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]=(121−1∣∣30)R2→R2−2R1(101−3∣∣3−6). The second row reads −3y=−6⇒y=2; back-substituting into the first row, x+2=3⇒x=1 -- matching the Cramer's-rule answer above.
The general shape for a 3×3 system. Row-reduce [A∣B] to 100∗10∗∗1∣∣∣∗∗∗-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.
Balancing a chemical equation. Writing a reaction x1(reactant1)+x2(reactant2)→x3(product1)+x4(product2) and demanding the atom count of each element balance on both sides gives one homogeneous linear equation per element in the unknowns x1,…,x4. Gaussian elimination on this homogeneous system typically leaves one unknown free; choosing the smallest value that makes every xi a positive integer gives the balanced equation.
Tip
Because Gaussian elimination never needs A to be square, it is the natural tool whenever a word problem produces more equations than unknowns (an over-determined system) or fewer (an under-determined one) -- situations where matrix inversion and Cramer's rule simply do not apply.
Reduce each augmented matrix to upper-triangular (row-echelon) form using row operations, then back-substitute.
x=−1,y=4,z=4.
x=3,y=1,z=2.
✓Final answer
x=−1,y=4,z=4.
x=3,y=1,z=2.
For each system we build the augmented matrix, use row operations to reach upper-triangular form, and then back-substitute to find z, then y, then x.
Step 1. (i) Write the augmented matrix.
213−22−13−12231
Step 2. (i) Eliminate x from R2,R3 using R1. Apply R2→2R2−R1 and R3→2R3−3R1: