Determine elementary matrices E1, E2, E3 of Type III such that E3E2E1A = U with U an upper triangular matrix. The matrix E1 should turn the element in position (2,1) into a 0. Enter this matrix in MATLAB as E1 using commands similar to the ones in Example 1. The matrix E2 should turn the element in position (3,1) into a zero. Enter this matrix in MATLAB as E2. Note that to zero out the entries in column 1, you need to add or subtract a multiple of row 1. Once you have found the matrices E1 and E2, compute the product E2E1A in MATLAB. Use format rat so that the entries will be given as fractions. Based on the result, determine the matrix E3 that turns the element in position (3,2) into a zero. Enter this matrix as E3 in MATLAB and compute U=E3*E2*E1*A.

the correct answer is

c. 4

step-by-step explanation:

the given polynomial is

the degree of the polynomial is the highest exponent which is 4.

the -4 is called the leading coefficient.

this polynomial has only one term so it is a monomial.

the correct answer is option c.

4.5%

step-by-step explanation:

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.

what steps are you having us

step-by-step explanation: