. Solve x+y+2z=1, 3x+2y+z=7 2x+y+3z=2 by matrix inversion method.

1.	. Solve x+y+2z=1, 3x+2y+z=7 2x+y+3z=2 by matrix inversion method.
Answer» ANSWER: 1 Answer +1 vote answered Nov 12, 2020 by Darshee (49.1k points) selected Nov 16, 2020 by Aanchi (i) The given system can be written as = 1(0 – 2) – (-1)(2 – 3) + 2(4 – 0) = -2 – (-1)(-1) + 2(4) = -2 – 1 + 8 = 5 ∴ adj A x = -1, y = 2, z = 3. (ii) The given system can be written as = 2(3 – 0) – 0(15 – 0) – 1(5 – 0) = 2(3) – 0(15) – 1(5) = 6 – 0 – 5 = 1 Read more on Sarthaks.com - https://www.sarthaks.com/977720/solve-by-matrix-inversion-method-i-x-y-2z-3-2x-z-1-3x-2y-z-4-ii-2x-z-0-5x-y-4-y-3z-5

Answer»

1 Answer +1 vote answered Nov 12, 2020 by Darshee (49.1k points) selected Nov 16, 2020 by Aanchi (i) The given system can be written as = 1(0 – 2) – (-1)(2 – 3) + 2(4 – 0) = -2 – (-1)(-1) + 2(4) = -2 – 1 + 8 = 5 ∴ adj A x = -1, y = 2, z = 3. (ii) The given system can be written as = 2(3 – 0) – 0(15 – 0) – 1(5 – 0) = 2(3) – 0(15) – 1(5) = 6 – 0 – 5 = 1 Read more on Sarthaks.com - https://www.sarthaks.com/977720/solve-by-matrix-inversion-method-i-x-y-2z-3-2x-z-1-3x-2y-z-4-ii-2x-z-0-5x-y-4-y-3z-5

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