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Therefore by equating them we get,3x + 4y = 2 …… (1)x – 2y = 4 …… (2)a + b = 5 …… (3)2a – b = – 5 …… (4)Multiplying equation (2) by 2 and adding to equation (1), we get3x + 4y + 2x – 4y = 2 + 8⇒ 5x = 10⇒ x = 2Now, substituting the value of x site link equation (1)3 × 2 + 4y = 2⇒ 6 + 4y = 2⇒ 4y = 2 – 6⇒ 4y = – 4⇒ y = – 1Now by adding equation (3) and (4)a + b + 2a – b = 5 + (– 5)⇒ 3a = 5 – 5 = 0⇒ a = 0Now, again by substituting the value of a in equation (3), we get0 + b = 5⇒ b = 5∴ a = 0, b = 5, x = 2 and y = – 19. Now it is an era of multiple choice questions. H. Students preparing for IIT JEE and other engineering entrance exams as well as students appearing useful site board exams should read this everyday, especially to master Algebra and Probability.
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We say that f is concave down on I if and only if the function f is concave up on I . H. SSo,
(i) A + B)T = AT + BT(ii) (AB)T = BT AT(iii) (2A)T = 2 ATSolution:(i) GivenConsider,L. So equation is y = 4x 14. Show that AB ≠ BA in each of the following cases:Solution:(i) Consider,Again consider,From equation (1) and (2), it is clear thatAB ≠ BA(ii) Consider,Now again consider,From equation (1) and (2), it is clear thatAB ≠ BA(iii) Consider,Now again consider,From equation (1) and (2), it is clear thatAB ≠ BA3. In case any user is found misusing our services, the user’s account will be immediately terminated.
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e. 1 Page No: 5. 19. Sir please can you give some theories and concept for relations and function. e.
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I will help you online for any doubt / clarification. H. 21. If A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4), find(i) A – 2B(ii) B + C 2A(iii) 2A + 3B – 5CSolution:(i) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag news 3 4)(ii) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)We have to find B + C 2AHere,Now we have to compute B + C 2A(iii) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)Now we have to find 2A + 3B – 5CHere,Now consider 2A + 3B – 5C6.
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Here are standard definitions of what it means for a function to be concave up/down, withoutassuming differentiability (and not even continuity!). SSo,
(iii) GivenConsider,L. e. A (B + C) = AB + AC. We have excellent notes prepared by Ex-IITian to best match the requirement of the exam. H.
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Show that AB ≠ BA in each of the following cases:Solution:(i) Consider,Again consider,From equation (1) and (2), it is clear thatAB ≠ BA(ii) Consider,Again consider,From equation (1) and (2) it is clear that,AB ≠ BA5. Therefore by equating them we get,2a + b = 4 …… (1)And a – 2b = – 3 …… (2)And 5c – d = 11 …… (3)4c + 3d = 24 …… (4)Multiplying equation (1) by 2 and adding to equation (2)4a + 2b + a – 2b = 8 – 3⇒ 5a = 5⇒ a = 1Now, substituting the value of a in equation (1)2 × 1 + b = 4⇒ 2 + b = 4⇒ b = 4 – 2⇒ b = 2Multiplying equation (3) by 3 and adding to equation (4)15c – 3d + 4c article 3d = 33 + 24⇒ 19c = 57⇒ c = 3Now, substituting the value of c in equation (4)4 × 3 + 3d = 24⇒ 12 + 3d = 24⇒ 3d = 24 – 12⇒ 3d = 12⇒ d = 4∴ a = 1, b = 2, c = 3 and d = 4Exercise 5. .