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Chapter 4: Problem 298

If $$ \mathrm{A}=\left|\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right| $$ then \(\mathrm{A}^{2013}=\) (a) \(2^{2012} \mathrm{~A}\) (b) \(2^{1006} \mathrm{~A}\) (c) \(-2^{2013} \mathrm{~A}\) (d) I

Short Answer

Expert verified
The correct answer is \( A^{2013} = 2^{2012} A \), which matches answer option (a).

Step by step solution

01

Calculate A^2

To calculate A^2, multiply matrix A by itself: \(A^2 = A \cdot A = \left|\begin{array}{ll} 1 & 1 \\\ 1 & 1 \end{array}\right| \cdot \left|\begin{array}{ll} 1 & 1 \\\ 1 & 1 \end{array}\right|\) To multiply two matrices, we multiply the elements of row i of the first matrix by the elements of column j of the second matrix and sum the products: \(A^2 = \left|\begin{array}{ll} (1)(1)+(1)(1) & (1)(1)+(1)(1) \\\ (1)(1)+(1)(1) & (1)(1)+(1)(1) \end{array}\right| = \left|\begin{array}{ll} 2 & 2 \\\ 2 & 2 \end{array}\right|\)
02

Calculate A^3

To calculate A^3, multiply matrix A^2 by matrix A: \(A^3 = A^2 \cdot A= \left|\begin{array}{ll} 2 & 2 \\\ 2 & 2 \end{array}\right| \cdot \left|\begin{array}{ll} 1 & 1 \\\ 1 & 1 \end{array}\right|\) Following the same matrix multiplication rules as before: \(A^3 = \left|\begin{array}{ll} (2)(1)+(2)(1) & (2)(1)+(2)(1) \\\ (2)(1)+(2)(1) & (2)(1)+(2)(1) \end{array}\right| = \left|\begin{array}{ll} 4 & 4 \\\ 4 & 4 \end{array}\right|\)
03

Identify the pattern

We can see a pattern emerging from these calculations. For each power n of matrix A, the elements of the resulting matrix A^n are equal to \(2^{n-1}\). Therefore, we can directly calculate A^2013 following this pattern:
04

Calculate A^2013

Using the pattern we have identified, we can find the elements of A^2013: \( A^{2013} = \left|\begin{array}{ll} 2^{2012} & 2^{2012} \\\ 2^{2012} & 2^{2012} \end{array}\right|\) Now we can see that A^2013 matches with answer option (a):
05

Choose the correct answer

After calculating A^2013 using the pattern, we find that the correct answer is (a): $$A^{2013} = 2^{2012} \mathrm{ ~A}$$

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