Let v1 2 3 4 4 v2 2 1 0 2 and v3 2 1 3 4 Determine whethe

Let v_1 = (2, -3, 4, -4), v_2 = (-2, -1, 0, -2) and v_3 = (2, 1, 3, -4). Determine whether (4, 10, 7, -14) is in span{v_1, v_2, v_3}; if it is, write it as a linear combination of v_1, v_2, and v_3.

Solution

let v1=(2,-3,4,-4,4),v2=(-2,-1,0,-2) and v3=(2,1,3,-4)detrmin whether (4,10,7,-14) is in span{v1,v2,v3}

We see that the first two columns are pivot columns, so the first two column of the ORIGINAL MATRIX A, namely, {(2, 2, 3)T ,(4, 6, 8)T }, form a basis for Col A. The last two columns are free, and we can easily read the general solution from the echelon form: x2 = 5 2 x3 3 2 x4, x1 = 2x2 x3 2x4 = 6x3 5x4, x3, x4 free Written in vector form, x = x1 x2 x3 x4 = 6x3 5x4 5 2 x3 3 2 x4 x3 x4 = 6x3 5 2 x3 x3 0 + 5x4 3 2 x4 0 x4 = x3 6 5 2 1 0 + x4 5 3 2 0 1 Thus, {(6, 5 2 , 1, 0)T ,(5, 3 2 , 0, 1)T } form a basis for Nul A.