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Question
Solve the system of linear equations using the matrix method.
2x + 3y + 3z = 5
x − 2y + z = −4
3x − y − 2z = 3
Solution
`[(2,3,3),(1,-2,1),(3,-1,-2)] [(x),(y),(z)] = [(5),(-4),(3)]` AX = B
A = `[(2,3,3),(1,-2,1),(3,-1,-2)], X = [(x),(y),(z)]` or B = `[(5),(-4),(3)]`
Now, `abs A = [(2,3,3),(1,-2,1),(3,-1,-2)]`
`= 2 (4 + 1) - 3 (-2 - 3) + 3 (-1 + 6)`
`= 10 + 15 + 15 = 40 ne 0`
`therefore A^-1` exists and hence the given equation has a unique solution.
`A_11 = (-1)^(1 + 1) abs ((-2,1),(-1,-2)) = 4+1 = 5`
`A_12 = (-1)^(1 + 2) abs ((1,1),(3,-2)) = (-2 - 3) = 5`
`A_13 = (-1)^(1 + 3) abs ((1,-2),(3,-1)) = -1 + 6 = 5`
`A_21 = (-1)^(2 + 1) abs ((3,3),(-1,-2)) = -(-6 + 3) = 3`
`A_22 = (-1)^(2 + 2) abs ((2,3),(3,-2)) = -4 - 9 = -13`
`A_23 = (-1)^(2 + 3) abs((2,3),(3,-1)) = -(-2 - 9) = 11`
`A_31 = (-1)^ (3 + 1) abs ((3,3),(-2,1)) = 3 + 6 = 9`
`A_32 = (-1)^(3 + 2) abs ((2,3),(1,1)) = -(2 - 3) = 1`
`A_33 = (-1)^(3 + 3) abs ((2,3),(1, -2)) = -4-3 = -7`
`therefore A^-1 = 1/abs A (Adj A)`
`= 1/40 [(5,5,5),(3,-13,11),(9,1,-7)] = 1/40 [(5,3,9),(5,-13,1),(5,11,-7)]`
`X = A^-1 B`
`=> [(x),(y),(z)] = 1/40 [(5,3,9),(5,-13,1),(5,11,-7)] [(5),(-4),(3)]`
`= 1/40 [(25 - 12 + 27),(25 + 52 + 3),(25 - 44 - 21)]`
`= 1/40 [(40),(80),(-40)] = [(1),(2),(-1)]`
so, x = `1, y =2, or z = -1.`
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