Find basis and dimension{xeR^4|xA=0} where A=[-1,1,2,1,1,0,2,3]^T

BolkowN

BolkowN

Answered question

2021-01-23

Find basis and dimension {xR4xA=0} where A=[1,1,2,1,1,0,2,3]T

Answer & Explanation

Theodore Schwartz

Theodore Schwartz

Skilled2021-01-24Added 99 answers

Let S = {xR4Ax=0} where A=[1,1,2,1,1,0,2,3].

Now
S={(x1,x2,x3,x4)R4:[1,1,2,1,1,0,2,3](x1,x2,x3,x4)T=0}

={(x1,x2,x3,x4)R4:x+2x2+x3+2x4=0,x1+x2+x4=0}

={(x1,x2,x3,x4)R4:x3=3x25x4}

={(x1,x2,3x25x4,x4):xiR}

={x1(1,0,0,0)+x2(0,1,3,0)+x4(0,0,5,1):xiR}

={(1,0,0,0),(0,1,3,0),(0,0,5,1)}
Again {(1,0,0,0),(0,1,3,0),(0,0,5,1)} is a linearly independent set. Hence {(1,0,0,0),(0,1,3,0),(0,0,5,1)} is a basis of S.

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