Q.A man is appointed in a job with a monthly salary of certain amount and a fixed amount of annual increment. If his salary was Rs.,19{,}800 per month at the end of the first month after 3 years of service and Rs.,23{,}400 per month at the end of the first month after 9 years of service, find his starting salary and his annual increment. (Use matrix inversion method to solve the problem.)

Package a system of $n$ linear equations in $n$ unknowns as $AX=B$, with $A$ the $n\times n$ coefficient matrix, $X$ the column of unknowns, and $B$ the column of constants. Matrix inversion method applies exactly when $A$ is square and non-singular ($|A|\ne0$).

Derivation. Starting from $AX=B$, pre-multiply both sides by $A^{-1}$:

Worked illustration. For $2x+y=8,\ x-y=1$: $A=\begin{pmatrix}2&1\\1&-1\end{pmatrix}$, $B=\begin{pmatrix}8\\1\end{pmatrix}$. Here $|A|=-2-1=-3\ne0$, so $A^{-1}=\dfrac1{-3}\begin{pmatrix}-1&-1\\-1&2\end{pmatrix}=\dfrac13\begin{pmatrix}1&1\\1&-2\end{pmatrix}$. Then $X=A^{-1}B=\dfrac13\begin{pmatrix}1&1\\1&-2\end{pmatrix}\begin{pmatrix}8\\1\end{pmatrix}=\dfrac13\begin{pmatrix}9\\6\end{pmatrix}=\begin{pmatrix}3\\2\end{pmatrix}$, i.e. $x=3,y=2$ -- check: $2(3)+2=8$ and $3-2=1$, both correct.

Practical shape for a $3\times3$ system. Write the three equations, read off $A$ (rows = equations, columns = coefficients of $x,y,z$ in that fixed order) and $B$, compute $|A|$, then $\operatorname{adj}A$ (transpose of the cofactor matrix), then $A^{-1}=\frac1{|A|}\operatorname{adj}A$, and finally multiply $A^{-1}B$ to read off $x,y,z$ from the resulting column. …