User:Egm4313.s12.team8.kyle/R5.7

Problem Statement
Given:

$$ \mathbf v=4\mathbf {e_1}+2\mathbf {e_2}=c_1\mathbf {b_1}+c_2\mathbf {b_2} $$

$$ \mathbf {b_1}=2\mathbf {e_1}+7\mathbf {e_2} $$

$$ \mathbf {b_2}=1.5\mathbf {e_1}+3\mathbf {e_2} $$

1. Find the components $$ c_1, c_2 $$ using the Gram matrix.

2. Verify the result by using $$ \mathbf {b_1}, \mathbf {b_2}$$ in $$ \mathbf {v}$$.

Solution
1. Since $$\mathbf {b_1}$$ and $$\mathbf {b_2}$$ are linearly independent we conclude:

$$\Gamma \neq 0$$ so Inverse of Gram exists so $$ \mathbf{c} = \mathbf{\Gamma}^{-1}\mathbf{d} $$

Where $$\mathbf {c} = [c_1, c_2]^T$$

and $$\mathbf{d} = [<\mathbf {b_1}, \mathbf {v}>, <\mathbf {b_2}, \mathbf {v}>]^T$$

Solving for the Gram Matrix gives:

$$\mathbf\Gamma=\left[ \begin{array}{cc} 4+49 & 3+21 \\ 3+21 & 2+(\frac14)+9 \end{array} \right] = \left[ \begin{array}{cc} 53 & 24 \\ 24 & 11+(\frac14) \end{array}\right]$$

Inverting the matrix using Wolfram Alpha yields:

$$ \mathbf{\Gamma}^{-1}=(\frac{1}{81})\left[ \begin{array}{cc} 45 & -96 \\ -96 & 212 \end{array}\right]$$

Solving for $$ \mathbf{d} $$ gives:

$$ \mathbf{d}=\left[ \begin{array}{c} 8+14 \\ 6+6 \end{array}\right]=\left[ \begin{array}{c} 22 \\ 12 \end{array}\right]$$

Using the relation to solve for $$\mathbf {c}$$:

$$ \mathbf{c} = \mathbf{\Gamma}^{-1}\mathbf{d}=(\frac{1}{81})\left[ \begin{array}{cc} 45 & -96 \\ -96 & 212 \end{array}\right]\left[ \begin{array}{c} 22 \\ 12 \end{array}\right]=\left[ \begin{array}{c} -2 \\ \frac{16}{3} \end{array}\right]$$

$$ c_1=-2, \, c_2=\frac{16}{3} $$

2. Verifying the result by using the calculated $$c_1,c_2$$ and $$ \mathbf {b_1}, \mathbf {b_2}$$ in $$ \mathbf {v}$$ gives:

$$ 4\mathbf {e_1}+2\mathbf {e_2}=c_1\mathbf {b_1}+c_2\mathbf {b_2} $$

$$ 4\mathbf {e_1}+2\mathbf {e_2}=(2c_1+\frac{3}{2}c_2)\mathbf {e_1}+(7c_1+3c_2)\mathbf {e_2}$$

Simplifying to:

$$4\mathbf {e_1}+2\mathbf {e_2}=4\mathbf {e_1}+2\mathbf {e_2}$$

Yielding equal expressions and thereby verifying the results for $$c_1$$ and $$c_2$$.