Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Form matrix \(B\) from a matrix \(A\) by writing the columns of \(A\) in reverse order. Express det \(B\) in terms of \(\operatorname{det} A\).

Short Answer

Expert verified
det \(B = (-1)^{n-1} \times \operatorname{det} A\) where \(n\) is the number of columns.

Step by step solution

01

Understand the problem

The task is to reverse the order of the columns in matrix \(A\) to form matrix \(B\), and then express the determinant of \(B\) in terms of the determinant of \(A\).
02

Conceptualize reversing columns

Reversing the columns of matrix \(A\) means that if \(A = [C_1, C_2, \ldots, C_n]\), where \(C_i\) are columns of \(A\), then \(B = [C_n, C_{n-1}, \ldots, C_1]\).
03

Understand the effect on determinants

Reversing the columns of a matrix can be seen as applying a sequence of column swaps. Each pair of column swaps multiplies the determinant by \(-1\). If there are \(n\) columns, reversing them entails \(n-1\) swaps.
04

Calculate the sign change

Reversing \(n\) columns entirely takes \(n/2\) (rounded down) swaps, since each pair reverses a position. Therefore, the sign of the determinant changes \(n-1\) times: \[\operatorname{det} B = (-1)^{n-1} \cdot \operatorname{det} A\]
05

Conclusion

Based on the number of columns, the determinant of matrix \(B\) is either \(\operatorname{det} A\) if \(n\) is odd, or \(-\operatorname{det} A\) if \(n\) is even: \[\operatorname{det} B = (-1)^{n-1} \cdot \operatorname{det} A\].

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 Vaia!

Key Concepts

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

Column Reversal
When transforming a matrix by reversing its columns, you might wonder what actually happens. Visualization can help here. Imagine you have a matrix \( A \) with columns \([C_1, C_2, \ldots, C_n]\). Reversing the columns means that the order is flipped to become \([C_n, C_{n-1}, \ldots, C_1]\).
Your matrix starts with the first column on the left, and reversing the columns switches the position of every column until the last is first and the first is last. This operation directly impacts how the data in your matrix is visually and structurally represented.
Understanding column reversal is crucial, as this action lays the groundwork for further operations, particularly when dealing with determinants. This operation by itself doesn't change the content of each individual column, just the order in which they appear.
Matrix Operations
Matrix operations are fundamental actions that can transform matrices in various ways, such as addition, subtraction, multiplication, and transposition. Each operation has unique rules and consequences. However, in this case, we are focusing on the effect of column reversal as an operation.
  • **Column Swap**: Reversing columns involves a series of column swaps. Each swap swaps two neighboring columns, temporarily flipping their positions.
  • **Multiplication by -1**: An important property of the determinant is that swapping two columns multiplies the determinant by \(-1\). Thus, reversing all columns in a matrix relates back to determinant properties.
Through these swaps, you're essentially performing a more complex matrix operation, with each swap tweaking the matrix's determinant. It's essential to understand these operations to appreciate how a simple idea, like reordering columns, affects the determinant's magnitude.
Determinant Properties
Determinants have several innate properties that are useful when performing matrix operations. A key feature is how column swaps affect the determinant's value.
Each time you swap two columns in a matrix, the determinant is multiplied by \(-1\). If a matrix has \( n \) columns, fully reversing the column order involves making multiple swaps. Specifically, you would need to perform \( n-1 \) swaps to reverse an \( n \)-column matrix, resulting in the calculations for determinant changes:
\[ \operatorname{det}(B) = (-1)^{n-1} \cdot \operatorname{det}(A) \]
Interestingly, the final sign of the determinant of the reversed matrix, \( B \), depends on \( n \):
  • If \( n \) is odd, there's an even number of swaps making \((-1)^{n-1}\) positive, so \( \operatorname{det}(B) = \operatorname{det}(A) \).
  • If \( n \) is even, there's an odd number of swaps making \((-1)^{n-1}\) negative, so \( \operatorname{det}(B) = -\operatorname{det}(A) \).
These determinant properties are critical in linear algebra, showing how simple operations like column reversal can affect a matrix's determinants in predictable ways.

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

Show that $$ \text { det }\left[\begin{array}{ccccc} 0 & 0 & \cdots & 0 & a_{1} \\ 0 & 0 & \cdots & a_{2} & 0 \\ \vdots & \vdots & & \vdots & \vdots \\ 0 & a_{n-1} & \cdots & * & * \\ a_{n} & \+ & \cdots & \+ & + \end{array}\right]=(-1)^{k} a_{1} a_{2} \cdots a_{n} $$ where either \(n=2 k\) or \(n=2 k+1,\) and \(+\) -entries are artitrary.

Let \(A^{2}=I,\) and assume that \(A \neq I\) and \(A \neq-I\) a. Show that the only eigenvalues of \(A\) are \(\lambda=1\) and \(\lambda=-1\) b. Show that \(A\) is diagonalizable. [Hint: Verify that \(A(A+I)=A+I\) and \(A(A-I)=-(A-I)\), and then look at nonzero columns of \(A+I\) and of \(A-I .]\) c. If \(Q_{m}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) is reflection in the line \(y=m x\) where \(m \neq 0\), use (b) to show that the matrix of \(Q_{m}\) is diagonalizable for each \(m\). d. Now prove (c) geometrically using Theorem 3.3.3.

Let \(A\) be any \(n \times n\) matrix and \(r \neq 0\) a real number. a. Show that the eigenvalues of \(r A\) are precisely the numbers \(r \lambda,\) where \(\lambda\) is an eigenvalue of \(A\). b. Show that \(c_{r A}(x)=r^{n} c_{A}\left(\frac{x}{r}\right)\).

Evaluate the determinant of each matrix by reducing it to upper triangular form. $$ \begin{array}{l} \text { a. }\left[\begin{array}{rrr} 1 & -1 & 2 \\ 3 & 1 & 1 \\ 2 & -1 & 3 \end{array}\right] & \text { b. }\left[\begin{array}{rrr} -1 & 3 & 1 \\ 2 & 5 & 3 \\ 1 & -2 & 1 \end{array}\right] \\ \text { c. }\left[\begin{array}{rrrr} -1 & -1 & 1 & 0 \\ 2 & 1 & 1 & 3 \\ 0 & 1 & 1 & 2 \\ 1 & 3 & -1 & 2 \end{array}\right] & \text { d. }\left[\begin{array}{rrrr} 2 & 3 & 1 & 1 \\ 0 & 2 & -1 & 3 \\ 0 & 5 & 1 & 1 \\ 1 & 1 & 2 & 5 \end{array}\right] \end{array} $$

If \(P^{-1} A P\) and \(P^{-1} B P\) are both diagonal, show that \(A B=B A .\) [Hint: Diagonal matrices commute.]

See all solutions

Recommended explanations on Economics Textbooks

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.

Sign-up for free