An algebra problem by Rocco Dalto

Algebra Level pending

Let P 2 \mathbb P_{2} denote the set of all polynomials of degree two and R 3 \mathbb R^3 denote the set of all vectors in three space.

Let f : P 2 → R 3 f: \mathbb P_{2} \rightarrow \mathbb R^3 be linear transform defined by

f ( p 0 + p 1 x + p 2 x 2 ) = ( p 0 + p 1 p 1 + p 2 p 0 + p 2 ) f(p_{0} + p_{1} x + p_{2} x^2) = \left( \begin{array}{ccc} p_{0} + p_{1} \\ p_{1} + p_{2} \\ p_{0} + p_{2} \\ \end{array} \right)

Let
A = { 1 + x , x + x 2 , x 2 } A = \{1 + x, x + x^2, x^2 \} be a basis for P 2 \mathbb P_{2} and
B = { ( 1 − 1 1 ) , ( − 1 1 1 ) , ( 1 0 1 ) } B = \left \{ \left( \begin{array}{ccc} 1 \\ -1 \\ 1 \\ \end{array} \right), \left( \begin{array}{ccc} -1 \\ 1 \\ 1 \ \end{array} \right), \left( \begin{array}{ccc} 1 \\ 0 \\ 1 \ \end{array} \right) \right \} be a basis for R 3 . \mathbb R^3. .

Let the matrix M = [ a i j ] 3 x 3 M = [a_{ij}]_{3 \: x \: 3} represent the linear transform above.

If [ p ] A = ( x 1 x 2 x 3 ) [p]_{A} = \left( \begin{array}{ccc} x_{1} \\ x_{2} \\ x_{3} \ \end{array} \right) , [ f ( p ) ] B = ( 1 2 3 ) \: [f(p)]_{B} = \left( \begin{array}{ccc} 1 \\ 2 \\ 3 \ \end{array} \right) , S = ∑ j = 1 3 x j \: S = \sum_{j = 1}^{3} x_{j} , p = b 0 + b 1 x + b 2 x 2 \: p = b_{0} + b_{1} x + b_{2} x^2 , and T = ∑ j = 0 2 b j \: T = \sum_{j = 0}^{2} b_{j} ,

Find: S + T S + T .

Refer to previous problem. . .


The answer is 10.5.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Dec 29, 2016

f ( 1 + x ) = ( 2 1 1 ) = α 1 ( 1 − 1 1 ) + α 2 ( − 1 1 1 ) + α 3 ( 1 0 1 ) = ( α 1 − α 2 + α 3 − α 1 + α 2 α 1 + α 2 + α 3 ) f(1 + x) = \left( \begin{array}{ccc} 2 \\ 1 \\ 1 \\ \end{array} \right) = \alpha_{1} \left( \begin{array}{ccc} 1 \\ -1 \\ 1 \\ \end{array} \right) + \alpha_{2} \left( \begin{array}{ccc} -1 \\ 1 \\ 1 \\ \end{array} \right) + \alpha_{3} \left( \begin{array}{ccc} 1 \\ 0 \\ 1 \\ \end{array} \right) = \left( \begin{array}{ccc} \alpha_{1} - \alpha_{2} + \alpha_{3} \\ -\alpha_{1} + \alpha_{2} \\ \alpha_{1} + \alpha_{2} + \alpha_{3} \\ \end{array} \right)

f ( x + x 2 ) = ( 1 2 1 ) = ( β 1 − β 2 + β 3 − β 1 + β 2 β 1 + β 2 + β 3 ) f(x + x^2) = \left( \begin{array}{ccc} 1 \\ 2 \\ 1 \\ \end{array} \right) = \left( \begin{array}{ccc} \beta_{1} - \beta_{2} + \beta_{3} \\ -\beta_{1} + \beta_{2} \\ \beta_{1} + \beta_{2} + \beta_{3} \\ \end{array} \right)

f ( x 2 ) = ( 0 1 1 ) = ( γ 1 − γ 2 + γ 3 − γ 1 + γ 2 γ 1 + γ 2 + γ 3 ) f(x^2) = \left( \begin{array}{ccc} 0 \\ 1 \\ 1 \\ \end{array} \right) = \left( \begin{array}{ccc} \gamma_{1} - \gamma_{2} + \gamma_{3} \\ -\gamma_{1} + \gamma_{2} \\ \gamma_{1} + \gamma_{2} + \gamma_{3} \\ \end{array} \right)

Solving the system:

α 1 ∗ − α 2 ∗ + α 3 ∗ = a \alpha_{1}^{*} - \alpha_{2}^{*} + \alpha_{3}^{*} = a − α 1 ∗ + α 2 ∗ = b -\alpha_{1}^{*} + \alpha_{2}^{*} = b α 1 ∗ + α 2 ∗ + α 3 ∗ = c \alpha_{1}^{*} + \alpha_{2}^{*} + \alpha_{3}^{*} = c

