/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 If \(M\) is a \(3 \times 3\) mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 7

If \(M\) is a \(3 \times 3\) matrix, where \(M^{\prime} M=I\) and \(\operatorname{det}(M)=1\), then \(\operatorname{det}(M-I)=\) (A) 1 (B) 0 (C) \(-1\) (D) None of these

Short Answer

Expert verified
The determinant \(\text{det}(M-I)\) is 0. (Option B)

Step by step solution

01

Understanding the Properties of Matrix M

We are given a matrix \(M\) such that \(M'M = I\). This indicates that \(M\) is an orthogonal matrix, since the transpose of \(M\) multiplied by \(M\) results in the identity matrix \(I\). Furthermore, we are told that \(\text{det}(M) = 1\). For orthogonal matrices, the determinant can be either \(1\) or \(-1\), but here it is specifically \(1\).
02

Calculating the Determinant \(\text{det}(M-I)\)

To find the determinant of \(M-I\), consider the properties of orthogonal matrices. Since \(M\) is orthogonal with determinant \(1\), it behaves like a rotation matrix. For a \(3 \times 3\) orthogonal matrix with determinant \(1\), the eigenvalues are typically \(1, \cos \theta + i \sin \theta, \cos \theta - i \sin \theta\). At least one of these eigenvalues is equal to \(1\).
03

Using Eigenvalues to Determine \(\text{det}(M-I)\)

Now, since one of the eigenvalues of \(M\) is \(1\), the eigenvalues of \(M - I\) will be \(1 - 1 = 0\), and the others will be \(\cos \theta + i \sin \theta - 1\) and \(\cos \theta - i \sin \theta - 1\). Because one eigenvalue of \(M - I\) is \(0\), this directly implies that the determinant of \(M-I\) is the product of the eigenvalues, which includes \(0\).
04

Conclusion

Since one of the eigenvalues of \(M-I\) is \(0\), the determinant \(\text{det}(M-I)\) must be \(0\). None of the other potential non-zero eigenvalues can affect this, as their product with \(0\) results in \(0\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Determinant
When we talk about the determinant of a matrix, like our matrix \(M\), we're looking at a particular value that gives us insights into certain properties of the matrix.
  • Interpretation: For a \(3 \times 3\) matrix, the determinant can indicate whether the matrix is invertible. If \(\operatorname{det}(M) eq 0\), the matrix is invertible. For orthogonal matrices such as \(M\), they're always invertible because their determinant is either 1 or -1.
  • Geometric Meaning: In the context of transformations, the determinant tells us about volume scaling. Specifically, a determinant of 1, like in our case, means the transformation preserves volume.
  • Calculation: Determining \(\operatorname{det}(M)\) involves specific formulas when using numerical entries of the matrix or leveraging properties like orthogonality, which are used extensively in our example given \(M'M = I\) and \(\operatorname{det}(M)=1\).
Eigenvalues
Eigenvalues play a crucial role in understanding the properties of matrices, including our orthogonal matrix \(M\). They are scalars that give us insights about the transformation effects of \(M\).
  • Definition: Eigenvalues are values \(\lambda\) such that there exists a non-zero vector \(v\) where \(Mv = \lambda v\). Here, \(\lambda\) describes how the transformation scales the vector \(v\).
  • Orthogonal Matrices: For orthogonal matrices with a determinant of 1, like \(M\), one eigenvalue is always \(1\), ensuring it preserves vector lengths. The other eigenvalues describe possible rotations, typically expressed in the form \(\cos \theta + i \sin \theta\).
  • Determinant and Eigenvalues: The product of the eigenvalues of a matrix equals its determinant. Therefore, for \(M-I\), one of the eigenvalues being 0 immediately indicates that \(\operatorname{det}(M-I) = 0\).
Rotation Matrix
A rotation matrix is a special kind of orthogonal matrix predominantly used in transformations that involve rotations in space.
  • Structure: For a \(3 \times 3\) rotation matrix, entries are often dependent on the angle of rotation. This structure ensures that the matrix can rotate a point around an axis without changing the point's distance from the origin.
  • Properties: They have a determinant of 1, and their eigenvalues typically include 1, \(\cos \theta + i \sin \theta\), and \(\cos \theta - i \sin \theta\). These eigenvalues capture the essence of the rotation, especially for axes not aligned with the axis of rotation.
  • Application in \(M\): In our matrix \(M\), the rotation matrix behavior contributes to understanding \(\operatorname{det}(M-I)\). Since \(M\) is orthogonal and behaves like a rotation matrix, at least one eigenvalue of \(M\) is 1, leading to the determination that \(\operatorname{det}(M-I) = 0\).

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If \(P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) and \(Q=P A P^{\prime}\), then \(p^{\prime} Q^{2005} P\) is (A) \(\left[\begin{array}{cc}1 & 1 \\ 2005 & 1\end{array}\right]\) (B) \(\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]\) (C) \(\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\) (D) \(\left[\begin{array}{cc}1 & 2005 \\ 2005 & 1\end{array}\right]\)

If \(f_{j}=\sum_{i=0}^{2} a_{i j} x^{i}, j=1,2,3\) and if \(f_{j}^{\prime}, f_{j}^{\prime \prime}\) denote \(\frac{d f_{j}}{d x}, \frac{d^{2} f_{j}}{d x^{2}}\) respectively, then \(g(x)=\left|\begin{array}{lll}f_{1} & f_{2} & f_{3} \\\ f_{1}^{\prime} & f_{2}^{\prime} & f_{3}^{\prime} \\ f_{1}^{\prime \prime} & f_{2}^{\prime \prime} & f_{3}^{\prime \prime}\end{array}\right|\) is (A) a cubic in \(x\) (B) a quadratic in \(x\) (C) linear in \(x\) (D) a constant

The values of \(m\) for which the system of equations \(3 x+m y=m\) and \(2 x-5 y=20\) has a solution satisfying the condition \(x>0, y>0\), are (A) \(m \in\left(-\infty, \frac{-15}{2}\right) \cup(0, \infty)\) (B) \(m \in\left(-\infty, \frac{-15}{2}\right) \cup(30, \infty)\) (C) \(m \in\left(-\infty, \frac{-15}{2}\right) \cup(0,30)\) (D) None of these

If \(f(x)=\left|\begin{array}{ccc}(1+x)^{a} & (1+2 x)^{b} & 1 \\ 1 & (1+x)^{a} & (1+2 x)^{b} \\ (1+2 x)^{b} & 1 & (1+x)^{a}\end{array}\right|, a, b\) being positive integers, then (A) constant term of \(f(x)\) is 4 (B) coefficieent of \(x\) in \(f(x)\) is 0 (C) constant term in \(f(x)\) is \(a-b\) (D) constant term in \(f(x)\) is \(a+b\)

Let \(a, b, c\) be such that \(b(a+c) \neq 0\). If \(\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|\) \(+\left|\begin{array}{rrr}a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n+1} b & (-1)^{n} c\end{array}\right|=0\), then the value of ' \(n\) 'is (A) zero (B) any even integer (C) any odd integer (D) any integer
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.