Q.A boy is walking along the path $y=ax^2+bx+c$ through the points $(-6,8), (-2,-12)$, and $(3,8)$. He wants to meet his friend at $P(7,60)$. Will he meet his friend? (Use Gaussian elimination method.)

Gaussian elimination solves $AX=B$ by row-reducing the augmented matrix $[A\,|\,B]$ to row-echelon form using elementary row operations, then reading off the unknowns from the bottom equation upward -- the method of back substitution. Unlike matrix inversion or Cramer's rule, it works even when the coefficient matrix is singular or non-square (any number of equations, any number of unknowns), which is exactly why it is the most general of the three methods.

Worked illustration. For $x+y=3,\ 2x-y=0$: $[A\,|\,B]=\begin{pmatrix}1&1&|&3\\2&-1&|&0\end{pmatrix}\xrightarrow{R_2\to R_2-2R_1}\begin{pmatrix}1&1&|&3\\0&-3&|&-6\end{pmatrix}$. The second row reads $-3y=-6\Rightarrow y=2$; back-substituting into the first row, $x+2=3\Rightarrow x=1$ -- matching the Cramer's-rule answer above.

The general shape for a $3\times3$ system. Row-reduce $[A\,|\,B]$ to $\begin{pmatrix}1&*&*&|&*\\0&1&*&|&*\\0&0&1&|&*\end{pmatrix}$-type echelon form; the bottom row gives $z$ directly, the middle row (with $z$ substituted) gives $y$, and the top row (with $y,z$ substituted) gives $x$.

Word problems -- a quadratic/cubic passing through given points, a polynomial's remainders under the Remainder Theorem, money split across interest-bearing bonds, a projectile's path -- all translate into a linear system whose augmented matrix is then row-reduced exactly as above; the method is identical whether the unknowns are prices, rates, or polynomial coefficients. …