Matrices module, 2011, Exam 1, Question 8
The assessors report does not cover this module because it was completed 'reasonably well' which leaves illiterates like me, helpless
Can someone show a worked solution, my cas does not want to solve for x with these types of questions
So basically:
1: Using CAS, find the inverse of the matrix in the question.
2: This gives you a matrix with all inverse values in terms of k.
3: The one that we are concerned with is the one in the corresponding position to k in the original matrix; that is: (-9/4(4k-9)) -1/4.
4: W know that these two are equal because the question tells us that the matrix equals it's inverse.
5: So now, put these into the CAS and solve for k.
6: CAS will give you k=0 or k=2.
7: Try both of them in the matrix and find their inverses. Only k=2 will give you the exact same matrix as the inverse.
8: Therefore, answer is D.
Hope that helps!