ધારો કે $A$ એ $3\times3$ એ સામાન્ય શ્રેણીક છે અને $(A - 3I) (A- 5I)\, = 0$, કે જ્યાં $I\,= I_3$ અને $O\,= O_3$. જો $\alpha A + \beta A^{- 1}\, = 4I$, તો $\alpha + \beta = . .. $
JEE MAIN 2018, Difficult
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we have
$(A-3I)(A-5I)=0$
$ \Rightarrow {A^2} - 8A + 15I = 0$
Multiplying both sides by ${A^{ - 1}}$, we get;
${A^{ - 1}}A.A - 8{A^{ - 1}}A + 15{A^{ - 1}}I = {A^{ - 1}}0$
$ \Rightarrow A - 8I + 15{A^{ - 1}} = 0$
$A + 15{A^{ - 1}} = 8I$
$\frac{A}{2} + \frac{{15{A^{ - 1}}}}{2} = 4I$
$\therefore \alpha + \beta = \frac{1}{2} + \frac{{15}}{2} = \frac{{16}}{2} = 8$
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