Problem 64
Question
Given a square matrix \(A^{-1}\), find matrix \(A\). $$A^{-1}=\left[\begin{array}{lll} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right]$$
Step-by-Step Solution
Verified Answer
The matrix \(A\) is \(\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}\).
1Step 1: Understanding the Matrix Relationship
The problem provides the inverse of matrix \(A\), denoted as \(A^{-1}\). Our goal is to find the original matrix \(A\) such that when it is multiplied with \(A^{-1}\), it results in the identity matrix \(I\). This can be expressed as: \(A \cdot A^{-1} = I\).
2Step 2: Understand the Properties of Transpose in the Matrix
The given inverse matrix \(A^{-1}\) is a permutation matrix due to its single 1 per row and column. By observing the structure, notice that \(A^{-1}\) is actually the transpose of itself, i.e., \((A^{-1})^T = A\). This means the roles of row and column are swapped for the identity property.
3Step 3: Transpose the Given Matrix
To find matrix \(A\), transpose \(A^{-1}\). The transpose operation involves flipping the matrix over its diagonal, switching the row and column indices. For our matrix: \[A^{-1} = \begin{bmatrix} 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \end{bmatrix}\] After transposition, matrix \(A\) becomes: \[A = \begin{bmatrix} 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \end{bmatrix}\].
4Step 4: Verify the Solution
To ensure correctness, multiply \(A\) by \(A^{-1}\) and verify if it results in the identity matrix \(I\). Calculate \(A \cdot A^{-1}\) as follows: \[A \cdot A^{-1} = \begin{bmatrix} 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}\] This confirms that \(A\) is correctly identified, as it results in the 3x3 identity matrix.
Key Concepts
Matrix MultiplicationMatrix TranspositionIdentity Matrix
Matrix Multiplication
When dealing with matrices, one of the most essential operations is "Matrix Multiplication". Every time we multiply two matrices together, we create a new matrix that's a blend of the original two. This operation is not as straightforward as multiplying single numbers but involves a precise process.
Here's how it works:
Here's how it works:
- The number of columns in the first matrix must equal the number of rows in the second matrix. This is a crucial requirement for multiplication to occur.
- To find each element of the resulting matrix, multiply the elements of the rows of the first matrix by the elements of the columns of the second matrix and add up the products. For example, the first element in the product matrix is the sum of the products of the first row in the first matrix and the first column in the second matrix.
Matrix Transposition
Matrix Transposition is an operation that flips a matrix over its diagonal. This means that the rows become columns and columns become rows. It’s a simple yet powerful concept in linear algebra, critical for transformations and solving systems of equations.
When you transpose a matrix:
When you transpose a matrix:
- The element at position (i, j) in the original matrix is moved to position (j, i) in the transposed matrix.
- For any square matrix (same number of rows and columns), the transposed version retains its size.
Identity Matrix
The "Identity Matrix" is a crucial concept when studying matrices. Think of it as the matrix equivalent of the number "1" in arithmetic operations. Multiplying any number by one leaves the number unchanged, and similarly, multiplying any matrix by an identity matrix leaves it unchanged.
Here are key features of an identity matrix:
Here are key features of an identity matrix:
- It is a square matrix, meaning that it has the same number of rows and columns.
- The diagonal elements (from the top left to the bottom right) are all ones, while all other elements are zeros.
Other exercises in this chapter
Problem 63
Solve each system graphically. Check your solutions. Do not use a calculator. $$\begin{aligned}&x^{2}+y^{2}=5\\\&x+y=3\end{aligned}$$
View solution Problem 64
Graph the solution set of each system of inequalities by hand. $$\begin{array}{l}-2
View solution Problem 64
Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to complete the solution. $$\begin{aligned}x+9 y &=-15 \\\3 x+2 y &=5\end{al
View solution Problem 64
Find each matrix product if possible. $$\left[\begin{array}{rrr}-1 & 2 & 0 \\ 0 & 3 & 2 \\ 0 & 1 & 4\end{array}\right]\left[\begin{array}{rrr}2 & -1 & 2 \\ 0 &
View solution