Problem 59
Question
Prove each. If \(A\) and \(B\) are two invertible matrices of order \(n,\) then \((A B)^{-1}=B^{-1} A^{-1} .\)
Step-by-Step Solution
Verified Answer
Let A and B be two invertible matrices of order n. To prove (AB)^(-1) = B^(-1)A^(-1), we multiply AB by B^(-1)A^(-1):
1. \((AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AA^{-1} = I\)
Similarly, we multiply B^(-1)A^(-1) by AB:
2. \((B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I\)
Since both products result in the identity matrix, we can conclude that (AB)^(-1) = B^(-1)A^(-1).
1Step 1: State the given information
Let A and B be two invertible matrices of order n.
2Step 2: Define inverse of a matrix
Recall that a matrix M is invertible if there exists a matrix M^(-1) such that MM^(-1)=M^(-1)M=I, where I is the identity matrix of the same order as M.
3Step 3: Show that (AB)(B^(-1)A^(-1)) is the identity matrix
We need to prove that (AB)^(-1)=B^(-1)A^(-1). To do this, we will multiply AB by B^(-1)A^(-1). Note that we are allowed to do this because matrix multiplication is associative. So, we get:
\( (AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1}\)
Since B and its inverse B^(-1) multiply to the identity matrix, we can simplify the equation:
\(A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1}\)
Now, using the properties of the inverse matrix, we have:
\( AA^{-1} = I\)
4Step 4: Show that (B^(-1)A^(-1))(AB) is the identity matrix
Similarly, we can multiply B^(-1)A^(-1) by AB and simplify:
\( (B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B \)
Since B and its inverse B^(-1) multiply to the identity matrix, we get:
\( B^{-1}B = I\)
5Step 5: Conclusion
Since both (AB)(B^(-1)A^(-1)) and (B^(-1)A^(-1))(AB) result in the identity matrix, we can conclude that (AB)^(-1) = B^(-1)A^(-1).
Key Concepts
Invertible MatricesMatrix MultiplicationIdentity Matrix
Invertible Matrices
Invertible matrices are special types of matrices that possess an inverse. The concept is similar to the inverse of a number, such as how multiplying any number by its inverse results in 1. For a matrix, the inverse is another matrix that, when multiplied together, yields the identity matrix. This property makes invertible matrices extremely important in linear algebra.
To dive a bit deeper:
To dive a bit deeper:
- A matrix is said to be invertible or non-singular if it has a square shape, meaning the same number of rows and columns.
- If you have a matrix \(M\), its inverse is often denoted as \(M^{-1}\).
- The criterion for \(M\) being invertible is that \(MM^{-1} = I\), where \(I\) is the identity matrix.
Matrix Multiplication
Matrix multiplication is an operation that takes two matrices and produces another matrix. Unlike adding or subtracting matrices, matrix multiplication involves combining rows and columns in a specific way, which even allows for the mixing of matrices of different dimensions under certain rules. Understanding how to properly multiply matrices is a key skill in mathematics:
- The number of columns in the first matrix must equal the number of rows in the second matrix to perform multiplication.
- If you have matrices \(A\) and \(B\), and you multiply them to get \(C = AB\), \(C\) usually has different values than \(BA\) unless they are special kinds of matrices.
- Matrix multiplication is associative, so \(A(BC) = (AB)C\), which is a versatile property in matrix operations.
Identity Matrix
The identity matrix acts like the number 1 in matrix operations. It has a property known as multiplicative identity, which means that when any matrix is multiplied by an identity matrix, it remains unchanged. This makes the identity matrix a crucial component in the study of matrices:
- The identity matrix is always square-shaped, having the same number of rows and columns.
- Along its main diagonal, from the top left to the bottom right, the components are 1’s, and all other components are zero.
- In notation, an identity matrix is represented by \(I\). For example, a 2x2 identity matrix looks like \(\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\).
Other exercises in this chapter
Problem 58
If \(A\) is an invertible matrix, then \(\left(A^{-1}\right)^{-1}=A\).
View solution Problem 59
Prove each, where \(x \in \mathbb{R}\) and \(n \in \mathbf{Z}.\) $$\lceil x\rceil=-\lfloor- x\rfloor$$
View solution Problem 59
Prove each, where \(x \in \mathbb{R}\) and \(n \in \mathbf{Z}\) $$\lceil x\rceil=-\lfloor-x\rfloor$$
View solution Problem 59
If \(A\) and \(B\) are two invertible matrices of order \(n,\) then \((A B)^{-1}=\) \(B^{-1} A^{-1}\).
View solution