WebDiagonalize the matrix, where the non-zero elements will be the matrix eigenvalues, then the same matrix raised to any power will have the same eigenvalues raised to the required power associated with the same eigenvectors of the original matrix, thus yielding your desired matrix. WebGiven: a 3 + b 3 + c 3 = 3abc Formula used: a 3 + b 3 + c 3 - 3abc = (a + b + c) (a 2 + b 2 + c 2 - ab - bc - ca) Calculations: We have a 3 + b 3 + c 3 = 3abc ---- (i) Now from the formula used a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca) From equation (i), we have ⇒ 3abc - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)
[Solved] If a3 + b3 + c3 = 3abc . Find the value of a - Testbook
WebGiven A=[34 4−3] and B=[247], find the matrix X such that AX=B. A [−4−3] B [43] C [−43] D [ 4−3] Easy Solution Verified by Toppr Correct option is B) Given, AX=B To perform matrix multiplication, AX the number of columns in A must equal the number of rows in X Hence, X should be matrix of order 2×1 Say, X=[xy] Consider, AX=B [34 4−3][xy]=[247] WebNov 19, 2024 · You can't apply AM-GM, as we need that a, b ≥ 0 and not a + b > 0. Let's use the a + b > 0 condition. Dividing by a + b > 0 gives that it's enough to prove a2 − ab … buy rogaine in thailand
SOLUTION: a=3, b=4, c=? - algebra.com
WebMay 24, 2016 · First, find tan A and tan B. #cos A = 3/5#--> #sin^2 A = 1 - 9/25 = 16/25#--> #cos A = +- 4/5# #cos A = 4/5# because A is in Quadrant I #tan A = sin A/(cos A) = (4/5)(5/3) = 4/3.# #sin B = 5/13#--> #cos^2 B = 1 - 25/169 = 144/169#--> #sin B = +- 12/13.# WebAssuming a and b are the short sides of a right triangle: c = 5 This is the 3, 4, 5 right triangle that we see a lot of. Carpenters use this to see if 2 walls are at a right angle: they measure 3' from the corner on one wall, 4' on the other, and check to see if the diagonal is 5 feet. WebAnswer (1 of 5): a-b=3…………..(1) and ab=4…. ……..(2) Squaring both side eq.(1) a2+b^2–2ab=9 a^2+b^2–2×4=9 a^2+b^2=17………….(3) a^3-b^3=(a-b)(a ... buy rogue bucks