(a) Solve the linear system. 3x + y + 6z = 0 −6x − 16z = 4 8y − 17z = 17 using the following steps: (i) find the LU-factorization of the coefficient matrix, (ii) solve the lower triangular system Ly = b, and (iii) solve the upper triangular system Ux = y
Answers
Step-by-step explanation:
can enter the "plus-minus" equations into your graphing calculator to verify this:
x2 – xy + y2 = 21
y2 – xy + (x2 – 21) = 0
y = [x ± sqrt(84 - 3x^2)]/2
x2 + 2xy – 8y2 = 0
0 = 8y2 – 2xy – x2
y = (x ± 3abs(x))/8
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As you can see, I used the Quadratic Formula in each of the original equations, in order to solve for y in terms of x. This gave me equations that I could graph. This technique doesn't come up that often, but it can be a life-saver when you can't seem to solve things any other way.
Oh, and where did those absolute-value bars come from? Recall that, technically, the square root of x2 is the absolute value of x. That's how I did that simplification in the next-to-last line above. And this absolute value will matter shortly.
The absolute value of x in the second equation above gives two cases for the values of y:
If x < 0, then | x | = –x, so y = (x ± 3x)/8 = x/2, – x/4
If x > 0, then | x | = x, so y = (x ± 3x)/8 = – x/4, x/2.
In either case, y = – x/4 or y = x/2.
Since I derived these "y=" solution-equations from the second of the original equations, I will plug them into the first equation to solve for some actual numerical values:
If y = – x/4: Copyright © 2002-2011 Elizabeth Stapel All Rights Reserved
x = ± 4
Then, plugging into the "y=" solution-equations above, I get:
x = -4 then y = 1; x = 4 then y = -1
If y = x/2:
x = ± 2sqrt(7)
Then:
x = -2sqrt(7) then y = -sqrt(7); x = 2sqrt(7) then y = sqrt(7)
Then the four solutions are:
(-2sqrt(7), -sqrt(7)), (2sqrt(7), sqrt(7)), (-4, 1), (4, -1)
Warning: Do not try to write the solution points as "(± 4, ± 1)" or "(± 2sqrt(7), ± sqrt(7))", because this is not correct. Not all combinations of these x-value and y-values are solution points. Don't be sloppy; write the solution out correctly.
By the way, the graph of the system looks like this:
graph of equations