⟹ ( α 1 ∗ α 2 ∗ α 3 ∗ ) = ( c − a − 2 b 2 c − a 2 a + b ) ⟹ \implies \left( \begin{array}{ccc} \alpha_{1}^{*} \\ \alpha_{2}^{*} \\ \alpha_{3}^{*} \\ \end{array} \right) = \left( \begin{array}{ccc} \dfrac{c - a - 2b}{2} \\ \dfrac{c - a}{2} \\ a + b \\ \end{array} \right) \implies

( α 1 α 2 α 3 ) = ( − 3 2 − 1 2 3 ) , ( β 1 β 2 β 3 ) = ( − 2 0 3 ) , ( γ 1 γ 2 γ 3 ) = ( − 1 2 1 2 1 ) ⟹ \left( \begin{array}{ccc} \alpha_{1} \\ \alpha_{2} \\ \alpha_{3} \\ \end{array} \right) = \left( \begin{array}{ccc} -\dfrac{3}{2} \\ -\dfrac{1}{2} \\ 3 \\ \end{array} \right), \: \left( \begin{array}{ccc} \beta_{1} \\ \beta_{2} \\ \beta_{3} \\ \end{array} \right) = \left( \begin{array}{ccc} -2 \\ 0 \\ 3 \\ \end{array} \right), \: \left( \begin{array}{ccc} \gamma_{1} \\ \gamma_{2} \\ \gamma_{3} \\ \end{array} \right) = \left( \begin{array}{ccc} -\dfrac{1}{2} \\ \dfrac{1}{2} \\ 1 \\ \end{array} \right) \implies

Matrix M = ∣ − 3 2 − 2 − 1 2 − 1 2 0 1 2 3 3 1 ∣ M = \left| \begin{array}{ccc} -\dfrac{3}{2} & -2 & -\dfrac{1}{2} \\ -\dfrac{1}{2} & 0 & \dfrac{1}{2} \\ 3 & 3 & 1 \\ \ \end{array} \right|

d e t ( M ) = − 1 < > 0 ⟹ det(M) = -1 <> 0 \implies [ f ( p ) ] B = ( 1 2 3 ) = ∣ − 3 2 − 2 − 1 2 − 1 2 0 1 2 3 3 1 ∣ ( x 1 x 2 x 3 ) \: [f(p)]_{B} = \left( \begin{array}{ccc} 1 \\ 2 \\ 3 \ \end{array} \right) = \left| \begin{array}{ccc} -\dfrac{3}{2} & -2 & -\dfrac{1}{2} \\ -\dfrac{1}{2} & 0 & \dfrac{1}{2} \\ 3 & 3 & 1 \\ \ \end{array} \right| \left( \begin{array}{ccc} x_{1} \\ x_{2} \\ x_{3} \ \end{array} \right) has a unique solution.

Rewriting the system M [ p ] A = [ f ( p ) ] B ⟹ M [p]_{A} = [f(p)]_{B} \implies 3 x 1 + 4 x 2 + x 3 = − 2 3 x_{1} + 4 x_{2} + x_{3} = -2 − x 1 + x 3 = 4 -x_{1} + x_{3} = 4 3 x 1 + 3 x 2 + x 3 = 3 3 x_{1} + 3 x_{2} + x_{3} = 3

⟹ x 3 = 4 + x 1 ⟹ x 2 = − 5 , x 1 = 7 2 ⟹ x 3 = 15 2 \implies x_{3} = 4 + x_{1} \implies x_{2} = -5, \: x_{1} = \dfrac{7}{2} \implies x_{3} = \dfrac{15}{2}

⟹ [ p ] A = ( 7 2 − 5 15 2 ) ⟹ S = ∑ j = 1 3 x j = 6 \implies \: [p]_{A} = \left( \begin{array}{ccc} \dfrac{7}{2} \\ -5 \\ \dfrac{15}{2} \ \end{array} \right) \implies S = \sum_{j = 1}^{3} x_{j} = \boxed{6} ,

a n d [ p ] A = ( 7 2 − 5 15 2 ) ⟹ and \: [p]_{A} = \left( \begin{array}{ccc} \dfrac{7}{2} \\ -5 \\ \dfrac{15}{2} \ \end{array} \right) \implies

p = 7 5 ( 1 + x ) + − 5 ( x + x 2 ) + 15 2 x 2 = 7 2 − 3 2 x + 5 2 x 2 ⟹ T = ∑ j = 0 2 b j = 9 2 p = \dfrac{7}{5} (1 + x) + -5 (x + x^2) + \dfrac{15}{2} x^2 = \dfrac{7}{2} - \dfrac{3}{2} x + \dfrac{5}{2} x^2 \implies T = \sum_{j = 0}^{2} b_{j} = \boxed{\dfrac{9}{2}}

∴ S + T = 21 2 \therefore \: S + T = \boxed{\dfrac{21}{2}} or 10.5 \boxed{10.5}

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