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Worked Examples · Example 7

Q.The total cost of one unit each of items A, B and C is ₹180. Twice the cost of A plus the cost of B, minus the cost of C, is ₹110. The cost of A minus the cost of B plus twice the cost of C is ₹100. Using Cramer's Rule, find the cost of each item.

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Setting up the equations

Let x,y,zx,y,z be the costs (in ₹) of A, B, C respectively:

x+y+z=180x+y+z=180

2x+y−z=1102x+y-z=110

x−y+2z=100x-y+2z=100

Forming Δ\Delta

Δ=∣11121−11−12∣=1(2−1)−1(4+1)+1(−2−1)=1−5−3=−7\Delta=\begin{vmatrix}1&1&1\\2&1&-1\\1&-1&2\end{vmatrix}=1(2-1)-1(4+1)+1(-2-1)=1-5-3=-7

Since Δ≠0\Delta\neq0, a unique solution exists.

Forming Δx\Delta_x

Δx=∣180111101−1100−12∣=180(2−1)−1(220+100)+1(−110−100)=180−320−210=−350\Delta_x=\begin{vmatrix}180&1&1\\110&1&-1\\100&-1&2\end{vmatrix}=180(2-1)-1(220+100)+1(-110-100)=180-320-210=-350

Forming Δy\Delta_y

Δy=∣118012110−111002∣=1(220+100)−180(4+1)+1(200−110)=320−900+90=−490\Delta_y=\begin{vmatrix}1&180&1\\2&110&-1\\1&100&2\end{vmatrix}=1(220+100)-180(4+1)+1(200-110)=320-900+90=-490

Forming Δz\Delta_z

Δz=∣11180211101−1100∣=1(100+110)−1(200−110)+180(−2−1)=210−90−540=−420\Delta_z=\begin{vmatrix}1&1&180\\2&1&110\\1&-1&100\end{vmatrix}=1(100+110)-1(200-110)+180(-2-1)=210-90-540=-420

Solving …

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