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Meghna Thapar 5 years ago

Elementary transformations of a matrix are: 1) rearrangement of two rows (columns); 2) multiplication of all row (column) elements of a matrix to some number, not equal to zero; 3) addition of two rows (columns) of the matrix multiplied by the same number, not equal to zero. Elementary row operations are used in Gaussian elimination to reduce a matrix to row echelon form. They are also used in Gauss-Jordan elimination to further reduce the matrix to reduced row echelon form.

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Jaspreet Singh 5 years ago

Sin¤ = P/ H

Rosy Panda 5 years ago

perpendicular/ hypotaneous
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Vinayak Avasthi 5 years, 1 month ago

5/13
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Dev Kumar 5 years, 1 month ago

Mathematics analysis youtube channel
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Vinayak Avasthi 5 years, 1 month ago

X= 1/5
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Shree Ram Faujdar 5 years, 1 month ago

Write easy in beginning of continuity and differentiability
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Rajan Kumar Pasi 5 years, 1 month ago

Sol: |adj ( adjA )|=(|A|n1)2=>|adj ( adjA )|=(231)2=>|adj ( adjA )|=(22)2=>|adj ( adjA )|=(4)2=>|adj ( adjA )|=16Ans:)

Diksha Garg 5 years, 1 month ago

|adj.(adj.A)|=|A|^(n-1) ^2 It means |A|^(3-1) ^2 =2^4=16
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Anuj Kaushik 5 years, 1 month ago

Sir, if i don't getting it wrong we do not want show to symmetry here. If so, please elaborate the full answer.

Rajan Kumar Pasi 5 years, 1 month ago


Here is the answer. For symmetricity, you have to include all the possible pairs which shows symmetry.
Only one pair does not make it symmetric.

 

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Jaspreet Singh 5 years ago

Let sin-¹ [-√3/2] = ¤ ¤ = -π/3 A. T. Q = Sin[π/2 + π/3] = Sin [5π/6] = sin 150• = sin (90 + 60)• = sin 90 + cos 60 + cos 90 + sin 60 = 1×1/2. + 0 × √3/2 = 1/2 = ANWSER...........!!!!!!!
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Ramkishor Saini 5 years, 1 month ago

X=1,y=4
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Tannu Rao 5 years, 1 month ago

-cos(2π)-(-cos(0)) = -1-(-1) = 0
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Jaspreet Singh 5 years ago

Let cos -¹ 4/5 = ¤ B/H = 4/5 = cos ¤ Sin ¤/2 = √1-cos ¤ / √2 = 1/√10

Vinod Varma 5 years, 1 month ago

Let, cos inverse 4/5 = A where, A belongs to [0,π] So, cosA = 4/5 .............(1) We know, cos2A = 1 - 2sin^2A cosA = 1- 2sin^2 A/2 2sin^2 A/2 = 1 - cosA sin^2 A/2 = (1 - cosA)/2 Put cosA = 4/5 in equation (1) = (1-4/5)/2 = (5-4)/5/2 = 1/5 . 1/2 = 1/10 sin^A/2 = 1/10 sin A/2 = +1/√10 (sin A/2 is +ve and 0 <A<π/2) A/2 = sin inverse (1/√10) sin (1/2 .A) = sinA/2 sinA/2 = sin (sin inverse 1/√10) So, sin(sin inverse 1/√10) = 1/√10
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Pritika Yadav 5 years, 1 month ago

Question kaha par h
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Aman Shree 5 years, 1 month ago

1. Write cos x = sin(90-x). 2.Then, Use formula to expand Sin x + Sin(90-x) 3. We will get x= (nπ+3π/4)

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