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Linear Algebra Assignment | Mathematics Assignment

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Added on  2020-05-28

Linear Algebra Assignment | Mathematics Assignment

   Added on 2020-05-28

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Linear AlgebraA32: Matricesa)R2∗3Assuming that v is a vector space then a linearly independent spanning set for V is called the basis.For a matrix in Rn then the standard basis isei=(1,0,0,0−−−,0,0,0),ise2=(0,1,0,0,0−−−,0,0,0),isen=(0,0,0,0,0−−−,0,0,0,1)but the Rn is either an n∗1foracolumn¿1∗nforarow¿.Then the R2∗3 will be a 2*3. For a 2*3 the basis ise1=(1,0,0),e2=(0,1,0)b)A=(−121)B=(31−1−21−2)C=[1010]calculatingallthepossible¿ABCA*ABACBBA*BCCCACB*The areas marked by * have a matrix appearing twice hence are ignored as per the requirements.Now applying the matrix multiplication rule that is; to multiply two matrices, then thenumber of columns in the first matrix should be equal to the number of rows in the second matrix.We now check the matrices for compatibility.1st; AB¿(−121)∗(31−1−21−2) the matrices are compatible hence the multiplication gives [−4−3]2ndAC=(−121)∗[1010] in this matrices matrix A has 3 columns while C has 2 rows from the guiding rule these matrices are not compatible hence multiplication is not possible.
Linear Algebra Assignment | Mathematics Assignment_1
Linear Algebra3rdBA;Bhas 2 columns while A has 1 row. Matrices are therefore not compatible.4th; BC=(31−1−21−2)∗[1010]=[2010−10]5th; CA; C has 2 columns while A has 1 row hence not compatible.6th; CB, C has 2 columns while B 3 rows therefore not compatible.Afterwards we now check matrices ABC∧BCAABC=[−4−3]∗[1010]=[−10]BCA=[2010−10]∗(−121)In this case BC has 2 columns while a has only one row hence incompatibility.The obtained multiplications are the only possible matrix product from matrix A, B and Cc)Calculating AnA=[412−4−2−51−2−51]First, we begin by computing A1,A2,A3−−−¿A1 as for any number gives the same number so in our case we obtain [412−4−2−51−2−51]now moving toA2=A∗A¿[412−4−2−51−2−51]∗[412−4−2−51−2−51]=[08−80−440−44]A3=A∗A∗A¿[412−4−2−51−2−51]∗[412−4−2−51−2−51]∗[412−4−2−51−2−51]=0¿herewe can conclude that An=0forallthen>2To prove this, we apply the diagonalization formulaSince A is 3 *3 matrix we try to put A into a diagonal matrix
Linear Algebra Assignment | Mathematics Assignment_2
Linear AlgebraSuppose D=P−1APTheDn=[P−1AP]n= P−1AP∗P−1AP−−−−¿P−1AP=P−1AnPFrom here we can derive An=P−1DnPThe aim is to put A in a diagonal format therefore there is need to obtain a basis of eigenvectors.Obtaining the eigenvalues from the characteristic polynomial[4−λ12−4−2−5−λ1−2−51−λ]¿−λ3=(−1)(λ)(λ)(λ)This gives λ1=λ2=λ3=0Now obtaining the eigenvectors from the eigenvalueA−λ∗E=[412−4−2−51−2−51]A−λ∗E=0 which gives a homogenous system of linear equations. Solving by gaussian elimination we get[412−4−2−51−2−51] gives [10201−1000]From here x1+2x3=0x2−x3=0Obtaining the variables x2 the system 2 gives x2=x3The vector X=[−2x3x3x3]Now x3[−211] which is the basis for the eigenvector for AIf this gives us P which means that P−1 does not exist.The number of eigenvectors is not equal to the matrix dimensions hence matrix A is not diagonalizable as it only has one eigenvector which is less than 3.
Linear Algebra Assignment | Mathematics Assignment_3

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