Answer to Question #157252 in Linear Algebra for Emerald

Question #157252

Given the vector u = (−1, 1, 2, 1) and the vectors v1 = (1, −1, 2, −1), v2 =

(−2, 2, 3, 2), v3 = (1, 2, 0, −1), v4 = (1, 0, 0, 1) which form an orthogonal ba-

sis for IR 4

, find the coordinate vector (u)S for the vector u. 


1
Expert's answer
2021-01-22T11:50:55-0500

Coordinates of u in basis v1, v2, v3, v4 are numbers s1, s2, s3, s4, such as:


u=s1v1+s2v2+s3v3+s4v4u=s_1v_1+s_2v_2+s_3v_3+s_4v_4


We can write this using matrix:

(v11v21v31v41v12v22v32v42v13v23v33v43v14v24v34v44)(s1s2s3s4)=(u1u2u3u4)\begin{pmatrix} v_{11} & v_{21} & v_{31} & v_{41} \\ v_{12} & v_{22} & v_{32} & v_{42} \\ v_{13} & v_{23} & v_{33} & v_{43} \\ v_{14} & v_{24} & v_{34} & v_{44} \\ \end{pmatrix} \begin{pmatrix} s_1 \\ s_2 \\ s_3 \\ s_4 \end{pmatrix} = \begin{pmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \end{pmatrix}


We've got the linear equation. We can solve it using matrix transformations (Gaussian method):


(v11v21v31v41u1v12v22v32v42u2v13v23v33v43u3v14v24v34v44u4)=\begin{pmatrix} v_{11} & v_{21} & v_{31} & v_{41} &|& u_1\\ v_{12} & v_{22} & v_{32} & v_{42} &|& u_2 \\ v_{13} & v_{23} & v_{33} & v_{43} &|& u_3 \\ v_{14} & v_{24} & v_{34} & v_{44} &|& u_4 \\ \end{pmatrix} =

=(12111122012300212111)= \begin{pmatrix} 1 & -2 & 1 & 1 &|& -1\\ -1 & 2 & 2 & 0 &|& 1 \\ 2 & 3 & 0 & 0 &|& 2 \\ -1 & 2 & -1 & 1 &|& 1 \\ \end{pmatrix}


Add first row to the second and fourth, and subtract 2 times first row from third row:


(12111003100722400020)\begin{pmatrix} 1 & -2 & 1 & 1 &|& -1\\ 0 & 0 & 3 & 1 &|& 0 \\ 0 & 7 & -2 & -2 &|& 4 \\ 0 & 0 & 0 & 2 &|& 0 \\ \end{pmatrix}


Swap second and third rows:


(12111072240031000020)\begin{pmatrix} 1 & -2 & 1 & 1 &|& -1\\ 0 & 7 & -2 & -2 &|& 4 \\ 0 & 0 & 3 & 1 &|& 0 \\ 0 & 0 & 0 & 2 &|& 0 \\ \end{pmatrix}


After transformations, we've got this system of equations:

s12s2+s3+s4=17s22s32s4=43s3+s4=02s4=0s_1-2s_2+s_3+s_4=-1 \\ 7s_2-2s_3-2s_4=4 \\ 3s_3+s_4=0 \\ 2s_4=0


s12s2+s3=17s22s3=43s3=0s4=0s_1-2s_2+s_3=-1 \\ 7s_2-2s_3=4 \\ 3s_3=0 \\ s_4=0


s12s2=17s2=4s3=0s4=0s_1-2s_2=-1 \\ 7s_2=4 \\ s_3=0 \\ s_4=0


s18/7=1s2=4/7s3=0s4=0s_1-8/7=-1 \\ s_2=4/7 \\ s_3=0 \\ s_4=0


s1=1/7s2=4/7s3=0s4=0s_1=1/7 \\ s_2=4/7 \\ s_3=0 \\ s_4=0


Check the solution. If solution is correct, we'll get (-1, 1, 2, 1):

u=1/7v1+4/7v2=1/7(v1+4v2)==1/7(18,1+8,2+12,1+8)==1/7(7,7,14,7)=(1,1,2,1)u=1/7v_1+4/7v_2=1/7(v_1+4v_2)=\\ =1/7(1-8, -1+8, 2+12, -1+8)=\\ =1/7(-7, 7, 14, 7)=(-1,1,2,1)


Solution is correct.


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