Algebra - Speech or Presentation Example

Algebra Answer Part Expand along the first row, In terms of symbol the cofactor is, Now, we put all the quantities for the entries. For the cofactors we will put the minor and a (“+1” or a “-1”) which depends on sign required for each minor. This signs will be by going to our “sign matrix” above beginning at the 1st entry in the particular row or column we are expanding along (Larson, & Falvo 110). After that as we move on that column or row we will put the sign required in that case.

Part 2Expand along the first column,In this case symbol form of the cofactor is,Here the signs of “a” are similar to the above case the first row and column have identical signs;Part 3In this case we will cofactor expansion along 3rd column, i.e.As a33 and a43 are zero, therefore to find C33 and C43 is useless. The cofactors C13 and C23 will be necessary, soAnd,Therefore C13= -20 and C23 = -16 so, the determinant A is,Answer # 2Part 1Let A= To find the determinant by using the column elementary method here we use -4R1+R3So the matrix will become,Now we use -1R1+ R2Now we can find the determinant by using 1st column,Part 2Let A= To find the determinant by using the column elementary method here we use -1R2+R3So the matrix will become,Now –2R2+ R1Now finding determinant using 1st column than,Part 3Let To find the determinant by using the column elementary method here we use 2R2+ R1So the matrix will become,Now we use –4R2+ R1 then,Now finding determinant using 3rd column than,Answer # 3Part 1Expand along the first column to find the determinant,So,Here for the cofactors we will put the minor and a (“+1” or a “-1”) which depends on sign required for each minor.

This signs will be by going to our “sign matrix” above beginning at the 1st entry in the particular row or column we are expanding along.Therefore, Part 2Expanding using 1st column we get the determinant,So the determinant is,Part 3Therefore Part 4Expand along the first column to find the determinant,So,Here for the cofactors we will put the minor and a (“+1” or a “-1”) which depends on sign required for each minor. This signs will be by going to our “sign matrix” above beginning at the 1st entry in the particular row or column we are expanding along.

Answer # 4Part 1So the transpose of matrix A is,Now we can find the determinant of above matrix,Part 2So, Now we find the determinant,Part 3AAT= Now we find the determinant of above matrix,Part 4So, Now, Part 5Now first we find the inverse of matrix A i.e. A-1,For that first we find Adjoint of A from above matrix,Now in second step we find determinant of A,Now, Now we find the determinant of above matrix,So the determinant is,Answer # 5 To find whether the matrix is singular or nonsingular we have to find the determinant, here we use the cofactor method to find the determinant,In this case we will cofactor expansion along 1st column, i.e.As a21, a31 and a41 are zero, therefore to find C21, C31 and C41 is useless.

The cofactors C11 will be necessary, soTherefore C11= 0 so, the determinant A is,Since the determinant of the matrix is Zero hence the given matrix is a singular matrix.Answer # 6Part 1The given equation can be converted into matrix form as,Now finding the determinant Using Cramer’s ruleThereforeNowNow for x2Now Similarly Solving both Hence the solution can be written asPart 2Writing the given equations in matrix form,Finding the determinant of D,Now we find the determinant of to find the value of fist variable Now we find the determinant of to find the value of second variable,And now we will find out the determinant of in order to find the value of third variable i.e. x3Now Similarly And Now first finding out the value of x1Now we find the value of the second variable i.e. x2Now we will find the value of third variable i.e. x3Hence the solution set of the given equation is as follows;Work CitedLarson, R & Falvo, D.C. Elementary Linear Algebra. 7th ed. Cengage Learning, 2012. Print.