Problem 190
Question
If \(A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\), then find \(A^{11}\).
Step-by-Step Solution
Verified Answer
The 11th power of matrix \( A \), \( A^{11} = A \).
1Step 1: Observe the Given Matrix
First look at the given matrix \( A = \left[\begin{array}{ll}0 & 1 \ 1 & 0\end{array}\right] \). Note that interchanging the numbers in the rows or the columns results in the same matrix. Now, let's see what happens when the matrix is squared.
2Step 2: Square the Given Matrix
Calculate the square of the given matrix to see the pattern it forms. To square a matrix, the matrix is multiplied by itself as follows: \( A^2 = A \cdot A = \left[\begin{array}{ll}0 & 1 \ 1 & 0\end{array}\right]\cdot\left[\begin{array}{ll}0 & 1 \ 1 & 0\end{array}\right] = \left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}\right] \). When calculated, we obtain the identity matrix. This leads us to the conclusion that every even power of the matrix \( A \) will result in the identity matrix.
3Step 3: Determine the Power of the Matrix
Now, to find \( A^{11} \), we can use the property that every even power of \( A \) equals the identity matrix. The number 11 can be written as 10 + 1, so we can rewrite \( A^{11} \) as \( A^{10}A^{1} = I \cdot A =A \). Hence, \( A^{11} = A \).
Key Concepts
Matrix MultiplicationIdentity MatrixMatrix Patterns
Matrix Multiplication
Matrix multiplication is a fundamental operation in linear algebra. It involves multiplying rows by columns to produce a new matrix. When multiplying two matrices, every element of the product matrix is derived from the sum of the products of rows and columns of the two matrices involved. Let's break it down more simply:
Once you grasp this concept, it becomes quite clear why pattern recognition in matrix multiplication can greatly save time with operations such as determining higher powers of matrices.
- Each element in the resulting matrix is calculated by taking the dot product of corresponding rows and columns from the matrices you are multiplying.
- This operation is not commutative, meaning that in general, \( AB eq BA \).
- The number of columns in the first matrix must equal the number of rows in the second matrix.
Once you grasp this concept, it becomes quite clear why pattern recognition in matrix multiplication can greatly save time with operations such as determining higher powers of matrices.
Identity Matrix
The identity matrix, often denoted by \( I \), is a special form of matrix with ones on the diagonal and zeros elsewhere. Think of it like the number 1 in matrix operations—not altering any matrix it multiplies, effectively acting as the multiplicative identity.
- If \( I \) is multiplied by any matrix \( A \), the result is \( A \), symbolically written as \( AI = IA = A \).
- For a 2x2 matrix, the identity matrix looks like \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \).
Matrix Patterns
Matrix patterns refer to the recognitions and repetitions within matrix operations that simplify computational processes. Observing these patterns can make predicting future computations easier without actually performing all the operations.
Take the example in the exercise. By observing the pattern when calculating matrix powers of \( A = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \), we find:
Take the example in the exercise. By observing the pattern when calculating matrix powers of \( A = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \), we find:
- \( A^1 \) results in the original matrix \( A \).
- \( A^2 \) results in the identity matrix \( I \).
Other exercises in this chapter
Problem 188
If \(A=\left[\begin{array}{cc}0.8 & 0.6 \\ -0.6 & 0.8\end{array}\right]\), find \(A^{3} .\)
View solution Problem 189
If \(X=\left[\begin{array}{cc}3 & -4 \\ 1 & -1\end{array}\right]\), then find \(X^{3}\).
View solution Problem 191
If \(J_{1}=\left[\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right]\) and \(J_{2}=\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]\), then find \(J_{1}^
View solution Problem 192
If \(A=\left[\begin{array}{ccc}1 & -3 & 2 \\ 2 & 1 & -3 \\ 4 & -3 & -1\end{array}\right], B=\left[\begin{array}{cc}1 & 4 \\ 2 & 1 \\ 1 & -2\end{array}\right]\)
View solution