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 March 14th, 2018, 01:15 AM #1 Newbie   Joined: Mar 2018 From: Pakistan Posts: 1 Thanks: 0 Modifying Solution of System of Linear Equation Consider that we have a linear system of equations resulting in the following matrix equation $Ax=b$ where $A$ is a $3\times 3$ matrix, $x$ and $b$ are $3\times 1$ vectors. Let's assume that we know the solution of this system of equation and represent it with $y$ whose entries are given by $y_i$. Now I want to change the matrix $A$ such that the new solution is vector $z$ in which $z_1>y_1$, $z_2=y_2$, and $z_3  March 22nd, 2018, 12:44 PM #2 Global Moderator Joined: Dec 2006 Posts: 20,484 Thanks: 2041 Sorry about delay.  March 23rd, 2018, 06:15 AM #3 Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 Write out Az=b Put in the z you want, b, and all of A except one unknown column. Solve for that column.  March 23rd, 2018, 06:50 AM #4 Senior Member Joined: Sep 2016 From: USA Posts: 599 Thanks: 366 Math Focus: Dynamical systems, analytic function theory, numerics This is a fairly cute exercise. Obviously there are a number of free choices which can be made which will work. Here is one way. Given$Ay = b$, you want to find a matrix$A'$so that$(A + A')(y+v) = b$where$A'$is the perturbation of your current matrix and$v = z-y$is the perturbation of your current solution. In your case, you are interested in$v = (v_1,v_2,v_3)$satisfying$v_1 > 0, v_2 = 0, v_3 < 0$. Expanding the above linear equation and using the fact that$Ay = b$you get $(A + A')(y+v) = b \implies A'z = -Av$ Notice that the only unknown in this equation is the matrix$A'$. Now, expand$z,v$in a basis (in particular, the basis for which$A$is written) as$z = z_1e_1 + z_2e_2 + z_3e_3$and$v = v_1e_1 + v_2e_2 + v_3e_3$and observe the action of$A,A'$on each basis vector. For example, with$e_1$you have $A'(z_1e_1) = -A(v_1e_1)$ and applying linearity you conclude that $\frac{-v_1}{z_1} A(e_1) = A'(e_1)$ and recall that$A(e_1)$is nothing more than the first column of$A$. So this says you may take the first column of$A'$to be the vector$\frac{-v_1}{z_1} A(e_1)$i.e. you should scale the first column of$A$by$\frac{-v_1}{z_1}$. As long as you choose$v_1 > 0$your condition that$z_1 > y_1\$ will be satisfied. Repeat this to obtain the 2nd and 3rd columns which will satisfy your remaining conditions.

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