JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
Let \(A=\left[a_{i j}\right]\) be a square matrix of order \(3\) such that \(a_{i j}=2^{j-i}\), for all \(i, j=1,2,3\). Then, the matrix \(A ^{2}+ A ^{3}+\ldots+ A ^{10}\) is equal to
- A \(\left(\frac{3^{10}-3}{2}\right) A\)
- B \(\left(\frac{3^{10}-1}{2}\right) A\)
- C \(\left(\frac{3^{10}+1}{2}\right) A\)
- D \(\left(\frac{3^{10}+3}{2}\right) A\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{3^{10}-3}{2}\right) A\)
Step-by-step Solution
Detailed explanation
\(A=\left(\begin{array}{lll}1 & 2 & 2^{2} \\ 1 / 2 & 1 & 2 \\ 1 / 2^{2} & 1 / 2 & 1\end{array}\right)\) \(A^{2}=3 A\) \(A ^{3}=3^{2} A\) \(A ^{2}+ A ^{3}+\ldots A ^{10}\) \(=3 A +3^{2} A +\ldots+3^{9} A =\frac{3\left(3^{9}-1\right)}{3-1} A\) \(=\frac{3^{10}-3}{2} A\)
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