# Write the vector v = (1;-2; 5) as linear combination of the vectorse1 = (1; 1; 1), e2 = (1; 2; 3) and e3 = (2;-1;-1).can any one please give a hand in this ? Is fe1; e2; e3g a basis for R3?

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You may write the given vector `bar v`  as the following linear combination such that:

`bar v = a*bar e_1 + b*bar e_2 + c*bar e_3`

`(1,-2,5) = a*(1,1,1) + b*(1,2,3) + c*(2,-1,-1)`

`bar i - 2bar j + 5 bar k = a bar i + a bar j + a bar k + b bar i + 2 b bar j + 3 b bark + 2c bar i - c bar j - c bar k`

`bar i - 2bar j + 5 bar k = bar i (a + b + 2c) + bar j(a + 2b - c) + bar k(a + 3b - c)`

Equating coefficients of like vectors yields:

`{(a + b + 2c = 1),(a + 2b - c = -2),(a + 3b - c = 5):}`

`a + 2b - c - a - b - 2c= -2 - 1`

`b - 3c = -3`

`a + 3b - c- a - b - 2c = 5 - 1 => 2b - 3c = 4`

`2b - 3c - b + 3c= 4 + 3 => b = 7`

`7 - 3c = -3 => -3c = -10 => c = 10/3`

`a +7 + 20/3 = 1 => 3a = 3 - 21 - 20 => 3a = -38 => a = -38/3`

Hence, you may write the vector `bar v`  as the following linear combination `bar v = -38/3 bar e_1 + 7 bar e_2 + 10/3 bar e_3.`

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