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I'm having trouble with this problem. I'm not sure how to solve it the "calculus way". My professor gave us the formula "$b = c + Dt$" to work with (y = mx + b), and then dealt with partial derivatives. I'm not entirely sure how to start it though. Thanks.

Find the best solution for the given system in two ways (a) using calculus (b) using linear algebra.

$\left\{ \begin{array}{c} x+2y=3 \\ 3x+2y=5 \\ x+y=2.09 \\ \end{array} \right.$

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    $\begingroup$ We can find x=1 and y=1 from first two equations. But as you can see x=1 and y=1 not satisfy third equation. $\endgroup$ Mar 22, 2017 at 7:19

2 Answers 2

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Consider the problem to be $$\left\{ \begin{array}{c} f_1=x+2y-3 \\ f_2=3x+2y-5 \\ f_3=x+y-2.09 \\ \end{array} \right.$$

and define $$\Phi=f_1^2+f_2^2+f_3^2$$ and say that you want this norm to be minimum.

So, compute the derivatives to get $$\Phi'_x=22x+18y-40.18$$ $$\Phi'_y=18x+18y-36.18$$ Set them equal to $0$. Solve for $x,y$.

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Solving by Linear Algebra-

There is no solution to these equations. You write the coefficients of the equation and its right had side in as an augmented matrix and you do row reduction. $$\left[ \begin{array}{cc|c} 1&2&3\\ 3&2&5\\ 1&1&2.09 \end{array} \right]$$ Consider each row as R1,R2,R3. Now perform $${R2-3R1}$$ $$\left[ \begin{array}{cc|c} 1&2&3\\ 0&-4&-4\\ 1&1&2.09 \end{array} \right]$$ $${-R2\over4}$$ $$\left[ \begin{array}{cc|c} 1&2&3\\ 0&1&1\\ 1&1&2.09 \end{array} \right]$$ $${R3-R1}$$ $$\left[ \begin{array}{cc|c} 1&2&3\\ 0&1&1\\ 0&-1&-0.91 \end{array} \right]$$

$${R3+R1}$$ $$\left[ \begin{array}{cc|c} 1&2&3\\ 0&1&1\\ 0&0&0.09 \end{array} \right]$$

Last row R3 denotes that 0=0.09. So there is no solution for the equations.

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