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  • Let  <img alt="\mathrm{A=\left(\begin{array}{cc} 2 & -1 \\ 0 & 2 \end{array}\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BA%3D%5Cleft%28%5Cbegin%7Barr

Let  \mathrm{A=\left(\begin{array}{cc} 2 & -1 \\ 0 & 2 \end{array}\right)}. If  \mathrm{B=I-{ }^{5} C_{1}(\operatorname{adj} A)+{ }^{5} C_{2}(\operatorname{adj} A)^{2}-\ldots-{ }^{5} C_{5}(\operatorname{adj} A)^{5}} then the sum of all elements of the matrix \mathrm{B} is

Option: 1

-5


Option: 2

-6


Option: 3

-7


Option: 4

-8


Answers (1)

best_answer

As \mathrm{I\: and \: A} are commutative in multiplication \mathrm{\text { (IA }=A I \text { ), }} so they follow all identities.

\mathrm{B=(I-\operatorname{adj} A)^{5}} \\

\mathrm{=\left(\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{cc} 2 & 1 \\ 0 & 2 \end{array}\right]\right)^{5}} \\

\mathrm{=\left[\begin{array}{cc} -1 & -1 \\ 0 & -1 \end{array}\right]^{5}} \quad \cdots \text { (i) }

\mathrm{Now\: let \left[\begin{array}{rr}-1 & -1 \\ 0 & -1\end{array}\right]=P }

\mathrm{\therefore P^{2} =\left[\begin{array}{rr} -1 & -1 \\ 0 & -1 \end{array}\right]\left[\begin{array}{rr} -1 & -1 \\ 0 & -1 \end{array}\right]} \\

          \mathrm{=\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]} \\

\mathrm{P^{4} =\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 4 \\ 1 & 1 \end{array}\right] }

\mathrm{B=P^{5} =P^{4} \cdot P=\left[\begin{array}{ll} 1 & 4 \\ 1 & 1 \end{array}\right]\left[\begin{array}{rr} -1 & -1 \\ 0 & -1 \end{array}\right]} \\

                                 \mathrm{=\left[\begin{array}{rr} -1 & -5 \\ 0 & -1 \end{array}\right]}

Sum of its elements \mathrm{=-7}

Hence answer is option 3

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