Three pizza outlets A, B and C sell three types of pizza namely cheese pizza, veg pizza and paneer pizza. In a day, A can sell 40 cheese pizza, 30 veg pizza and 20 paneer pizza; B can sell 20 cheese pizza, 40 veg pizza and 60 paneer pizza; C can sell 60 cheese pizza, 20 veg pizza and 30 paneer pizza. If the revenue generated in a day by A is ₹6000, by B is ₹9000 and by C is ₹7000. If x denotes selling price of cheese pizza, y is selling price of veg pizza and z be the selling price of Paneer pizza then based on this information, answer the following question:

The matrix representation of the above problem is :

$\left[\begin{array}{lll}4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3\end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]=\left[\begin{array}{c}600 \\ 900 \\ 700\end{array}\right]$

The correct answer is Option 2: $\left[\begin{array}{lll}4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3\end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]=\left[\begin{array}{c}600 \\ 900 \\ 700\end{array}\right]$

Based on the passage, we assign variables: $x$ = price of cheese pizza, $y$ = price of veg pizza, and $z$ = price of paneer pizza. The total revenue for each outlet is:

• Outlet A: $40x + 30y + 20z = 6000$ = $4x + 3y + 2z = 600$

• Outlet B: $20x + 40y + 60z = 9000$ = $2x + 4y + 6z = 900$

• Outlet C: $60x + 20y + 30z = 7000$ =$6x + 2y + 3z = 700$

Using the simplified values: