# Show that the set of four polynomials 1, 1-t, 2-4t+(t^2), 6-18t+9(t^2)-(t^3) is linearly independentShow that the set of four polynomials 1, `1-t`, `2-4t+t^(2)`, `6-18t+9t^(2)-t^(3)` is linearly...

Show that the set of four polynomials 1, 1-t, 2-4t+(t^2), 6-18t+9(t^2)-(t^3)

is linearly independent

Show that the set of four polynomials 1, `1-t`, `2-4t+t^(2)`, `6-18t+9t^(2)-t^(3)`

is linearly independent

You need to study the following equation that helps you to tell that the four polynomials are linearly independent such that:

`c_1(1) + c_2(1 - t) + c_3(2 - 4t + t^2) + c_4(6 - 18t + 9t^2 - t^3) = 0`

Notice that the equation is equal to zero if `c_1=c_2=c_3=c_4=0` .

`c_1 + c_2 - c_2*t + 2c_3 - 4c_3t + c_3t^2 + 6c_4 - 18tc_4 + 9c_4t^2 - c_4t^3 = 0`

You need to combine the terms that contain `t, t^2` and `t^3 ` such that:

`t(-c_2 - 4c_3 - 18c_4) + t^2(c_3 + 9c_4) - c_4t^3 + c_1 + c_2 + 2c_3 + 6c_4 = 0`

Equating coefficients of like terms both sides, yields:

`{(-c_2 - 4c_3 - 18c_4 = 0),(c_3 + 9c_4 = 0),(- c_4 = 0),(c_1 + c_2 + 2c_3 + 6c_4 = 0):}`

`-c_2 - 4c_3 = 0`

`c_3 + 9*0 = 0 => c_3 = 0`

`-c_2 - 4*0 = 0 => c_2 = 0`

`c_1 + 0 + 2*0 + 6*0 = 0 => c_1 = 0`

**Since `c_1=c_2=c_3=c_4=0` yields that the four given polynomials are linearly independent.**

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