ધારોકે $A =\left(\begin{array}{cc}2 & -1 \\ 0 & 2\end{array}\right)$. જો $B = I -{ }^{5} C _{1} (\operatorname{adj} A )+{ }^{5} C _{2}$ $(\operatorname{adjA})^{2}-\ldots-{ }^{5} C _{5} (\operatorname{adj} A )^{5}$,તો શ્રેણીક $B$નાં તમામ ઘટકોનો સરવાળો $\dots\dots\dots$ છે.
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