Math Problem Statement

Can you find A^7 ?

Solution

Let's find A7A^7 given that A2=(2111)A^2 = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} and A3=(3221)A^3 = \begin{pmatrix} 3 & 2 \\ 2 & 1 \end{pmatrix}.

Step 1: Identify the Pattern or Use the Recurrence Relation

To find A7A^7, we can use the Cayley-Hamilton theorem or look for a recurrence relation involving the powers of AA.

Let's denote AnA^n as the matrix raised to the power nn.

Given: A2=(2111),A3=(3221)A^2 = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}, \quad A^3 = \begin{pmatrix} 3 & 2 \\ 2 & 1 \end{pmatrix}

Step 2: Express Higher Powers Using the Given Matrices

We can find the matrices A4A^4 and so on by multiplying the known powers:

A4=A3A=(3221)(2111)A^4 = A^3 \cdot A = \begin{pmatrix} 3 & 2 \\ 2 & 1 \end{pmatrix} \cdot \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}

Perform the matrix multiplication:

A4=(3(2)+2(1)3(1)+2(1)2(2)+1(1)2(1)+1(1))=(8553)A^4 = \begin{pmatrix} 3(2) + 2(1) & 3(1) + 2(1) \\ 2(2) + 1(1) & 2(1) + 1(1) \end{pmatrix} = \begin{pmatrix} 8 & 5 \\ 5 & 3 \end{pmatrix}

Continue with:

A5=A4A=(8553)(2111)A^5 = A^4 \cdot A = \begin{pmatrix} 8 & 5 \\ 5 & 3 \end{pmatrix} \cdot \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}

A5=(8(2)+5(1)8(1)+5(1)5(2)+3(1)5(1)+3(1))=(2113138)A^5 = \begin{pmatrix} 8(2) + 5(1) & 8(1) + 5(1) \\ 5(2) + 3(1) & 5(1) + 3(1) \end{pmatrix} = \begin{pmatrix} 21 & 13 \\ 13 & 8 \end{pmatrix}

And similarly:

A6=A5A=(2113138)(2111)A^6 = A^5 \cdot A = \begin{pmatrix} 21 & 13 \\ 13 & 8 \end{pmatrix} \cdot \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}

A6=(21(2)+13(1)21(1)+13(1)13(2)+8(1)13(1)+8(1))=(55343421)A^6 = \begin{pmatrix} 21(2) + 13(1) & 21(1) + 13(1) \\ 13(2) + 8(1) & 13(1) + 8(1) \end{pmatrix} = \begin{pmatrix} 55 & 34 \\ 34 & 21 \end{pmatrix}

Finally, for A7A^7:

A7=A6A=(55343421)(2111)A^7 = A^6 \cdot A = \begin{pmatrix} 55 & 34 \\ 34 & 21 \end{pmatrix} \cdot \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}

A7=(55(2)+34(1)55(1)+34(1)34(2)+21(1)34(1)+21(1))=(144898955)A^7 = \begin{pmatrix} 55(2) + 34(1) & 55(1) + 34(1) \\ 34(2) + 21(1) & 34(1) + 21(1) \end{pmatrix} = \begin{pmatrix} 144 & 89 \\ 89 & 55 \end{pmatrix}

Final Result

Thus, the matrix A7A^7 is: A7=(144898955)\boxed{A^7 = \begin{pmatrix} 144 & 89 \\ 89 & 55 \end{pmatrix}}

If you have any questions or need further clarification, feel free to ask!


Here are 5 related questions to explore:

  1. How would you find A8A^8 given the result for A7A^7?
  2. What general pattern can you observe from the powers of matrix AA?
  3. How can you express AnA^n for any nn using a closed-form formula?
  4. What are the eigenvalues of matrix AA given the patterns in its powers?
  5. How does the Fibonacci sequence relate to the powers of matrix AA?

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Math Problem Analysis

Mathematical Concepts

Matrix Algebra
Matrix Powers

Formulas

Matrix multiplication

Theorems

Cayley-Hamilton theorem

Suitable Grade Level

Advanced Mathematics