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Inverse of a Matrix

If A3 = 0 t...

Question

If $A^{3} = 0$ then $1 + A + A^{2}$ is equal to

A

I + A

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B

(I + A)^{- 1}

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C

I - A

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D

(I - A)^{- 1}

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Solution

The correct option is D $(I - A)^{- 1}$
since the eigenvalue of matrix $A$ is zero

$(I - A)$ is invertible we have

$(I - A) (I + A + A^{2}) = I - A^{3} = I - 0 = I$

$(I - A) (I + A + A^{2}) = I$

$∴ I + A + A^{2} = (I - A)^{- 1}$

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Similar questions

A^{3} = 0

I + A + A^{2}

A^{3} = 0

1 + A + A^{2}

A^{3} = 0

I + A + A^{2}

a_{1}, a_{2}, a_{3}, . . . . a_{n}

are in A.P., where

a_{1} > 0

i

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}}

Q. A square matrix A is said to be nilpotent of index m. If

A^{m} = 0

, now, if for this A,

(I - A)^{n} = I + A + A^{2} + A^{3} + . . . . + A^{m - 1}

, then n is equal to?

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