Solving Systems of Linear Equations using Matrices

A system of linear equations can be written as a single matrix equation $AX=B$, where $A$ is the matrix of coefficients, $X$ the column of unknowns, and $B$ the column of constants. For example, $\begin{aligned} a_1x+b_1y&=c_1\\ a_2x+b_2y&=c_2 \end{aligned}\quad\Longleftrightarrow\quad \underbrace{\begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix}}_{A}\underbrace{\begin{pmatrix} x \\ y \end{pmatrix}}_{X}=\underbrace{\begin{pmatrix} c_1 \\ c_2 \end{pmatrix}}_{B}.$ Two matrix methods solve this system; both are standard Std XII commerce techniques.

Matrix inversion method. If $A$ is non-singular ($|A|\neq 0$), multiply the equation $AX=B$ on the left by $A^{-1}$: $A^{-1}(AX)=A^{-1}B\ \Rightarrow\ (A^{-1}A)X=A^{-1}B\ \Rightarrow\ IX=A^{-1}B\ \Rightarrow\ X=A^{-1}B.$ So the solution is simply $X=A^{-1}B$: find $A^{-1}$ (by either method of the previous section) and multiply it by the constant column. Because multiplication is not commutative, $A^{-1}$ must pre-multiply $B$ (i.e. $A^{-1}B$, never $BA^{-1}$).

Reduction method (Gaussian elimination). Instead of inverting $A$, form the augmented matrix $[\,A\mid B\,]$ and apply elementary row operations to reduce the left side to upper-triangular (row-echelon) form. Reading the reduced rows back as equations gives the value of the last unknown immediately, and the rest follow by back-substitution. This method avoids computing an inverse and is often faster for a $3\times 3$ system. …