If you manipulated a matrix so all the entries above the diagonal entries are 0s, how would the solution differ from when made the entries below the diagonal entries 0s?

(Exact Answer) When all the entries above the diagonal entries are 0, the solution will not be different from when all the entries below the diagonal entries are 0. This is because in both cases, there will be one row that would be left with just one unknown element, thus giving us the solution to the other unknowns through back-substitution. For instance, in a linear system with three unknowns, x, y, and z, the traditional triangulated form will give us the value of z first; however, if the entries above the diagonal entries in the matrix are 0, then we will get the value of x first. In either case, the solution will be the same.

Step-by-step explanation:

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions

1.true 2.false? ? i think. .

step-by-step explanation: