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IfA = [[0 , 1, -1],[4, -3, 4],[3, -3, 4]...

If`A = [[0 , 1, -1],[4, -3, 4],[3, -3, 4]]` find the transpose of A matrix

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The correct Answer is:
2
` because A = [[0,1,-1],[4,-3,4],[3,-3,4]]`
`therefore A^(2) = A cdot A = [[0,1,-1],[4,-3,4],[3,-3,4]] [[0,1,-1],[4,-3,4],[3,-3,4]] = [[1,0,0],[0,1,0],[0,0,1]]= I `
`rArr A^(2) = I rArr A^(4) = A^(6) = A^(8) = ... = I`
Now, `A^(x) = I`
`rArr x = 2, 4, 6, 8...`
`therefore sum (cos ^(x) theta + sin ^(x) theta ) = ( cos^(2) theta + sin ^(2) theta) + (cos ^(4) theta + sin^(4) theta ) + (cos^(6) theta + sin ^(6) theta) + ...`
`=(cos^(2) theta + cos^(4) theta + cos ^(6) theta +...) `
`+ (sin^(2) theta + sin^(4) theta + sin^(6) theta + ...)`
`= (cos^(2) theta)/(1- cos^(2) theta) + (sin ^(2) theta)/(1- sin ^(2) theta)`
`= cot^(2) theta + tan ^(2) theta ge 2`
hences, minimum value of `sum (cos^(x) theta+ sin ^(x) theta)` is 2.
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