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The Determinant of Where L is the determinant of Where L is the determinant of Where L is the determinant of Where L is the determinant of Where L is the determinant

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That’s Solution of the system The reduced row echelon form of vector A is therefore, the value of x’s is Problem 3 a subspace of v then Since then meaning w is a subspace of v For any value of x then hence w is a subspace of v the span of W is 3 Problem 4 True; by definition a set of vectors has only the trivial solution.

The Determinant of Where L is the determinant of Where L is the determinant of Where L is the determinant of Where L is the determinant of Where L is the determinant

   Added on 2019-10-30

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Problem 1Proof by inductionn≥0,then2n≥nwherenisanaturalnumberp(n);2n>n,forn=1wehanep(1);21>1 hence p (1) is trueSuppose p(k)=truefork>+0 then p(k+1)=truewheneverpk=true2k>k[n=k]2∗2k>2k2k+1>2kbut2kthats(k+k)Being that (k+k)>k+1Then 2k+1>k+1 therefore p(k+1)istruewheneverp(k)istrueHence by principal of mathematical induction p(n)istrueforalln≥0Problem 2A=[1232014156060011]a)A−1A−1=1det(A)∗adj(A)det(A)=(1∗1∗0∗1)+(1∗4∗6∗0)+(1∗1∗6∗1)+(2∗0∗6∗1)+(2∗4∗5∗1)+(2∗1∗0∗0)∗(3∗0∗6∗1)+(3∗1∗6∗0)+(3∗1∗5∗0)+(2∗0∗0∗0)+(2∗1∗5∗1)+(2∗4∗6∗0)−(1∗1∗6∗1)−(1∗4∗6∗1)−(1∗1∗0∗0)−(2∗0∗0∗1)−(2∗4∗6∗0)−(2∗1∗5∗1)−(3∗0∗6∗0)−(3∗1∗5∗1)−(3∗1∗6∗0)−(2∗0∗6∗1)−(2∗1∗0∗0)−(2∗4∗5∗0)This gives6+40+10−6−24−10−15=1Hence the determinant is 1Adj(A)=[b11b12b13b14b21b22b23b24b31b32b33b34b41b42b43b44]nowb11=(1∗0∗1)+(4∗1∗0)+(1∗6∗1)−(1∗6∗1)−(4∗6∗1)−(1∗0∗0)=−23b12=(2∗1∗1)+(3∗6∗1)+(0∗0∗0)−(2∗0∗1)−(3∗6∗0)−(2∗6∗1)=8
The Determinant of Where L is the determinant of Where L is the determinant of Where L is the determinant of Where L is the determinant of Where L is the determinant_1
b13=(2∗4∗1)+(3∗1∗0)+(2∗1∗1)−(2∗1∗1)−(3∗1∗1)−(2∗4∗0)=5b14=(2∗1∗0)+(3∗1∗6)+(2∗4∗6)−(2∗4∗6)−(3∗1∗6)−(2∗1∗0)=0b21=(0∗6∗1)+(4∗5∗1)+(1∗0∗0)−(0∗0∗1)−(4∗6∗0)−(1∗5∗1)=15b22=(1∗0∗1)+(3∗6∗0)+(2∗5∗1)−(1∗6∗1)−(3∗5∗1)−(2∗0∗0)=−11b23=(1∗1∗1)+(3∗0∗1)+(2∗4∗0)−(1∗4∗1)−(3∗1∗0)−(2∗0∗1)=−3b24=(1∗4∗6)+(3∗1∗5)+(2∗0∗0)−(1∗1∗0)−(3∗0∗6)−(2∗4∗5)=−1b31=(0∗6∗1)+(1∗6∗0)+(1∗5∗0)−(0∗6∗0)−(1∗5∗1)−(1∗6∗0)=−5b32=(1∗6∗0)+(2∗5∗1)+(2∗6∗0)−(1∗6∗1)−(2∗6∗0)−(2∗5∗0)=4b33=(1∗1∗1)+(2∗1∗0)+(2∗0∗0)−(1∗1∗0)−(2∗0∗1)−(2∗1∗0)=1b34=(1∗1∗6)+(2∗0∗6)+(2∗1∗5)−(1∗1∗6)−(2∗1∗5)−(2∗0∗0)=0b41=(0∗0∗0)¿+(1∗5∗1)+(4∗6∗0)−(0∗6∗1)−(1∗0∗0)−(4∗5∗0)=5b42=(1∗6∗1)+(2∗0∗0)+(3∗5∗0)−(1∗0∗0)−(2∗5∗1)−(3∗6∗0)=−4b43=(1∗4∗0)+(2∗0∗1)+(3∗1∗1)−(1∗1∗1)−(2∗4∗0)−(3∗0∗0)=2b44=(1∗1∗0)+(2∗4∗5)+(3∗0∗6)−(1∗4∗6)−(2∗0∗0)−(3∗1∗5)=1Now the Adj(A)=[−2385015−11−3−1−54105−421]sincethedeterminantofA=1thentheinverseofA=adj(A)whichis[−2385015−11−3−1−54105−421]b)det(A)=1 as calculated in part (a) aboveThe determinant of A−1A−1=[−2385015−11−3−1−54105−421]det(A−1)=(−23∗−11∗0)+(−23∗−3∗0∗4)+(−23∗−1∗4∗2)+(8∗15∗0∗2)+(8∗−3∗−5∗1)+(8∗0∗1∗5)+(5∗15∗4∗1)+(5∗−11∗0∗5)+(5∗−1∗−5∗−4)+(0∗15∗1∗−4)+(0∗−11∗−5∗2)+(0∗−3∗4∗5)−(−23∗−11∗0∗2)−(−23∗−3∗4∗1)−(−23∗−1∗1∗−4)−(8∗15∗1∗1)−(−8∗−3∗0∗5)−(8∗−1∗−5∗2)−(5∗15∗0∗−4)−(58−11∗−5∗1)−(5∗−1∗4∗5)−(0∗15∗4∗2)−(0∗−11∗1∗5)−(0∗−3∗−5∗−4)
The Determinant of Where L is the determinant of Where L is the determinant of Where L is the determinant of Where L is the determinant of Where L is the determinant_2

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