Linear Algebra Assignment | Mathematics Assignment
Added on - 28 May 2020
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Linear AlgebraA32: Matricesa)R2∗3Assuming that v is a vector space then a linearly independent spanning set for V iscalled the basis.For a matrix inRnthen 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 theRnis either ann∗1foracolumn¿1∗nforarow¿.Then theR2∗3will 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=calculatingallthepossible¿ABCA*ABACBBA*BCCCACB*The areas marked by * have a matrix appearing twice hence are ignored as per therequirements.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 thesecond matrix.We now check the matrices for compatibility.1st; AB¿(−121)∗(31−1−21−2)the matrices are compatible hence the multiplicationgives[−4−3]2ndAC=(−121)∗in this matrices matrix A has 3 columns while C has 2rows from the guiding rule these matrices are not compatible hence multiplication isnot possible.
Linear Algebra3rdBA;Bhas 2 columns while A has 1 row. Matrices are therefore not compatible.4th;BC=(31−1−21−2)∗=[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 matricesABC∧BCAABC=[−4−3]∗=[−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, Band Cc)CalculatingAnA=[412−4−2−51−2−51]First, we begin by computingA1,A2,A3−−−¿A1as 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 thatAn=0forallthen>2To prove this, we apply the diagonalization formulaSince A is 3 *3 matrix we try to put A into a diagonal matrix
Linear AlgebraSupposeD=P−1APTheDn=[P−1AP]n=P−1AP∗P−1AP−−−−¿P−1AP=P−1AnPFrom here we can deriveAn=P−1DnPThe aim is to put A in a diagonal format therefore there is need to obtain a basis ofeigenvectors.Obtaining the eigenvalues from the characteristic polynomial[4−λ12−4−2−5−λ1−2−51−λ]¿−λ3=(−1)(λ)(λ)(λ)This givesλ1=λ2=λ3=0Now obtaining the eigenvectors from the eigenvalueA−λ∗E=[412−4−2−51−2−51]A−λ∗E=0which gives a homogenous system of linear equations. Solving bygaussian elimination we get[412−4−2−51−2−51]gives[10201−1000]From herex1+2x3=0x2−x3=0Obtaining the variablesx2the system 2 givesx2=x3The vectorX=[−2x3x3x3]Nowx3[−211]which is the basis for the eigenvector for AIf this gives usPwhich means thatP−1does not exist.The number of eigenvectors is not equal to the matrix dimensions hence matrix A isnot diagonalizable as it only has one eigenvector which is less than 3.