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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
If $A$ and $ B$ are two square matrices such that $B = - {A^{ - 1}}BA$, then ${(A + B)^2} = $
  • A
    $0$
  • ✓
    ${A^2} + {B^2}$
  • C
    ${A^2} + 2AB + {B^2}$
  • D
    $A + B$
✓

Answer

Correct option: B.
${A^2} + {B^2}$
b
(b) Given, $B = - {A^{ - 1}}BA$

$\therefore$ $AB = - A{A^{ - 1}}BA = - IBA = - BA$

$\therefore$ $AB = - BA$

Now ${(A + B)^2} = (A + B)(A + B)$

= ${A^2} + AB + BA + {B^2}$

= ${A^2} + {B^2}$               [$\because$ $BA$= $-BA$]

Thus, ${(A + B)^2} = {A^2} + {B^2}.$

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