SSo,Solution:GivenNow we browse around these guys to find (AB)TSo,Exercise 5. Let us break down the proof of the latter into a few steps. Recall how continuity is defined using limits. Solution:(i) GivenConsider LHS,Now consider RHS,From equation (1) and (2), it is clear that A (B + C) = AB + AC(ii) Given,Consider the LHSNow consider RHS,Solution:Given,Consider the LHS,Now consider RHSFrom the above equations LHS = RHSTherefore, A (B – C) = AB – AC. 60Solution:GivenConsider, … (i) … (ii)From (i) and (ii) we can see thatA skew-symmetric matrix is a square matrix whose transpose equal to its negative, that is,X = – XTSo, A – AT is a skew-symmetric. We need differentiation when the rate of change is not constant.
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2) Let us restrict our attention to the case where I is an open interval of the form I = (a, b), wherea b are real numbers. v where u and v are functions of x Then i) y = (x+2)(ax2+bx)dy/dx = (x + 2)(2 ax + b)+(ax 2 + bx)ii) y = (4×3-3x+2)(2×2+4x)dy/dx = (4 x 3 – 3x +2)(2×2 + 4x)dy/dx =(4×3 – 3 x + 2)(4 x + 4) + (2 x 2 + 4 x)(12 x2 – 3)If y = u/v where u and v are functions of xIf y is a function of v, and v is a function of x, then y is a function of x and dy/dx = dy/dv . d/dx loga(x) = 1/x In(a). __mirage2 = {petok:”8df4577ef458b8597b3e67ed636f20a7e1b25c40-1664733614-31536000″};
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Exercise 5.
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181. The JEE Mathematics questions test a students acquired knowledge as well as his aptitude. H. S(ii) GivenConsider,L. Hence, X + Y = AExercise 5.
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Find the values of a, b, c and d from the following equations:Solution:GivenWe know that if two matrices are equal then the elements of each matrices are also equal. Securing Higher Grades Costing Your Pocket? Book Your Assignment Help Tutor at The Lowest Price Now!Differentiation is all about finding rates of change of one quantity compared to another. H. H. (AB) C = A (BC)Solution:(i) GivenConsider,Now consider RHS,From equation (1) and (2), it is clear that (AB) C = A (BC)(ii) Given,Consider the LHS,Now consider RHS,From equation (1) and (2), it is clear that (AB) C = A (BC)17.
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Of course the same holds for concave down functions! This isbecause f is concave down if and only if f is concave up, and f is continuous if and only f iscontinuous. All assignments available for free download on the website are developed by the best teachers having many years of teaching experience find more information CBSE schools all over the country. Solution:GivenSolution:GivenSolution:GivenSolution:GivenSolution:GivenSolution:GivenI is identity matrix, soAlso given,
Now, dig this have to find A2, we getNow, we will find the matrix for 8A, we getSo,Substitute corresponding values from eqn (i) and (ii), we getAnd to satisfy the above condition of equality, the corresponding entries of the matrices should be equalHence,Therefore, the value of k is 7Solution:GivenTo show that f (A) = 0Substitute x = A in f(x), we getI is identity matrix, soNow, we will find the matrix for A2, we getNow, we will find the matrix for 2A, we getSubstitute corresponding values from eqn (ii) and (iii) in eqn (i), we getSo,Hence ProvedSolution:GivenSoNow, we will find the matrix for A2, we getNow, we will find the matrix for λ A, we getBut given, A2 = λ A + μ ISubstitute corresponding values from equation (i) and (ii), we getAnd to satisfy the above condition of equality, the corresponding entries of the matrices should be equalHence, λ + 0 = 4 ⇒ λ = 4And also, 2λ + μ = 7Substituting the obtained value of λ in the above equation, we get2(4) + μ = 7 ⇒ 8 + μ = 7 ⇒ μ = – 1Therefore, the value of λ and μ are 4 and – 1 respectively39. .