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March 14th, 2018, 01:15 AM   #1
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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 $z_3<y_3$. Is there a systematic way to achieve this?

In other words, is there a systematic way of finding out what changes should I introduce in matrix $A$ such that some entries of new solution $z$ are greater than corresponding entries of old solution $y$, other entries of new solution $z$ are equal to corresponding entries of old solution $y$, and still other are less than the corresponding entries of old solution $y$. Is there a method or technique to achieve this? What is it called?

Thank you.
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March 22nd, 2018, 12:44 PM   #2
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March 23rd, 2018, 06:15 AM   #3
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Write out Az=b

Put in the z you want, b, and all of A except one unknown column.

Solve for that column.
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March 23rd, 2018, 06:50 AM   #4
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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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