(A) Given that $A$ is a non-singular matrix,so $|A| \neq 0$. Given equation: $(A+I)(A-I) = 0$. Expanding the product: $A^2 - AI + IA - I^2 = 0$. Since

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(A) Given that $A$ is a non-singular matrix,so $|A| \neq 0$. Given equation: $(A+I)(A-I) = 0$. Expanding the product: $A^2 - AI + IA - I^2 = 0$. Since $AI = IA = A$ and $I^2 = I$,we get $A^2 - I = 0$. Therefore,$A^2 = I$. Multiplying both sides by $A^{-1}$,we get $A^{-1}A^2 = A^{-1}I$. This simplifies to $A = A^{-1}$. Now,substituting $A^{-1} = A$ into the expression $A + A^{-1}$,we get $A + A = 2A$.

Class 12 Mathematics 3 and 4 .Determinants and Matrices Adjoint and inverse of matrices

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If $A$ is a non-singular matrix and $(A+I)(A-I)=0$,then $A+A^{-1} = \dots$

MHT CET 2019 All MHT CET

A
$2A$
B
$0$
C
$I$
D
$3I$

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3 and 4 .Determinants and Matrices

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Adjoint and inverse of matrices

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If $A$ is a non-singular matrix and $(A+I)(A-I)=0$,then $A+A^{-1} = \dots$

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