The coordinate vector of [p (t) = 6 + 3t −t2] relative to the basis {B} ={1 + t, 1 + t2, t + t2}

Question

The coordinate vector of [p (t) = 6 + 3t −t2] relative to the basis {B} ={1 + t, 1 + t2, t + t2}

Nathaniel Kramer · Accepted Answer

Any arbitrary vector in 𝟚P2 can be written as, [p (t) = a ( 1 + t2 ) + b ( t + t2 ) + c ( 1 + 2 t + t2 )] Thus, for the vector [p ( t )=6 + 3 t − t2] can be written as, 1) [a ( 1 + t ) + b ( 1 + t2 ) + c ( t + t2 )=6 + 3t − t2] On comparing the terms of both the sides in the above equation (1) gives the following equations. a + b=6 a + c=3 b + c=−1 The representation of the linear system in a matrix is, [110101011][acd]=[63−1] The solution of the above system gives as by solving the augmented matrix [11061013011−1] Use Elementary row transformations to reduce the augmented matrix to row reduced Echelon form. Step 1: [110  6101  3011−1]R2 → R2 − R1→[110   60−1   1−3011−1] Step 2 [110   60−1   1−3011−1]R3 → R3 + R2→[110   60−1   1−3002−4] Step 3: [110   60−1   1−3002−4]R1 → R1 + R2 −12 R3→[100   50−1   1−3002−4] Step 4:

The coordinate vector of \displaystyle{\left[{\mathbf{{{p}}}}

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