how-much-matrices-of-integers-number-A-a-b-c-d-if-A-2-A-2I-c-0-det-A-4- Tinku Tara June 3, 2023 Matrices and Determinants 0 Comments FacebookTweetPin Question Number 12046 by 7991 last updated on 10/Apr/17 howmuchmatricesofintegersnumberA=[abcd]ifA2+A=2I,c=0,det(A)=4 Answered by sma3l2996 last updated on 10/Apr/17 A2=[ab0d][ab0d]=[a2ab+db0d2]A2+A=[a2+aab+db+d0d2+d]=[2002]a2+a=2;d2+d=2;ab+db+d=0det(A)=ad=4(i):a2+a−2=0⇔a=1ora=−2(ii):ad=4⇔d=4ord=−2(iii):d2+d−2=0⇔d=1ord=−2sod=a=−2ab+db+d=0⇔−4b−2=0⇔b=−12soA=[−2−120−2] Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: how-much-matrices-of-integers-number-A-a-b-c-d-if-A-2-I-and-b-c-Next Next post: Let-a-b-0-1-and-a-b-1-Prove-that-1-1-a-1-1-b-1-2-2-a-b- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![how much matrices of integers number A= [(a,b),(c,d) ]if A^2 +A=2I, c=0, det(A)=4](https://www.tinkutara.com/question/Q12046.png)
![A^2 = [(a,b),(0,d) ] [(a,b),(0,d) ]= [(a^2 ,(ab+db)),(0,d^2 ) ] A^2 +A= [((a^2 +a),(ab+db+d)),(0,(d^2 +d)) ]= [(2,0),(0,2) ] a^2 +a=2 ; d^2 +d=2 ; ab+db+d=0 det(A)=ad=4 (i):a^2 +a−2=0 ⇔ a=1 or a=−2 (ii):ad=4⇔d=4 or d=−2 (iii):d^2 +d−2=0⇔d=1 or d=−2 so d=a=−2 ab+db+d=0⇔ −4b−2=0 ⇔ b=−(1/2) so A= [((−2),((−1)/2)),(0,(−2)) ]](https://www.tinkutara.com/question/Q12052.png)