(A) Given,$(A+B)(A-B)=A^{2}-B^{2}$ Expanding the left side: $A^{2}-AB+BA-B^{2}=A^{2}-B^{2}$ Subtracting $A^{2}-B^{2}$ from both sides: $-AB+BA=0$ Ther

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(A) Given,$(A+B)(A-B)=A^{2}-B^{2}$ Expanding the left side: $A^{2}-AB+BA-B^{2}=A^{2}-B^{2}$ Subtracting $A^{2}-B^{2}$ from both sides: $-AB+BA=0$ Therefore,$AB=BA$ Now,we evaluate $(ABA^{-1})^{2}$: $(ABA^{-1})^{2} = (ABA^{-1})(ABA^{-1})$ Since $AB=BA$,we can write $AB = BA$,so $ABA^{-1} = BAA^{-1} = BI = B$ Thus,$(ABA^{-1})^{2} = B^{2}$

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

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If $A$ and $B$ are square matrices of the same order such that $(A+B)(A-B)=A^{2}-B^{2}$,then $(A B A^{-1})^{2}$ is equal to

KCET 2009 All KCET

A
$B^{2}$
B
$I$
C
$A^{2} B^{2}$
D
$A^{2}$

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3-4. Matrices and Determinants

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Adjoint and inverse of matrices

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If $A$ and $B$ are square matrices of the same order such that $(A+B)(A-B)=A^{2}-B^{2}$,then $(A B A^{-1})^{2}$ is equal to

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