Problem 57
Question
The inverse of a square matrix \(A\) is unique. (Hint: Assume \(A\) has two inverses \(B\) and \(C .\) Show that \(B=C .\) )
Step-by-Step Solution
Verified Answer
To prove that the inverse of a square matrix A is unique, assume A has two inverses B and C. Using the definition of matrix inverses and the properties of matrix multiplication, we can show that B = C:
1. AB = I and AC = I
2. Multiply the equation AC = I on the left side by B: B(AC) = BI
3. Apply the associative property of matrix multiplication: (BA)C = BI
4. Substitute the inverse property BA = I into the equation: IC = BI
5. Use the identity matrix property: C = B
Thus, B = C, proving the inverse of a square matrix A is unique.
1Step 1: Define the problem and given information
We are given the following information:
1. A is a square matrix.
2. A has two inverses, B and C.
Our goal is to show that B = C.
2Step 2: Use the definition of matrix inverses
By the definition of matrix inverses, we know that the product of a matrix and its inverse is the identity matrix. So, for A, its inverses B and C, we have:
1. AB = BA = I
2. AC = CA = I
3Step 3: Multiply A and B to get the identity matrix
We know that A has an inverse B, which means that the product of A and B results in the identity matrix. Mathematically, this can be written as:
AB = I
4Step 4: Multiply A and C to get the identity matrix
Similarly, we know that A has another inverse, C, which also means that the product of A and C results in the identity matrix. Mathematically, this is represented as:
AC = I
5Step 5: Use the results from Step 3 and 4 to show that B = C
Now, let's use these two equalities to show that B must be equal to C.
Multiply the equation AC = I on the left side by B. Here's what we get:
B(AC) = BI
6Step 6: Apply the associative property of matrix multiplication
Now we can apply the associative property of matrix multiplication (which states that (AB)C = A(BC)), to the left side of the equation B(AC) = BI:
(BA)C = BI
7Step 7: Use the inverse property from Step 3
Recall that BA = I from Step 3. Now, let's substitute I for BA in the equation from Step 6:
IC = BI
8Step 8: Use the identity matrix property
Since multiplying any matrix by the identity matrix results in the original matrix, we can simplify the equality:
C = B
This shows that B = C, which means the inverses of a square matrix A are unique. So, the inverse of a square matrix A is unique.
Key Concepts
Square MatrixIdentity MatrixAssociative Property of Matrix Multiplication
Square Matrix
A square matrix is a type of matrix that has the same number of rows and columns. For example, a matrix with dimensions 2x2 or 3x3 is considered a square matrix. This is an essential feature when discussing matrix inverses because only square matrices can have inverses.
Think of a square matrix as a special type of grid where the number of vertical lines matches the number of horizontal lines. This balance is crucial for performing many advanced matrix operations.
Think of a square matrix as a special type of grid where the number of vertical lines matches the number of horizontal lines. This balance is crucial for performing many advanced matrix operations.
- Only square matrices are eligible for determining inverses because non-square matrices do not have a determinant, which is a key element in calculating an inverse.
- In linear algebra, square matrices often represent systems that can model real-world phenomena, such as transformations and rotations.
Identity Matrix
An identity matrix is a special type of square matrix where all elements on the main diagonal are ones, and all other elements are zeros. The main diagonal is the line that stretches from the top left to the bottom right of the matrix.
The identity matrix functions much like the number one in arithmetic operations. When you multiply any matrix by the identity matrix, it remains unchanged.
The identity matrix functions much like the number one in arithmetic operations. When you multiply any matrix by the identity matrix, it remains unchanged.
- The identity matrix is denoted as 'I' and is pivotal when discussing matrix inverses, as the product of a matrix and its inverse is the identity matrix.
- For example, a 2x2 identity matrix looks like this: \[ I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \]
Associative Property of Matrix Multiplication
The associative property of matrix multiplication states that when you multiply three or more matrices, the result is the same regardless of how the matrices are grouped. In other words, the equation \[ (AB)C = A(BC) \]always holds true.
This property is crucial in the realm of matrices as it enables flexibility in computation and simplification.
This property is crucial in the realm of matrices as it enables flexibility in computation and simplification.
- This allows us to rearrange matrices in a multiplication chain without affecting the outcome, which simplifies calculations and proofs.
- For example, we used the associative property in solving the exercise by showing that \[ B(AC) = (BA)C \]and later simplified to \[ IC = BI \]using the identity matrix property.
Other exercises in this chapter
Problem 57
Let \(f: X \rightarrow Y\) be bijective. Let \(S\) and \(T\) be subsets of \(Y .\) Prove each. $$f^{-1}(S \cup T)=f^{-1}(S) \cup f^{-1}(T)$$
View solution Problem 57
Prove each, where \(x \in \mathbb{R}\) and \(n \in \mathbf{Z}\) $$\left\lfloor\frac{n}{2}\right\rfloor+\left\lceil\frac{n}{2}\right\rceil=n$$
View solution Problem 58
Prove each. If \(A\) is an invertible matrix, then \(\left(A^{-1}\right)^{-1}=A.\)
View solution Problem 58
Let \(f: X \rightarrow Y\) be bijective. Let \(S\) and \(T\) be subsets of \(Y .\) Prove each. $$f^{-1}(S \cap T)=f^{-1}(S) \cap f^{-1}(T)$$
View solution