Method for solving system of linear equations

Question

I thought of this method:

Suppose the given equations are:

$$a_1x+a_2y+a_3z=d_1$$ $$b_1x+b_2y+b_3z=d_2$$ $$c_1x+c_2y+c_3z=d_3$$

Let $\vec{p}=x\hat{i}+y\hat{j}+z\hat{k},\:\:\:\: \vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k},\:\:\:\: \vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k},\:\:\:\: \vec{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$

Then, $d_1=\vec{p}\cdot \vec{a},\:\:\:\: d_2=\vec{p}\cdot \vec{b},\:\:\:\: d_3=\vec{p}\cdot \vec{c}$

Now, evaluate $\vec{1}=\vec{a}, \:\:\:\:\vec{2}=\vec{b}-\frac{(\vec{a}\cdot\vec{b})\vec{a}}{|\vec{a}|^2},\:\:\:\: \vec{3}=\vec{c}-\frac{(\vec{c}\cdot\vec{a})\vec{a}}{|\vec{a}|^2}-\frac{(\vec{c}\cdot \vec{b})\vec{b}}{|\vec{b}|^2}$. The vectors $\frac{\vec{1}}{|\vec{1}|}, \frac{\vec{2}}{|\vec{2}|}, \frac{\vec{3}}{|\vec{3}|}$ are perpendicular to each other. The components of $\vec{p}$ along these three vectors are $\frac{d_1}{|\vec{a}||\vec{1}|}$, $\frac{1}{|\vec{2}|}(d_2-\frac{(\vec{a}\cdot\vec{b})d_1}{|\vec{a}|^2})$, and $\frac{1}{|\vec{3}|}(d_3-\frac{(\vec{c}\cdot\vec{a})d_1}{|\vec{a}|^2}-\frac{(\vec{c}\cdot \vec{b})d_2}{|\vec{b}|^2})$ respectively.

Therefore, $$\vec{p}=d_1\frac{\vec{1}}{|\vec{1}|^2}+\left(d_2-\frac{(\vec{a}\cdot\vec{b})d_1}{|\vec{a}|^2}\right)\frac{\vec{2}}{|\vec{2}|^2}+\left(d_3-\frac{(\vec{c}\cdot\vec{a})d_1}{|\vec{a}|^2}-\frac{(\vec{c}\cdot \vec{b})d_2}{|\vec{b}|^2}\right)\frac{\vec{3}}{|\vec{3}|^2}$$

Is this method more or less expensive than Cramer's Rule? Or Does it involve the same calculations as in Cramer's rule?

Answer 1

Congrats, you rediscovered the Gram-Schmidt orthogonalization method, which generates an orthogonal base from a given set of vectors. Though your expressions are wrong (for instance, $\vec 1\cdot\vec 2\ne0$). But the intuition is correct.

As far as I know, this is not used for the numerical resolution of a system, as Gaussian elimination is more efficient. Anyway, for large system it is certainly better than Cramer (which as a cost proportional to $n!$).

For the case of $n=3$, Cramer takes $4$ determinant evaluations, each taking a sum of six triple products, and three divisions. For your method, it's you job to count the operations and compare. Be sure to write down all equations giving the final unknowns.