User:Egm6341.s10.team4.anandankala/HW6-3

Given

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 * $$\underbrace{\begin{bmatrix}

1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \end{bmatrix}.}_{A} \underbrace{\begin{bmatrix}

C_0 \\ C_1 \\ C_2 \\ C_3 \end{bmatrix}}_{C} = \begin{bmatrix}

z_i \\ z^'_i \\ z_i+1 \\ z^'_i+1 \end{bmatrix} $$
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Prove That

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 * Arrive at the third and fourth rows of the matrix A using the equations,

$$\displaystyle d_3=z_i+1=z(s=1).$$ $$\displaystyle d_4=z^'_{i+1}=z^'(s=1).$$
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Solution

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We have $$\displaystyle z(s)= \sum_{i=0}^{3} C_i s^i= C_0+ C_1s+ C_2s^2+ C_3s^3$$ $$
 * $$\displaystyle \longrightarrow (1)
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$$\displaystyle z^'(s)= \sum_{i=1}^{3} iC_i s^{i-1}= C_1+ 2C_2s+ 3C_3s^2$$ $$
 * $$\displaystyle \longrightarrow (2)
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We get the product of the third row of the matrix A and the column vector C by substituting s=1 in the equation(1). $$\displaystyle \Rightarrow z_{i+1}= z(s=1)= C_0+ C_1+ C_2+ C_3$$ $$
 * $$\displaystyle \longrightarrow (3)


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We get the product of the fourth row of the matrix A and the column vector C by substituting s=1 in the equation(2). $$\displaystyle \Rightarrow z^'_{i+1}= z^'(s=1)= C_1+ 2C_2+ 3C_3$$ $$
 * $$\displaystyle \longrightarrow (4)


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By looking at the equations (3) and (4), one can say that the third and fourth rows of the matrix A are [1 1 1 1] and [0 1 2 3].
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