Get Expert Help with Invertible Matrices: Tips and Techniques

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The discussion focuses on understanding the conditions for matrix invertibility, particularly regarding the expression (A + 2A^-1) and its determinant. It highlights that a matrix is invertible if its determinant is non-zero, prompting questions about the determinant of the given expression. Additionally, it examines the implications of the equation B^2 - 2B + I = 0, suggesting that B - I must be non-invertible. The conversation also explores the existence of a non-invertible matrix B that satisfies this equation. Overall, the thread seeks clarity on these matrix properties and their relationships.
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Hi all. I'm having a real tough time with this question. I don't know where to begin and how to go about the question. Can someone point me in the right direction please?

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haven't we seen this before?

a) If A is invertible, then (A+ 2A-1) is invertible.
When does there exist a matrix, B, such that (A+ 2A-1)B= I?
If I remember correctly, a matrix is invertible if and only if it's determinant is not 0. What is the determinant of A+ 2A-1?

b) If a real matrix B satisfies B2- 2B+ 1= 0, then B-I cannot be invertible.
B2- 2B+I= (B- I)2

c) There is a non-invertible matrix, B, satisfying B2- 2B+ I= 0.
As I said, B2-2B+ I= (B-I)2. Therefore B= ? or ?. Is either of those non-invertible?
 

Thread 'Find the polar form of a complex number'

The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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