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Adjoint of a Matrix

If A2=I and...

Question

If $A^{2} = I$ and |A|>0 then det(A-I)=

A

0

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B

-1

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C

1

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D

can not say anything

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Solution

The correct option is A 0
$A^{2} = I$
$A = \pm I$
Since, $| A | > 0$ then,
$A = I$
$A - I = 0$
$\Rightarrow$ $| A - I | = 0$
$∴$ $d e t (A - I) = 0$

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Similar questions

A^{2} = I

, then the value of

d e t (A - I)

A

3

)

Q. Fill in the blank with the most suitable option:

Unless he _____________ this office, I will not say anything.

Q. Assertion :Let

A

2 \times 2

matrix with real entries. Let

I

2 \times 2

identity matrix. Denote by

t r (A)

, the sum of diagonal entries of

A

A^{2} = I

.

A \neq I

A \neq - I

d e t (A) = - 1

A \neq I

A \neq - I

t r (A) \neq 0

.

A A^{T} = I

and det(A) = 1, then

A

2 \times 2

matrix with real entries, Let I be the

2 \times 2

identity matrix. Denote by

t r (A),

the sum of diagonal entries of

A

A^{2} = I

1 :

A \neq I

A \neq - I

A = - 1

2 :

A \neq I

A \neq - I

t r (A) \neq 0,

then which of the following is correct

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