The square matrix X is such that X^3 = 0. Show that the inverse of the matrix (I - X) is I + X + X^2.
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$\displaystyle (i-x)(i+x+x^{2}) = i^{2} + ix +ix^{2} - xi -x^{2} - x^{3} $ $\displaystyle = i + x + x^{2} - x - x^{2} - x^{3} = i - 0 = i $ EDIT : I don't why it keeps making it lowercase.
You should probably also show that (I+X+X^2)(I-X) = I
Originally Posted by Random Variable EDIT : I don't why it keeps making it lowercase. Type [tex]\mathcal{X}[/tex] to get $\displaystyle \mathcal{X} $.
Originally Posted by Plato Type [tex]\mathcal{X}[/tex] to get $\displaystyle \mathcal{X} $. Why is it converted to lowercase sometimes? I guess I could also use $\displaystyle \text{X}$
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