# Eigenvalues of a 3x3 matrix | Alternate coordinate systems (bases) | Linear Algebra | Khan Academy | Summary and Q&A

1.1M views
November 14, 2009
by
Eigenvalues of a 3x3 matrix | Alternate coordinate systems (bases) | Linear Algebra | Khan Academy

## TL;DR

This video explains how to find the eigenvalues of a 3x3 matrix by using the determinant of lambda times the identity matrix minus A.

## Install to Summarize YouTube Videos and Get Transcripts

### Q: What is the definition of an eigenvalue?

An eigenvalue is a scalar lambda that satisfies the equation A * v = lambda * v, where A is a matrix and v is a non-zero vector.

### Q: How can eigenvalues be found for a 3x3 matrix?

Eigenvalues can be found by solving the characteristic equation, which is the determinant of lambda times the identity matrix minus A equal to zero.

### Q: Why is it important to find the eigenvalues of a matrix?

Eigenvalues provide important information about a matrix, such as its determinant, trace, and eigenvectors, which can be used to solve systems of linear equations and analyze linear transformations.

### Q: How are the possible eigenvalues determined?

The possible eigenvalues are the values of lambda that make the determinant of lambda times the identity matrix minus A equal to zero.

## Summary & Key Takeaways

• Eigenvalues for a 3x3 matrix can be found by determining if the determinant of lambda times the identity matrix minus A is equal to zero.

• The characteristic polynomial for a 3x3 matrix is lambda - 3 * (lambda + 3) * (lambda - 3) = 0.

• The possible eigenvalues for the 3x3 matrix are lambda = 3 and lambda = -3.