Problem 73
Question
Let \(A\) be a square matrix all of whose entries are integers. Then which one of the following is true? [2008] (a) If det \(A=\pm 1\), then \(A^{-1}\) exists but all its entries are not necessarily integers (b) If det \(A \neq \pm 1\), then \(A^{-1}\) exists and all its entries are non integers (c) If det \(A=\pm 1\), then \(A^{-1}\) exists but all its entries are integers (d) If det \(A=\pm 1\), then \(A^{-1}\) need not exists
Step-by-Step Solution
Verified Answer
Option c is true: If det \( A = \pm 1 \), then \( A^{-1} \) exists and all its entries are integers.
1Step 1: Understanding the Determinant Condition
The determinant condition, \det A = \pm 1\, is crucial in determining whether the inverse exists and the nature of its entries. A determinant of \( \pm 1 \) indicates that matrix \( A \) is unimodular, which has specific implications for its inverse.
2Step 2: Determining the Existence of the Inverse
For square matrices, an inverse exists if and only if the determinant is not equal to zero. Here, since \det A = \pm 1\, the determinant is non-zero, indicating that \( A^{-1} \) does indeed exist.
3Step 3: Analyzing Integer Property of the Inverse
If \( A \) is a square matrix with integer entries and determinant \( \pm 1 \), then \( A \) is invertible with \( A^{-1} \) also having integer entries due to the property of unimodular matrices. This is because unimodular matrices are both integer matrices themselves and have integer inverses.
4Step 4: Conclusion Based on the Options
Given that \( A^{-1} \) exists and has integer entries when the determinant is \( \pm 1 \), option \( c \) (If det \( A = \pm 1 \), then \( A^{-1} \) exists but all its entries are integers) is correct.
Key Concepts
Integer MatrixMatrix InverseUnimodular Matrix
Integer Matrix
An integer matrix is a type of matrix where all the elements are integers. This means each entry in the matrix is a whole number, without any fractions or decimals. For example, a matrix like
Integer matrices are significant in various mathematical contexts because they simplify calculations and have unique properties. One of these is that under certain conditions, such as having a determinant of 1 or -1, an integer matrix can have an inverse that is also an integer matrix. This feature is not shared by matrices with non-integer entries, making integer matrices particularly useful in applications where maintaining integer properties is essential.
- \( \begin{bmatrix} 1 & 2 \ -3 & 4 \end{bmatrix} \)
Integer matrices are significant in various mathematical contexts because they simplify calculations and have unique properties. One of these is that under certain conditions, such as having a determinant of 1 or -1, an integer matrix can have an inverse that is also an integer matrix. This feature is not shared by matrices with non-integer entries, making integer matrices particularly useful in applications where maintaining integer properties is essential.
Matrix Inverse
In linear algebra, a matrix inverse is like the reciprocal of a number. If a matrix \( A \) has an inverse, denoted as \( A^{-1} \), then:
Not all matrices have inverses. For a matrix to be invertible, its determinant must not be zero. When dealing with integer matrices, if the determinant equals \( \pm 1 \), not only does the inverse exist, but it is also an integer matrix. This is because the calculation for the inverse preserves integer property due to the determinant condition.
Finding the inverse of a matrix involves using the formula:
- \( A \times A^{-1} = I \)
- \( A^{-1} \times A = I \)
Not all matrices have inverses. For a matrix to be invertible, its determinant must not be zero. When dealing with integer matrices, if the determinant equals \( \pm 1 \), not only does the inverse exist, but it is also an integer matrix. This is because the calculation for the inverse preserves integer property due to the determinant condition.
Finding the inverse of a matrix involves using the formula:
- \( A^{-1} = \frac{1}{ ext{det}(A)} \, \text{adj}(A) \)
Unimodular Matrix
A unimodular matrix is a special type of integer matrix with a determinant of 1 or -1. These matrices have some very interesting properties. The most notable property is that their inverses are also integer matrices.
This makes unimodular matrices extremely useful in number theory and integer programming, as they help maintain integer solutions across transformations. By containing all integer entries and possessing integer inverses, unimodular matrices maintain a "closed loop" of sorts within the integer domain.
The concept of unimodularity is directly tied to the practical applications of integer matrices, particularly in algorithm design and optimization. When a matrix is unimodular, it has a stable structure which is reliable for calculations that require precision and adherence to integer values. This is why they are often used in fields where exactness is key.
This makes unimodular matrices extremely useful in number theory and integer programming, as they help maintain integer solutions across transformations. By containing all integer entries and possessing integer inverses, unimodular matrices maintain a "closed loop" of sorts within the integer domain.
The concept of unimodularity is directly tied to the practical applications of integer matrices, particularly in algorithm design and optimization. When a matrix is unimodular, it has a stable structure which is reliable for calculations that require precision and adherence to integer values. This is why they are often used in fields where exactness is key.
Other exercises in this chapter
Problem 71
Consider the following relation \(\mathrm{R}\) on the set of real square matrices of order \(3 . \quad[2011 \mathrm{RS}]\) \(R=\left\\{(A, B) \mid A=P^{-1} B P\
View solution Problem 72
Let A be a \(2 \times 2\) matrix Statement \(-1: \operatorname{adj}(\operatorname{adj} \mathrm{A})=\mathrm{A}\) Statement \(-\mathbf{2}: \mid\) adj \(\mathrm{A}
View solution Problem 74
If \(A^{2}-A+I=0\), then the inverse of \(A\) is (a) \(A+I\) (b) \(\bar{A}\) (c) \(A-I\) (d) \(I-A\)
View solution Problem 75
Let \(A=\left(\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right)\) and \(B=\left(\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2
View solution