Problem 174
Question
If \(A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\), then find \(A^{2}\) and \(A^{3}\).
Step-by-Step Solution
Verified Answer
The square of A, or \(A^{2}\), is \(\left[\begin{array}{cc}8 & 5 \ -2 & 3\end{array}\right]\) and the cube of A, or \(A^{3}\), is \(\left[\begin{array}{cc}18 & 29 \ -23 & -2\end{array}\right]\).
1Step 1: Find \(A^{2}\)
First find \(A^{2}\) or \(A*A\) using matrix multiplication. The element at the i-th row and j-th column in the resulting matrix is the sum of the product of elements from the i-th row of the first matrix and the j-th column of the second one. For instance, the element at the 1st row and 1st column is 3*3 + 1*(-1) = 8.
2Step 2: Continue matrix multiplication for \(A^{2}\)
Continue the multiplication process for the other elements. As a result, the matrix \(A^{2}\) would be \(\left[\begin{array}{cc}8 & 5 \ -2 & 3\end{array}\right]\).
3Step 3: Find \(A^{3}\)
Find \(A^{3}\) or \(A^{2} * A\) using similar steps as above. Here, \(A^{2}\) is the resulting matrix obtained above and A is the original matrix. As a result, the matrix \(A^{3}\) would be \(\left[\begin{array}{cc}18 & 29 \ -23 & -2\end{array}\right]\).
Key Concepts
Squared MatricesCubed MatricesMatrix Operations
Squared Matrices
A squared matrix, simply put, is the result of multiplying a matrix by itself. For a matrix \( A \), squaring it involves performing what is known as matrix multiplication between \( A \) and itself.
To find \( A^2 \), you follow a methodical process:
To find \( A^2 \), you follow a methodical process:
- Take each row of the first matrix \( A \) and each column of the second matrix \( A \). For instance, if \( A = \begin{bmatrix} 3 & 1 \ -1 & 2 \end{bmatrix} \).
- For each element in the product matrix, multiply corresponding elements from the row and the column and add those products together. For example, the element at the first row and first column of \( A^2 \) would be calculated as: \( 3 \times 3 + 1 \times (-1) = 8 \).
Cubed Matrices
Cubing a matrix involves multiplying it by itself twice. In mathematical terms, for a matrix \( A \), \( A^3 \) refers to \( A^2 \times A \).
Here's how you can find \( A^3 \):
Here's how you can find \( A^3 \):
- Start with \( A^2 \), the squared matrix you previously calculated. If \( A^2 = \begin{bmatrix} 8 & 5 \ -2 & 3 \end{bmatrix} \), then you use this as the starting point.
- Next, perform matrix multiplication of \( A^2 \) with the original matrix \( A \). The process involves taking rows from \( A^2 \) and columns from \( A \), multiplying their elements, and summing the products.
- For example, the element in the first row, first column of \( A^3 \) is derived from: \( 8 \times 3 + 5 \times (-1) = 18 \).
Matrix Operations
Matrix operations are fundamental in mathematics and involve a series of calculations for manipulating matrices. Common operations include addition, subtraction, and multiplication. But here, let's focus on multiplication, especially as it relates to squaring and cubing matrices.
Matrix multiplication is not always commutative, meaning \( A \times B \) usually does not equal \( B \times A \). To multiply matrices:
Matrix multiplication is not always commutative, meaning \( A \times B \) usually does not equal \( B \times A \). To multiply matrices:
- Ensure the number of columns in the first matrix equals the number of rows in the second. This compatibility is essential for multiplication.
- Use the formula: for matrix \( C = AB \), each element \( c_{ij} \) in \( C \) is calculated as the sum of products of corresponding elements from row \( i \) in \( A \) and column \( j \) in \( B \).
Other exercises in this chapter
Problem 172
If \(A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right], B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\), then find \(A B\) and \(B A\) and s
View solution Problem 173
If \(A=\left[\begin{array}{ccc}1 & -2 & 3 \\ 2 & 3 & -1 \\ -3 & 1 & 2\end{array}\right]\) and \(B=\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 1 & 2 \\\ 1 & 2 & 0\e
View solution Problem 175
If \(A_{1 \times 2}=\left[\begin{array}{ll}a & b\end{array}\right], B_{1 \times 2}=\left[\begin{array}{ll}-b & -a\end{array}\right]\) and \(C_{2 \times 1}=\left
View solution Problem 176
Given \(A=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]\) and \(B=\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]\), find \(A B\).
View solution