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Exercise 2.3 · Q32

Q.The cost of 4 pencils, 3 pens and 2 erasers is Rs. 60. The cost of 2 pencils, 4 pens and 6 erasers is Rs. 90, whereas the cost of 6 pencils, 2 pens and 3 erasers is Rs. 70. Find the cost of each item by using matrices.

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Step 1: Let the costs of one pencil, one pen and one eraser be Rs. xx, Rs. yy, Rs. zz. The three statements give 4x+3y+2z=604x+3y+2z=60, 2x+4y+6z=902x+4y+6z=90, 6x+2y+3z=706x+2y+3z=70.

Step 2: In matrix form, [432246623][xyz]=[609070]\begin{bmatrix}4&3&2\\2&4&6\\6&2&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}60\\90\\70\end{bmatrix}, i.e. AX=BAX=B.

Step 3: ∣A∣|A|, expanding along row 1: 4(4⋅3−6⋅2)−3(2⋅3−6⋅6)+2(2⋅2−4⋅6)=4(0)−3(−30)+2(−20)=0+90−40=50≠04(4\cdot3-6\cdot2)-3(2\cdot3-6\cdot6)+2(2\cdot2-4\cdot6)=4(0)-3(-30)+2(-20)=0+90-40=50\neq0.

Step 4: Computing all nine cofactors and transposing gives adj A=[0−510300−20−201010]\text{adj}\,A=\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}, so A−1=150[0−510300−20−201010]A^{-1}=\dfrac{1}{50}\begin{bmatrix}0&-5&10\\30&0&-20\\-20&10&10\end{bmatrix}. …

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