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Exercise 2.3 · Q34

Q.The total cost of 3 T.V. sets and 2 V.C.R.s is Rs. 35000. The shop-keeper wants profit of Rs. 1000 per television and Rs. 500 per V.C.R. He can sell 2 T.V. sets and 1 V.C.R. and get the total revenue as Rs. 21,500. Find the cost price and the selling price of a T.V. set and a V.C.R.

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Step 1: Let Rs. xx and Rs. yy be the cost prices of one T.V. and one V.C.R. "Total cost of 3 T.V. and 2 V.C.R. is Rs. 35000": 3x+2y=350003x+2y=35000.

Step 2: The shopkeeper's selling price is (x+1000)(x+1000) per T.V. and (y+500)(y+500) per V.C.R. "Selling 2 T.V. and 1 V.C.R. gives total revenue Rs. 21500": 2(x+1000)+1(y+500)=215002(x+1000)+1(y+500)=21500, i.e. 2x+y+2500=215002x+y+2500=21500, i.e. 2x+y=190002x+y=19000.

Step 3: In matrix form, [3221][xy]=[3500019000]\begin{bmatrix}3&2\\2&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}35000\\19000\end{bmatrix}, i.e. AX=BAX=B.

Step 4: ∣A∣=3(1)−2(2)=−1≠0|A|=3(1)-2(2)=-1\neq0. adj A=[1−2−23]\text{adj}\,A=\begin{bmatrix}1&-2\\-2&3\end{bmatrix}, so A−1=1−1[1−2−23]=[−122−3]A^{-1}=\dfrac{1}{-1}\begin{bmatrix}1&-2\\-2&3\end{bmatrix}=\begin{bmatrix}-1&2\\2&-3\end{bmatrix}.

Step 5: X=A−1B=[−122−3][3500019000]=[−35000+3800070000−57000]=[300013000]X=A^{-1}B=\begin{bmatrix}-1&2\\2&-3\end{bmatrix}\begin{bmatrix}35000\\19000\end{bmatrix}=\begin{bmatrix}-35000+38000\\70000-57000\end{bmatrix}=\begin{bmatrix}3000\\13000\end{bmatrix}. …

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