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Generalizations: 2B The General Linear-Systems Problem

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Example

In algebra and precalculus, the forward problem we learn to solve is function evaluation and the backward problem is algebra based on the quadratic function.

Let $f (x) = x^{2}$ . (Remember that this is nonlinear, so this analogy only goes so far.)

Let's solve the forward problem by substituting in a value for $x$ . Let $x = 2$ :

\begin{aligned} ⟹ f (x) = f (2) = 2^{2} = 4 \\ ⟹ (2, 4) \in f \end{aligned}

To solve, the backward problem, we have three options.

Option 1: Unique Solution

\begin{aligned} f (x) = x^{2} & = 0 \\ ⟹ \sqrt{x^{2}} & = \sqrt{0} \\ ⟹ | x | & = 0 \\ ⟹ x & = 0 \end{aligned}

Option 2: Nonunique solution

\begin{aligned} f (x) = 4 & ⟹ x^{2} = 4 \\ ⟹ \sqrt{x^{2}} = \sqrt{4} \\ ⟹ | x | = + 2 \\ ⟹ x = + 2 or x = - 2 \end{aligned}

The analogy breaks down at the line because there are multiple solutions. In matrix-vector multiplication, if a solution exists and it is not unique, there will be many more than just two solutions.

Option 3: Output $b \in Codomain (f)$ but $b \notin Rng (f)$

f (x) = - 1 ⟹ x^{2} = - 1

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There is no solution in $R$ , but we can "minimize" the "distance" from outputs of $f (x)$ to our chosen value of $b \in Codomain (f)$ .