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Chapter 8: Problem 109

Let \(i=\sqrt{-1}\) and let \(A=\left[\begin{array}{ll}i & 0 \\ 0 & i\end{array}\right] \quad\) and \(\quad B=\left[\begin{array}{ll}0 & -i \\ i & 0\end{array}\right].\) (a) Find \(A^{2}, A^{3},\) and \(A^{4}\). Identify any similarities with \(i^{2}, i^{3},\) and \(i^{4}.\) (b) Find and identify \(B^{2}.\)

Short Answer

Expert verified
For part (a), we find that \(A^{2}=\[\begin{array}{ll}-1 & 0 \ 0 & -1\end{array}\]\), \(A^{3}=\[\begin{array}{ll}-i & 0 \ 0 & -i\end{array}\]\), and \(A^{4}=\[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\]\). On comparison with the powers of \(i\), we can observe that \(i^{2}=A^{2}, i^{3}=A^{3}, i^{4}=A^{4}\). \n\nFor part (b), we found \(B^{2}=\[\begin{array}{ll}-1 & 0 \ 0 & -1\end{array}\]\).

Step by step solution

01

Compute \(A^{2}, A^{3},\) and \(A^{4}\)

Recall that the multiplication of two matrices is performed element-wise. Therefore, we have the following computations: \n\n\(A^{2}=\left[\begin{array}{ll}i & 0 \ 0 & i\end{array}\right]\times \left[\begin{array}{ll}i & 0 \ 0 & i\end{array}\right] = \left[\begin{array}{ll}-1 & 0 \ 0 & -1\end{array}\right]\)\n\n\(A^{3}=A\times A^{2}=\left[\begin{array}{ll}i & 0 \ 0 & i\end{array}\right]\times \left[\begin{array}{ll}-1 & 0 \ 0 & -1\end{array}\right] = \left[\begin{array}{ll}-i & 0 \ 0 & -i\end{array}\right]\)\n\n\(A^{4}=A \times A^{3}=\left[\begin{array}{ll}i & 0 \ 0 & i\end{array}\right]\times\left[\begin{array}{ll}-i & 0 \ 0 & -i\end{array}\right] = \left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\right]\)
02

Compare with powers of \(i\)

Taking the powers of \(i\): \n\n \(i^{2}=-1, i^{3}=-i, i^{4}=1\)\n\nComparing the results of step 1 with the powers of \(i\), we are able to observe that \n\n\(i^{2}=A^{2}, i^{3}=A^{3}, i^{4}=A^{4}\)
03

Compute \(B^{2}\)

To solve for \(B^{2}\), use the rule of matrix multiplication:\nB^{2}=\[\begin{array}{ll}0 & -i \ i & 0\end{array}\]\times\[\begin{array}{ll}0 & -i \ i & 0\end{array}\] = \[\begin{array}{ll}-1 & 0 \ 0 & -1\end{array}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Numbers
Complex numbers are a form of mathematics where numbers comprise a real and an imaginary part. The imaginary unit, denoted as i, is defined by the property that i^2 = -1. This unit enables the creation of complex numbers in the form a + bi, where a and b are real numbers. Complex numbers can be manipulated using arithmetic operations much like real numbers, with the caveat that you must always consider i^2 = -1 during multiplication or exponentiation.

For instance, when you square i, you get i^2 = -1. Following the pattern, i^3, which is i * i^2, simplifies to -i, and i^4 (or i^2 * i^2) simplifies to 1. This cyclic pattern repeats every four powers, which is a crucial characteristic to understand when dealing with complex numbers in various mathematical contexts, including matrix exponentiation.
Matrix Multiplication
Matrix multiplication is an operation that produces a new matrix from two matrices with compatible dimensions. The standard method involves multiplying each row of the first matrix by each column of the second matrix and summing the products to yield an entry in the resulting matrix.

For example, to multiply a 2x2 matrix A by another 2x2 matrix B, you take the dot product of the row of A with the column of B to find each element of the resulting matrix. This operation is not commutative, meaning that A * B does not necessarily equal B * A. Matrix multiplication is fundamental in various fields of science and engineering, including computer graphics, physics, and systems theory.

When you have a matrix with complex numbers, as in the exercise above, the process remains the same; just apply the rules of complex number multiplication when calculating the dot product for each element in the resultant matrix.
Identity Matrix
The identity matrix is a special type of square matrix that plays a similar role to the number 1 in multiplication. An identity matrix is denoted by I and is a square matrix with ones on the diagonal and zeros elsewhere. For any matrix A, when you multiply A by the identity matrix (and the dimensions agree), A remains unchanged:

A * I = I * A = A.

For instance, the result of raising the matrix A to the fourth power in the exercise produced an identity matrix, showing that the matrix A, when raised to the fourth power, essentially performs a 'reset' to the identity matrix. This concept is pivotal in linear algebra, where the identity matrix is used as a multiplicative identity for matrix multiplication, and it reveals important properties about matrix transformations and systems of linear equations.

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Most popular questions from this chapter

Consider a company that specializes in potting soil. Each bag of potting soil for seedlings requires 2 units of sand, 1 unit of loam, and 1 unit of peat moss. Each bag of potting soil for general potting requires 1 unit of sand, 2 units of loam, and 1 unit of peat moss. Each bag of potting soil for hardwood plants requires 2 units of sand, 2 units of loam, and 2 units of peat moss. Find the numbers of bags of the three types of potting soil that the company can produce with the given amounts of raw materials. 500 units of sand 750 units of loam 450 units of peat moss

Use an inverse matrix to solve (if possible) the system of linear equations. $$\left\\{\begin{array}{l} 4 x-y+z=-5 \\ 2 x+2 y+3 z=10 \\ 5 x-2 y+6 z=1 \end{array}\right.$$

Consider the circuit in the figure. The currents \(I_{1}, I_{2},\) and \(I_{3},\) in amperes, are given by the solution of the system of linear equations \(\left\\{\begin{aligned} 2 I_{1} &+4 I_{3}=E_{1} \\ I_{2}+4 I_{3} &=E_{2} \\\ I_{1}+I_{2}-I_{3} &=0 \end{aligned}\right.\) where \(E_{1}\) and \(E_{2}\) are voltages. Use the inverse of the coefficient matrix of this system to find the unknown currents for the given voltages. \(E_{1}=10\) volts, \(E_{2}=10\) volts

The sums have been evaluated. Solve the given system for \(a\) and \(b\) to find the least squares regression line for the points. Use a graphing utility to confirm the results. $$\left\\{\begin{aligned} 5 b+10 a &=11.7 \\ 10 b+30 a &=25.6 \end{aligned}\right.$$

Use an inverse matrix to solve (if possible) the system of linear equations. $$\left\\{\begin{array}{l} 4 x-2 y+3 z=-2 \\ 2 x+2 y+5 z=16 \\ 8 x-5 y-2 z=4 \end{array}\right.$$
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