Q.A family of 3 people went out for dinner in a restaurant. The cost of two dosai, three idlies and two vadais is Rs.,150. The cost of two dosai, two idlies and four vadais is Rs.,200. The cost of five dosai, four idlies and two vadais is Rs.,250. The family has Rs.,350 in hand and they ate 3 dosai and six idlies and six vadais. Will they be able to manage to pay the bill within the amount they had?

For a system of 3 linear equations in 3 unknowns $a_{i1}x_1+a_{i2}x_2+a_{i3}x_3=b_i$ ($i=1,2,3$) whose coefficient determinant $\Delta=\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}$ is non-zero, Cramer's rule gives each unknown directly as a ratio of two determinants:

where $\Delta_k$ is $\Delta$ with its $k$th column replaced by the constants column $(b_1,b_2,b_3)^T$, everything else unchanged. (The same pattern extends to 2 equations in 2 unknowns: $\Delta=\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}$, $x=\Delta_1/\Delta$, $y=\Delta_2/\Delta$.)

Why it works. Multiplying $\Delta$ by $x_1$ and using the linearity-in-a-column property of determinants (splitting the first column $a_{i1}x_1$ into the sum $a_{i1}x_1+a_{i2}x_2+a_{i3}x_3$ using the original equations, then subtracting off the $x_2,x_3$ multiples of the identical columns 2 and 3, which vanish) collapses the first column to exactly the constants $b_i$ -- giving $x_1\Delta=\Delta_1$, and dividing by $\Delta\ne0$ gives the rule.

Worked illustration. For $x+y=3,\ 2x-y=0$: $\Delta=\begin{vmatrix}1&1\\2&-1\end{vmatrix}=-3$, $\Delta_1=\begin{vmatrix}3&1\\0&-1\end{vmatrix}=-3$, $\Delta_2=\begin{vmatrix}1&3\\2&0\end{vmatrix}=-6$. So $x=\Delta_1/\Delta=1$, $y=\Delta_2/\Delta=2$ -- check: $1+2=3$ and $2(1)-2=0$, correct.

Word problems that produce equations like $y=ax^2+bx+c$ through three given points, or rate/mixture/scoring problems, translate to a $3\times3$ system in the unknown constants exactly as for matrix inversion, then Cramer's rule reads off each unknown independently -- convenient when only one or two of the unknowns are actually needed. A system with fractional unknowns like $\dfrac{a}{x}+by=c$ is first turned linear by the substitution $u=\dfrac1x$ (or $\dfrac1y$, $\dfrac1z$), solved for $u,v,(w)$ by Cramer's rule, and only inverted back to $x,y,(z)$ at the very last step. …