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Let \mathrm{A}=\left[\mathrm{a}_{i j}\right] be a square matrix of order 3 such that \mathrm{a}_{i j}=2^{j-i}, for all i, j=1,2,3. Then, the matrix \mathrm{A}^{2}+\mathrm{A}^{3}+\ldots+\mathrm{A}^{10} is equal to :

Option: 1

\left(\frac{3^{10}-3}{2}\right) \mathrm{A}


Option: 2

\left(\frac{3^{10}-1}{2}\right) \mathrm{A}


Option: 3

\left(\frac{3^{10}+1}{2}\right) \mathrm{A}


Option: 4

\left(\frac{3^{10}+3}{2}\right) \mathrm{A}


Answers (1)

best_answer

Forming matrix A using given \mathrm{aij} formula

\mathrm{A=\left[\begin{array}{ccc} 1 & 2 & 2^{2} \\ 2^{-1} & 1 & 2 \\ 2^{-2} & 2^{-1} & 1 \end{array}\right]}

\mathrm{A^{2}=\left[\begin{array}{ccc} 1 & 2 & 2^{2} \\ 2^{-1} & 1 & 2 \\ 2^{-2} & 2^{-1} & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & 2^{2} \\ 2^{-1} & 1 & 2 \\ 2^{-2} & 2^{-1} & 1 \end{array}\right]}

\mathrm{=\left[\begin{array}{ccc} 3 & 6 & 12 \\ 3 \cdot 2^{-1} & 3 & 6 \\ 3 \cdot 2^{-2} & 3 \cdot 2^{-1} & 3 \end{array}\right]=3 \mathrm{~A}}

\mathrm{\therefore A^{2}=3 A} \\

\mathrm{\Rightarrow A^{3}=A^{2} \cdot A=3 A \cdot A=3 A^{2}=3 \cdot 3 A=3^{2} A} \\

\mathrm{\Rightarrow A^{4}=A^{2} \cdot A^{2}=3 A \cdot 3 A=3^{2} A^{2}=3^{2} \cdot 3 A=3^{3} A }\\

\mathrm{\text { Similarly } A^{5}=3^{4} A, \cdots \cdot A^{10}=3^{9} A} \\

\mathrm{\therefore A^{2}+A^{3}+\ldots+A^{10}=A\left(3+3^{2}+\cdots+3^{9}\right) }

\mathrm{=\frac{3\left(3^{9}-1\right)}{3-1} A} \\

\mathrm{=\left(\frac{3^{10}-3}{2}\right) A .}

Hence the correct answer is option 1.

Posted by

rishi.raj

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