Let U be a square matrix such that Upper U Superscript Upper TUequalsI. Show that det Uequalsplus or minus1. Assume that Upper U
Superscript Upper TUequalsI. Since the desired result is that det Uequalsplus or minus1, an intermediate step must be found which contains the expression det U. Which of the following can be applied to the assumption Upper U Superscript Upper TUequalsI to achieve the desired result? A. det (Upper U Superscript Upper TU)Superscript negative 1equalsdet I B. det (Upper U Superscript Upper TU)equalsI C. (Upper U Superscript Upper TU)Superscript negative 1equalsISuperscript negative 1 D. det (Upper U Superscript Upper TU)equalsdet I Simplify the right side of the equation found in the first step. nothing Which property can be used to simplify the left side of the equation found in the first step? Select all that apply.
1 answer:
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Step-by-step explanation:
We have to find the point in the feasible region which maximizes the objective function. To find that point first we need to graph the given inequalities to find the feasible region.
Steps to graph 3x + y ≤ 12:
First we graph 3x + y = 12 then shade the graph for ≤.
plug any value of x say x=0 and x=2 into 3x + y = 12 to find points.
plug x=0
3x + y = 12
3(0) + y = 12
0 + y = 12
y = 12
Hence first point is (0,12)
Similarly plugging x=2 will give y=6
Hence second point is (2,6)
Now graph both points and joint them by a straight line.
test for shading.
plug any test point which is not on the graph of line like (0,0) into original inequality 3x + y ≤ 12:
3(0) + (0) ≤ 12
0 + 0 ≤ 12
0 ≤ 12
Which is true so shading will be in the direction of test point (0,0)
We can repeat same procedure to graph other inequalities.
From graph we see that ABCD is feasible region whose corner points will result into maximum or minimu for objective function.
Since objective function is not given in the question so i will explain the process.
To find the maximum value of objective function we plug each corner point of feasible region into objective function. Whichever point gives maximum value will be the answer
