 \\-5x_1-10x_2 \\9x_1+18x_2 \end{bmatrix}\} \\~~~~~~~~~~~~~~~~~=\frac{1}{7}\begin{bmatrix} 3x_1-x_2 \\-17x_1-6x_2 \\9x_1+18x_2 \end{bmatrix} \\Which~gives: \\T(4,-5)=\frac{1}{7}\begin{bmatrix} 3(4)-(-5) \\-17(4)-6(-5) \\9(4)+18(-5) \end{bmatrix} \\~~~~~~~~~~~~~~~~~=\frac{1}{7}\begin{bmatrix} 12+5 \\-68+30 \\36-90 \end{bmatrix} \\~~~~~~~~~~~~~~~~~=\frac{1}{7}\begin{bmatrix} 17 \\-38 \\-54 \end{bmatrix} \\~~~~~~~~~~~~~~~~~=\begin{bmatrix} \frac{17}{7} \\-\frac{38}{7} \\-\frac{54}{7} \end{bmatrix} \\~~~~~~~~~~~~~~~~~=\begin{bmatrix} 2.428571 \\-5.428571 \\-7.71428571 \end{bmatrix}$