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Worked Examples · Example 47

Q.A school plans to award ₹6000 in total to its students to reward for certain values - honesty, regularity and hard work. When three times the award money for hard work is added to the award money given for honesty amounts to ₹11000. The award money for honesty and hard work together is double the award money for regularity. Use matrix method to find the prize money for each category of award.

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Solving the matrix equation gives prizes: honesty ₹500, regularity ₹2000, hard work ₹3500.

Let x,y,zx,y,z (₹) be the award for honesty, regularity, hard work. Translate the conditions to AX=BAX=B and solve by determinants: x=ΔxΔx=\dfrac{\Delta_x}{\Delta}, etc.

  1. Translate the statements:

x+y+z=6000,x+3z=11000,x+z=2y ⇒ x−2y+z=0.x+y+z=6000,\qquad x+3z=11000,\qquad x+z=2y\ \Rightarrow\ x-2y+z=0.

  1. Matrix form A=[1111031−21], B=[6000110000]A=\begin{bmatrix}1&1&1\\1&0&3\\1&-2&1\end{bmatrix},\ B=\begin{bmatrix}6000\\11000\\0\end{bmatrix}.
  2. Δ=1(0⋅1−3⋅(−2))−1(1⋅1−3⋅1)+1(1⋅(−2)−0⋅1)=1(6)−1(−2)+1(−2)=6+2−2=6.\Delta=1(0\cdot1-3\cdot(-2))-1(1\cdot1-3\cdot1)+1(1\cdot(-2)-0\cdot1)=1(6)-1(-2)+1(-2)=6+2-2=6.
  3. Δx=∣60001111000030−21∣=6000(6)−1(11000)+1(−22000)=36000−11000−22000=3000⇒x=30006=500.\Delta_x=\begin{vmatrix}6000&1&1\\11000&0&3\\0&-2&1\end{vmatrix}=6000(6)-1(11000)+1(-22000)=36000-11000-22000=3000\Rightarrow x=\dfrac{3000}{6}=500. …

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