Let A = [-7 -3 3 9 5 -3 -9 -3 5] (i) The characteristic polynomial of A is p(x)
ID: 3035095 • Letter: L
Question
Let A = [-7 -3 3 9 5 -3 -9 -3 5] (i) The characteristic polynomial of A is p(x) = x^3 + x^2 + x + = (x-)(x-)(x-) Press the "Submit Answers" button before attempting part (ii). The blanks for entering answers to (ii) and (iii) will appear only if your answers to previous parts were correct. In that case, please ignore the message stating that some of your answers were incorrect! (ii) The distinct eigenvalues of A are: (You'll have to solve part (i) correctly before you can attempt part (ii)) (iii) The eigenspaces of A are: (You'll have to solve parts (i) and (ii) correctly before you can attempt part (iii))Explanation / Answer
-9
-3
3
9
3
-3
-9
-3
3
Multiply the 1st row by -1/9
Add -9 times the 1st row to the 2nd row
Add 9 times the 1st row to the 3rd row
Then the RREF of A-2I is
1
1/3
-1/3
0
0
0
0
0
0
Thus, if x = (x1,x2,x3)T, then the equation (A-2I) x = 0 is equivalent to x1 +x2/3 –x3/3 = 0 . Let x2 = 3r and x3 = 3t. Then x1 =-r + t so that x = (-r+t, 3r,3t)T = r(-1,3,0)T +t (1,0,3)T. Hence the eigenspace of A corresponding to the eigenvector 2 is span{ (-1,3,0)T, (1,0,3)T}.
Similarly, tttheeigenvector of vA corresponding to the eigenvalue -1 is solution to the equation (A+I)x = 0. Thev RREF of A+I is
1
0
-1
0
1
1
0
0
0
Thus,the equation (A+I)x = 0 is equivalent to x1 -x3 = 0 and x2 +x3 = 0. Let x3 = t. Then x = (t,-t,t)T = t(1,-1,1)T. Hence the eigenspace of A corresponding to the eigenvector -1 is span{ (1,-1,01T}.
-9
-3
3
9
3
-3
-9
-3
3
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.