User:Egm4313.s12.team11.Suh/R5.7

Problem Statement
$$v=4e_{1}+2e_{2}=c_{1}b_{1}+c{2}b_{2}\!$$

$$b_{1}=2e_{1}+7e_{2}\!$$

$$b_{2}=1.5e_{1}+3e_{2}\!$$ 1. Find the components $$c_{1},c_{2}\!$$ using the Gram matrix. 2. Verify the result by using $$b_{1}\!$$ and $$b_{2}\!$$, and rely on the non-zero determinant matrix of $$b_{1}\!$$ and $$b_{2}\!$$ relative to the bases of $$e_{1}\!$$ and $$e_{2}\!$$.

Gram Matrix
$$ T(b_{1},b_{2})=\begin{bmatrix}  &  \\  &  \end{bmatrix}$$ $$ = \!$$ Thus,

$$ =  = <(2)(2) + (7)(7)> = 53\!$$

$$ =  = <(1.5)(1.5) + (3)(3)> = 11.25\!$$

$$ =  =  = <(2)(1.5) + (7)(3)> = 24\!$$

$$ T(b_{1},b_{2})=\begin{bmatrix} 53 & 24 \\ 24 & 11.25 \end{bmatrix}$$

$$\Gamma = det[T] = (53)(11.25)-(24)(24) = 20.25\!$$

Defining c
Define: $$c =\begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix}$$ $$d = \begin{bmatrix}  \\  \end{bmatrix}= \begin{bmatrix} d_{1} \\ d_{2} \end{bmatrix}\!$$

If $$\Gamma \neq 0 \!$$, then $$\Gamma^{-1}\!$$ exists

$$c = \Gamma^{-1}d\!$$

Finding c
$$\Gamma = 20.25 \neq 0 \!$$ thus, $$\Gamma ^{-1}\!$$exists

$$d =\begin{bmatrix} d_{1} \\ d_{2} \end{bmatrix} = \begin{bmatrix} (2)(4)+(7)(2) \\ (1.5)(4)+(3)(2) \end{bmatrix} = \begin{bmatrix} 22 \\ 12 \end{bmatrix}\!$$

$$\begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix} = \begin{bmatrix} 53 & 24 \\ 24 & 11.25 \end{bmatrix}^{-1}\begin{bmatrix} 22 \\ 12 \end{bmatrix}\!$$

$$\begin{bmatrix} c_{1} \\ c_{2} \end{bmatrix} = \begin{bmatrix} -2 \\ 5.33 \end{bmatrix}\!$$

Part 2 Solution
$$v = 4e_{1} + 2e_{2} \equiv c_{1}b_{1} + c_{2}b_{2}\!$$

$$c_{1}b_{1} + c_{2}b_{2} = (-2)(2e_{1}+7e_{2}) + (5.33)(1.5e_{1} + 3e_{2})\!$$

$$c_{1}b_{1} + c_{2}b_{2} = -4e_{1} - 14e_{2} + 8e_{1} + 16e_{2}\!$$

$$c_{1}b_{1} + c_{2}b_{2} = 4e_{1} + 2e_{2}\!$$

$$4e_{1} + 2e_{2} \equiv c_{1}b_{1} + c_{2}b_{2}\!$$

Thus, the solution is correct.