My Math Forum Modifying Solution of System of Linear Equation

 Linear Algebra Linear Algebra Math Forum

 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 $z_3  March 22nd, 2018, 12:44 PM #2 Global Moderator Joined: Dec 2006 Posts: 19,178 Thanks: 1646 Sorry about delay.  March 23rd, 2018, 06:15 AM #3 Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,390 Thanks: 100 Write out Az=b Put in the z you want, b, and all of A except one unknown column. Solve for that column.  March 23rd, 2018, 06:50 AM #4 Senior Member Joined: Sep 2016 From: USA Posts: 395 Thanks: 211 Math Focus: Dynamical systems, analytic function theory, numerics This is a fairly cute exercise. Obviously there are a number of free choices which can be made which will work. Here is one way. Given$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.

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post JulieK Linear Algebra 1 June 15th, 2014 08:21 PM Primoz Linear Algebra 1 April 11th, 2014 06:50 AM evariste_jian Applied Math 0 September 10th, 2013 12:52 AM maxgeo Algebra 1 July 26th, 2012 08:29 PM evariste_jian Algebra 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top