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.
$\endgroup$ 1Find the best solution for the given system in two ways (a) using calculus (b) using linear algebra.
$\left\{ \begin{array}{c} x+2y=3 \\ 3x+2y=5 \\ x+y=2.09 \\ \end{array} \right.$
2 Answers
$\begingroup$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 $x,y$.
$\endgroup$ $\begingroup$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.
