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Worked Examples · Example 46

Q.Three shopkeepers A, B and C are using polythene bags, handmade bags and newspaper bags. A uses 20, 30 and 40 number of bags of respective type. B uses 30, 40 and 20 of each respective kind while C uses 40, 20 and 30 of each type. Each shopkeeper spent ₹250, ₹220 and ₹200 on the bags. Find the cost of each carry bag using matrix method.

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Solving AX=BAX=B by the matrix (determinant) method gives per-bag costs x=2227, y=8527, z=9427x=\tfrac{22}{27},\ y=\tfrac{85}{27},\ z=\tfrac{94}{27} rupees.

Let x,y,zx,y,z be the cost (₹) of a polythene, handmade and newspaper bag. Write AX=BAX=B and solve by determinants: x=ΔxΔx=\dfrac{\Delta_x}{\Delta}, etc.

  1. Form equations from the spending (in ₹), then divide each by 1010:

20x+30y+40z=250, 30x+40y+20z=220, 40x+20y+30z=20020x+30y+40z=250,\ 30x+40y+20z=220,\ 40x+20y+30z=200

⇒ 2x+3y+4z=25,3x+4y+2z=22,4x+2y+3z=20.\Rightarrow\ 2x+3y+4z=25,\quad 3x+4y+2z=22,\quad 4x+2y+3z=20.

  1. Matrix form AX=BAX=B with A=[234342423], B=[252220]A=\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 2 \\ 4 & 2 & 3 \end{bmatrix},\ B=\begin{bmatrix}25\\22\\20\end{bmatrix}.
  2. Coefficient determinant:

Δ=2(4⋅3−2⋅2)−3(3⋅3−2⋅4)+4(3⋅2−4⋅4)=2(8)−3(1)+4(−10)=16−3−40=−27.\Delta=2(4\cdot3-2\cdot2)-3(3\cdot3-2\cdot4)+4(3\cdot2-4\cdot4)=2(8)-3(1)+4(-10)=16-3-40=-27.

  1. Δx\Delta_x (constants in column 1):

Δx=∣253422422023∣=25(8)−3(66−40)+4(44−80)=200−78−144=−22⇒x=−22−27=2227.\Delta_x=\begin{vmatrix}25&3&4\\22&4&2\\20&2&3\end{vmatrix}=25(8)-3(66-40)+4(44-80)=200-78-144=-22\Rightarrow x=\dfrac{-22}{-27}=\dfrac{22}{27}.

  1. Δy\Delta_y (constants in column 2): …

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