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Question
Solve the following equations by the inversion method :
2x + 3y = - 5 and 3x + y = 3.
Solution
AX = B
`[(2, 3), (3, 1)][(x),(y)] = [(-5), (3)]`
`[(2, 3), (3, 1)]A^-1 = [(1, 0), (0, 1)]`
R2 - R1 gives
`[(2, 3), (1, -2)]A^-1 = [(1, 0), (-1, 1)]`
R1 ↔ R2 gives
`[(1, -2), (2, 3)]A^-1 = [(-1, 1), (1, 0)]`
R2 - 2R1 gives
`[(1, -2), (0, 7)]A^-1 = [(-1, 1), (3, -2)]`
`1/7`R2 gives
`[(1, -2), (0, 1)]A^-1 = [(-1, 1), (3/7, -2/7)]`
R1 + 2R2
`[(1, 0), (0, 1)]A^-1 = [(-1/7, 3/7), (3/7, -2/7)]`
`A^-1 = [(-1/7, 3/7), (3/7, -2/7)]`
AX = B
Pre multiplying by A-1
A-1(AX) = A-1B
IX = A-1B
X = A-1B
`[(x), (y)] = 1/7[(-1, 3), (3, -2)][(-5), (3)]`
`[(x), (y)] = [((-1/7) (3/7)), ((3/7) (-2/7))] [(-5), (3)]`
`[(x),(y)] = [ ((-1/7)(-5) + (3/7)(3)), ((3/7)(-5) + (-2/7)(3))]`
`[(x), (y)] = [(2), (-3)]`
x = 2, y = -3
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