Question

If $$ \mathrm{A}=\left|\begin{array}{lll} \mathrm{i} & 0 & 0 \\ 0 & \mathrm{i} & 0 \\ 0 & 0 & \mathrm{i} \end{array}\right| $$ \(\mathrm{i}=\sqrt{(-1)}\), then \(\mathrm{A}^{\mathrm{n}}=\) I where \(\mathrm{I}\) is unit matrix when \(\mathrm{n}=\ldots\) (a) \(4 \mathrm{p}+1\) (b) \(4 \mathrm{p}+3\) (c) \(4 \mathrm{p}\) (d) \(4 \mathrm{p}+2\)

Short answer

The correct answer is (c) \(4p\).

Step-by-step solution

Analyzing powers of i

Remember how the powers of i behave: \(i^1 = i\) \(i^2 = -1\) \(i^3 = -i\) \(i^4 = 1\) Notice that after the 4th power of i, the result starts to repeat, forming a cyclic pattern.

Calculate the powers of A

Now let's compute the powers of A. Since A is a diagonal matrix, we can calculate its powers by raising the individual diagonal elements to the given power. 1. A^1: \(\begin{bmatrix} i & 0 & 0\\ 0 & i & 0\\ 0 & 0 & i \end{bmatrix}\) 2. A^2: \(\begin{bmatrix} i^2 & 0 & 0\\ 0 & i^2 & 0\\ 0 & 0 & i^2 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & -1 \end{bmatrix}\) 3. A^3: \(\begin{bmatrix} i^3 & 0 & 0\\ 0 & i^3 & 0\\ 0 & 0 & i^3 \end{bmatrix} = \begin{bmatrix} -i & 0 & 0\\ 0 & -i & 0\\ 0 & 0 & -i \end{bmatrix}\) 4. A^4: \(\begin{bmatrix} i^4 & 0 & 0\\ 0 & i^4 & 0\\ 0 & 0 & i^4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\) Observe that A^4 gives us the identity matrix, I.

Confirm the pattern in the powers of A

Since there's a cyclic pattern in the powers of i, we can be confident that this pattern will continue for higher powers of A. For example, A^5 will be equal to A^1, and A^6 will be equal to A^2, and so on.

Finding the correct power equation

Now, we want to find an equation to represent the powers of A that are equal to the identity matrix. Let's look at our options: (a) 4p + 1 (b) 4p + 3 (c) 4p (d) 4p + 2 We found that when n = 4, A^n = I. So, let's try to plug in n = 4 into each of these options and see which one holds true: (a) 4p + 1 = 4 → p = 3/4 (not an integer, discard) (b) 4p + 3 = 4 → p = 1/4 (not an integer, discard) (c) 4p = 4 → p = 1 (integer, possible) (d) 4p + 2 = 4 → p = 1/2 (not an integer, discard) Only option (c) gives us an integer for p when n = 4. So n = 4p represents the correct powers that make A^n equal to the identity matrix I. Answer: (c) \(4p\)

Key concepts

Powers of Imaginary Numbers

Imaginary numbers come about when we deal with the square root of negative numbers. The basic unit of an imaginary number is \(i\) which is defined as \(i = \sqrt{-1}\). These numbers behave differently from real numbers, particularly when it comes to their powers. Here's a simple breakdown:

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)

Once we reach \(i^4\), the powers of \(i\) start repeating themself in a cycle every four steps. Understanding this cyclic nature is essential when performing calculations involving powers of \(i\). In mathematical problems, recognizing this pattern helps simplify complex expressions involving imaginary numbers.

Diagonal Matrices

A diagonal matrix is a specific kind of matrix where all entries outside the main diagonal are zero. In a 3x3 diagonal matrix, it is structured like this: \[ \begin{bmatrix} a & 0 & 0 \0 & b & 0 \0 & 0 & c \end{bmatrix} \] In these matrices, calculations such as determining powers of the matrix become straightforward. This is because raising a diagonal matrix to a power only requires raising each of the diagonal elements to that power. Therefore, if \(A\) is a diagonal matrix, computing \(A^n\) means finding \(\begin{bmatrix} a^n & 0 & 0\0 & b^n & 0\0 & 0 & c^n \end{bmatrix}\). In our specific homework problem, the matrix is entirely filled with \(i\), showing how diagonal matrices can focus more on individual elements and their specific outcomes.

Identity Matrix

An identity matrix is a special type of diagonal matrix. In an identity matrix, all the values on the main diagonal are 1, while all other entries are 0. In a 3x3 identity matrix, it looks like this: \[ \begin{bmatrix} 1 & 0 & 0 \0 & 1 & 0 \0 & 0 & 1 \end{bmatrix} \] An identity matrix functions similarly to the number 1 in multiplication. In matrix terms, any matrix \(A\) multiplied by an identity matrix \(I\) (of the same size) results in \(A\) again. So, \(AI = IA = A\). This matrix is crucial because it represents a neutral element in matrix multiplication. When raising our diagonal matrix from the homework to certain powers, we aim to reach this identity matrix to solve the equation.

Cyclic Patterns in Mathematics

Cyclic patterns frequently occur in mathematics and are particularly useful for simplifying computations. In our homework problem, we saw this with the powers of \(i\). These patterns can drastically reduce the amount of work needed by recognizing repetition. For example, we identified that the powers of \(i\) repeat every 4 terms (\(i^1, i^2, i^3, i^4\)). As a result, multiplying by \(i\) a number greater than 4 involves finding its remainder when divided by 4. This makes it easy to predict the outcome of powers of \(i\) without calculating each one. Cyclic patterns are useful in numerous mathematical fields, such as modular arithmetic, and help solve problems faster by recognizing these repeating sequences.