Problem 104 - Precalculus

Question

$$\text { For } A=\left[\begin{array}{ll}1 & 0 \\\0 & 1\end{array}\right] \text { find } A, A^{2}, A^{3}, \ldots . \text { What is } A^{n} ?$$

Step-by-Step Solution

Answer

\(A^n = A\) for any positive integer \(n\).

Step 1: Understand the Matrix A

The given matrix \(A\) is the identity matrix of order 2, which is \(\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\). This matrix is distinguished because it does not change any vector when used in multiplication and has a special property in powers.

Step 2: Calculate \(A^2\)

To find \(A^2\), multiply matrix \(A\) by itself:\[ A^2 = A \cdot A = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = A \]

Step 3: Validate that Matrix A is Idempotent

Notice that the multiplication \(A^2 = A\) indicates that \(A\) has been verified to be idempotent, meaning multiplying it by itself does not change it.

Step 4: Explore Higher Powers of A

Using the observation from \(A^2\), calculate \(A^3\) as follows:\[ A^3 = A^2 \cdot A = A \cdot A = A = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \]This pattern will repeat for any higher power of \(A\).

Step 5: Conjecture for \(A^n\)

Since \(A^2 = A\), \(A^3 = A\), and this pattern repeats, the conjecture is \(A^n = A\) for any positive integer \(n\). This means that raising the identity matrix to any power will result in the identity matrix itself.

Key Concepts

Matrix Multiplication

Matrix multiplication is an essential operation in linear algebra. It involves taking two matrices and producing another matrix. This process, however, is not as straightforward as regular number multiplication.
When multiplying two matrices, you multiply the corresponding entries and sum the results. For example, if you multiply a 2x2 matrix, like the identity matrix \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \), with itself, each entry in the resulting matrix is derived by performing these steps:

• Multiply the first row by the first column: \((1 \times 1) + (0 \times 0) = 1\)
• Multiply the first row by the second column: \((1 \times 0) + (0 \times 1) = 0\)
• Multiply the second row by the first column: \((0 \times 1) + (1 \times 0) = 0\)
• Multiply the second row by the second column: \((0 \times 0) + (1 \times 1) = 1\)

This results in another identity matrix, affirming that identity matrices do not change other matrices when multiplied by them. Thus, multiplication involving such matrices can simplify various computations.

Idempotent Matrix

An idempotent matrix is a matrix that, when multiplied by itself, yields the same matrix. It essentially "maintains" its original value through multiplication. The identity matrix \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \) is a great example of this.
The property of being idempotent means that any matrix multiplication with a similar identity matrix results in the matrix itself. This is because the identity matrix has a distinctive diagonal made entirely of ones, thereby preserving its structure through repeated operations.
Key characteristics of idempotent matrices include:

• Preservation in matrix multiplication, meaning \( A^2 = A \).
• They often simplify complex computations and confirm structural properties when verifying matrix equations or transformations.

Understanding this concept can greatly assist in reducing complexity in matrix operations.

Matrix Powers

Whenever you raise a matrix to a power, you are essentially multiplying the matrix by itself repeatedly. For instance, when computing powers of an identity matrix like \( A^n \), the result is surprisingly simple.
As demonstrated with the identity matrix \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \), raising it to any power results in the identity matrix itself. This pattern occurs because of the matrix's multiplicative identity property.

• For \( A^2 \), we calculated \( A \cdot A = A \).
• For \( A^3 \), \( A^3 = A^2 \cdot A = A \).
• This pattern continues for any power \( n \), resulting in \( A^n = A \).

When dealing with identity matrices, this implies that even as the exponent grows, the resulting matrix stays unchanged. In applications, this can lead to highly efficient solutions where you repeatedly require the same consistent result.