**Toolbox:**

- Consistency of a system of 2 ( or 3 ) linear equations in 2 ( or 3 ) unknowns. Case 1 : If $ \Delta \neq 0$ thesystem is consistent and has a unique solution. Cramer's rule can be used. Case 2 : If $ \Delta = 0$, there are two / ( four) possibilities:
- 2a: If $ \Delta = 0$, and atleast one of the values $ \Delta_ x, \Delta_ y ( or \: \Delta_ z )$ is nonzero, the system is inconsistent and has no solution.
- 2b: If $ \Delta = 0\: and \: \Delta_x=\Delta_y ( = \Delta_z)=0$ and at least one element ( second order minor ) of $ \Delta$ is nonzero, the system is consistent with infinitely many solution. The system reduces to one equation ( two equations )
- 2c: If $ \Delta = 0$ and $ \Delta_x=\Delta_y = \Delta_z=0$ and all their 2 x 2 minors are 0, and $\Delta$ has atleast one nonzero element, then the system is consistent with infinitely many solutions. The system reduces to one equation.
- 2d: If $ \Delta = 0, \Delta_x=\Delta_y = \Delta_z=0$, all 2 x 2 minors of $ \Delta$ are zero but there is at least one 2 x 2 minor of $\Delta_x, \Delta_y \: or \: \Delta_z$ that is nonzero, then the system is inconsistent and has no solution.

Step 1

Let $ x , y, z$ be the number of red, blue and green chairs, costing Rs. 240, Rs. 260 and Rs. 300 each respectively.

The total number of chairs = 100

$ x +y+z=100$

The total cost of the chairs = Rs.25000

$ 240x+260y+300z = 25000$

Step 2

These are two equations in three unknowns and have infinitely many solutions. Let $ z = k,\: k \in R$ ( since the numbers have to be whole numbers). The equations reduce to

$x+y=100-k$

$240x+260y=25000-300k$

$ \Delta = \begin{vmatrix} 1 & 1 \\ 240 & 260 \end{vmatrix} = 260-240=20 \neq 0.$ Cramer's rule can be used.

Step 3

$ \Delta_x = \begin{vmatrix} 100-k & 1 \\ 25000-300k & 260 \end{vmatrix} = 26000-260k-25000+300k$

$ = 1000+40k$

$ \Delta_y = \begin{vmatrix} 1 & 100-k \\ 240 & 25000-300k \end{vmatrix} = 25000-300k-24000+240k$

$ = 1000-60k$

$ x = \large\frac{\Delta_x}{\Delta} = \large\frac{1000+40k}{20} = 50+2k$

$ y = \large\frac{\Delta_y}{\Delta} = \large\frac{1000-60k}{20} = 50-3k$

$ (x, y, z) = (50+2k, \: 50-3k, \: k) $ For whole number values of all three.

Step 4

Three possible solutions

when $ k = 0\: \: \: \: (x,y,z) = (50, 50, 0)$

$\: \: \: \: \: \: \: \: \: k = 5\: \: \: \: (x,y,z) = (60, 35, 5)$

$\: \: \: \: \: \: \: \: \: k = 10\: \: \: \: (x,y,z) = (70, 20, 10)$