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JEE MAIN 2025
23-01-2025 SHIFT-1
Question
If A,B, and $\left( {{\mathop{\rm adj}\nolimits} \left( {{{\rm{A}}^{ - 1}}} \right) + {\mathop{\rm adj}\nolimits} \left( {{{\rm{B}}^{ - 1}}} \right)} \right)$ are non-singular matrices of same order, then the inverse of ${\rm{A}}{\left( {{\mathop{\rm adj}\nolimits} \left( {{{\rm{A}}^{ - 1}}} \right) + {\mathop{\rm adj}\nolimits} \left( {{{\rm{B}}^{ - 1}}} \right)} \right)^{ - 1}}\;{\rm{B}}$, is equal to
Select the correct option:
A
$\frac{1}{{|AB|}}({\mathop{\rm adj}\nolimits} (B) + {\mathop{\rm adj}\nolimits} (A))$
B
$\frac{{A{B^{ - 1}}}}{{|A|}} + \frac{{B{A^{ - 1}}}}{{|B|}}$
C
${\rm{A}}{{\rm{B}}^{ - 1}} + {{\rm{A}}^{ - 1}}\;{\rm{B}}$
D
${\mathop{\rm adj}\nolimits} \left( {{{\rm{B}}^{ - 1}}} \right) + {\mathop{\rm adj}\nolimits} \left( {{{\rm{A}}^{ - 1}}} \right)$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
${\left[ {{\rm{A}}{{\left( {{\mathop{\rm adj}\nolimits} \left( {\;{{\rm{A}}^{ - 1}}} \right) + {\mathop{\rm adj}\nolimits} \left( {{{\rm{B}}^{ - 1}}} \right)} \right)}^{ - 1}} \cdot \;{\rm{B}}} \right]^{ - 1}}$
$\;{{\rm{B}}^{ - 1}} \cdot \left( {{\mathop{\rm adj}\nolimits} \left( {\;{{\rm{A}}^{ - 1}}} \right) + {\mathop{\rm adj}\nolimits} \left( {{{\rm{B}}^{ - 1}}} \right)} \right) \cdot {{\rm{A}}^{ - 1}}$
$\;{{\rm{B}}^{ - 1}}{\mathop{\rm adj}\nolimits} \left( {\;{{\rm{A}}^{ - 1}}} \right){{\rm{A}}^{ - 1}} + {{\rm{B}}^{ - 1}}\left( {{\mathop{\rm adj}\nolimits} \left( {\;{{\rm{B}}^{ - 1}}} \right)} \right) \cdot {{\rm{A}}^{ - 1}}$
$\;{{\rm{B}}^{ - 1}}\left| {\;{{\rm{A}}^{ - 1}}} \right|{\rm{I}} + \left| {{{\rm{B}}^{ - 1}}} \right|{\rm{I}}{{\rm{A}}^{ - 1}}$
$\frac{{\;{{\rm{B}}^{ - 1}}}}{{|\;{\rm{A}}|}} + \frac{{{{\rm{A}}^{ - 1}}}}{{|\;{\rm{B}}|}}$
$ \Rightarrow \frac{{{\mathop{\rm adjB}\nolimits} }}{{|{\rm{B}}||{\rm{A}}|}} + \frac{{{\mathop{\rm adj}\nolimits} \mid }}{{|{\rm{A}}||{\rm{B}}|}}$
$ = \frac{1}{{|\;{\rm{A}}||{\rm{B}}|}}({\mathop{\rm adjB}\nolimits} + {\mathop{\rm adjA}\nolimits} )$
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