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SIT292 Linear Algebra Assignment

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Linear Algebra (SIT292)

   

Added on  2019-11-25

SIT292 Linear Algebra Assignment

   

Linear Algebra (SIT292)

   Added on 2019-11-25

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SIT292 LINEAR ALGEBRA2017ASSIGNMENT 2STUDENT ID [Pick the date]
SIT292 Linear Algebra Assignment_1
Question 1(i)Given matrix A=[321112−1−20]Cofactors Cijof the matrix Cij=¿C11=¿C12=¿C13=¿C21=¿C22=¿C23=¿C31=¿C32=¿C33=¿Hence, cofactors of the matrix A would be given below:1
SIT292 Linear Algebra Assignment_2
CofactorsA=[4−2−1−2143−51]AdjAof the matrix A would be given below:AdjA=[4−23−21−5−141](ii)Verification that computed adjAis correct. If A∗(adjA)detA=INow, LHS A∗(adjA)=[321112−1−20].[4−23−21−5−141]¿[3.4+2.(−2)+1.(−1)3.(−2)+2.1+1.43.3+2.(−5)+1.11.4+1.(−2)+2.(−1)1.(−2)+1.1+2.41.3+1.(−5)+2.1−1.4+(−2).(−2)+0.(−1)(−1).(−2)+(−2).1+0.4(−1).3+(−2).(−5)+0.1]¿[700070007]And A=[321112−1−20]2
SIT292 Linear Algebra Assignment_3
detA=|321112−1−20|¿3(0+4)−2(0+2)+1(−2+1)¿12−4−1¿7A∗(adjA)detA=[700070007]7=[100010001]RHS I =[100010001]LHS = RHSIt is apparent that both the sides are equal and therefore, the computed adjA is correct. 3
SIT292 Linear Algebra Assignment_4
Question 2 Two vectors would be orthogonal when there dot product becomes zero. ¿7∝+∝.∝+3∝.(−3)+1.1+(−4).4¿7∝+∝2−9∝+1−16¿∝2−2∝−15Hence, ∝2−2∝−15=0(∝−5)(∝+3)=0∝=5,−3Therefore, the for ∝=5,−3thevectorswouldbeorthogonal.4
SIT292 Linear Algebra Assignment_5
Question 3 Given equations Gaussian elimination method to reduce the following system of equations into row echelon form is applied below:5
SIT292 Linear Algebra Assignment_6
The row echelon form of the given system is given below:This system does not have any solution because 0≠−5.It is apparent from the above that the given system has three linear equations and 4 variables. Hence, the system is said to be inconsistent and does not have any solution as evident from the above. 6
SIT292 Linear Algebra Assignment_7
Question 4 Eigen values and eigenvectors of the given matrixes are as highlighted below:For matrix A Eigen valuesA=[101010101]Let λ is the eigenvalues of the given matrix in such a way that det(A−¿λI)=0¿Now, A−λI=[101010101]−λ[100010001]A−λI=[101010101]−[|λ000λ000λ|]A−λI=[1−λ0101−λ0101−λ]Determinant of A−λI7
SIT292 Linear Algebra Assignment_8

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