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The price of 1 kg onion is Rs 69 ½. how much quantity of onion can be purchased for Rs 444 Dont answer unless you know it and dont spam by saying "Please mark as brainlist" or something |
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Answer» -step explanation:Let the cost of 1 kg onion , 1 kg wheat and 1 kg of rice be Rsx,Rsy,RSZ respectively.So, 4x+3y+2z=602x+4y+6z=906x+2y+3z=70These equations can be written asAX=Bwhere A= ⎣⎢⎢⎡ 426 342 263 ⎦⎥⎥⎤ ,X= ⎣⎢⎢⎡ xyz ⎦⎥⎥⎤ ,B= ⎣⎢⎢⎡ 609070 ⎦⎥⎥⎤ Here, ∣A∣=4(12−12)−3(6−36)+2(4−24)⇒∣A∣=90−40=50Since, ∣A∣=0Hence, the system of equations is consistent and has a unique solution given by X==A −1 BA −1 = ∣A∣adjA and adjA=C T C 11 =(−1) 1+1 ∣∣∣∣∣∣ 42 63 ∣∣∣∣∣∣ ⇒C 11 =12−12=0C 12 =(−1) 1+2 ∣∣∣∣∣∣ 26 63 ∣∣∣∣∣∣ ⇒C 12 =−(6−36)=30C 13 =(−1) 1+3 ∣∣∣∣∣∣ 26 42 ∣∣∣∣∣∣ ⇒C 13 =4−24=−20C 21 =(−1) 2+1 ∣∣∣∣∣∣ 32 23 ∣∣∣∣∣∣ ⇒C 21 =−(9−4)=−5C 22 =(−1) 2+2 ∣∣∣∣∣∣ 46 23 ∣∣∣∣∣∣ ⇒C 22 =12−12=0C 23 =(−1) 2+3 ∣∣∣∣∣∣ 46 32 ∣∣∣∣∣∣ ⇒C 23 =−(8−18)=10C 31 =(−1) 3+1 ∣∣∣∣∣∣ 34 26 ∣∣∣∣∣∣ ⇒C 31 =18−8=10C 32 =(−1) 3+2 ∣∣∣∣∣∣ 42 26 ∣∣∣∣∣∣ ⇒C 32 =−(24−4)=−20C 33 =(−1) 3+3 ∣∣∣∣∣∣ 42 34 ∣∣∣∣∣∣ ⇒C 33 =16−6=10Hence, the co-factor MATRIX is C= ⎣⎢⎢⎡ 0−510 300−20 −201010 ⎦⎥⎥⎤ ⇒adjA=C T = ⎣⎢⎢⎡ 030−20 −5010 10−2010 ⎦⎥⎥⎤ ⇒A −1 = ∣A∣adjA = 501 ⎣⎢⎢⎡ 030−20 −5010 10−2010 ⎦⎥⎥⎤ Solution is given by ⎣⎢⎢⎡ xyz ⎦⎥⎥⎤ = 501 ⎣⎢⎢⎡ 030−20 −5010 10−2010 ⎦⎥⎥⎤ ⎣⎢⎢⎡ 609070 ⎦⎥⎥⎤ ⎣⎢⎢⎡ xyz ⎦⎥⎥⎤ = 501 ⎣⎢⎢⎡ 0−450+7001800+0−1400−1200+900+700 ⎦⎥⎥⎤ ⎣⎢⎢⎡ xyz ⎦⎥⎥⎤ = 501 ⎣⎢⎢⎡ 250400400 ⎦⎥⎥⎤ ⎣⎢⎢⎡ xyz ⎦⎥⎥⎤ = ⎣⎢⎢⎡ 588 ⎦⎥⎥⎤ Hence, x=5,y=8,z=8So, the cost of 1 kg onion is RS 5 , 1 kg wheat is Rs 8 , 1 kg rice is Rs 8. |
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