Eigenvalue of a 3×3 Matrix Notes

Published on: September 29, 2026

This Eigenvalue of a 3×3 Matrix Calculator finds the eigenvalues of a \(3 \times 3\) matrix and shows the working line by line. Subtracting \(\lambda\) from the main diagonal and setting the determinant to zero gives \(\det(A - \lambda I) = 0\), which is a cubic in \(\lambda\) rather than a quadratic. Its roots are the eigenvalues, and a repeated root counts more than once. Following the same pattern every time keeps them checkable.

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Subtract lambda from the main diagonal.
  4. Expand the determinant into a polynomial.
  5. Solve the polynomial for the eigenvalues.

Formula:

\(\det(A - \lambda I) = 0\)
See Example 1 Hide Example 1

Problem

\(A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 4 & 9 \end{pmatrix}\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Subtract lambda from the main diagonal.
  4. Expand the determinant into a polynomial.
  5. Solve the polynomial for the eigenvalues.

Step 1

Step 1 - Set up the problem.

In this problem: We are finding the values of \(\lambda\) that make \(A - \lambda I\) collapse, for the matrix \(\begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 4 & 9 \end{pmatrix}\).

\(A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 4 \\ 0 & 4 & 9 \end{pmatrix}\)

Step 2

Step 2 - Write the formula.

In this problem: An eigenvalue is a number that leaves \(A - \lambda I\) with no inverse, and a matrix has no inverse exactly when its determinant is zero.

\(\det(A - \lambda I) = 0\)

Step 3

Step 3 - Subtract lambda from the main diagonal.

In this problem: Subtracting \(\lambda\) from each entry on the main diagonal gives \(\begin{pmatrix} 2 - \lambda & 0 & 0 \\ 0 & 3 - \lambda & 4 \\ 0 & 4 & 9 - \lambda \end{pmatrix}\).

\(A - \lambda I = \begin{pmatrix} 2 - \lambda & 0 & 0 \\ 0 & 3 - \lambda & 4 \\ 0 & 4 & 9 - \lambda \end{pmatrix}\)

Step 4

Step 4 - Expand the determinant into a polynomial.

In this problem: Taking the determinant of that matrix turns it into a polynomial in \(\lambda\), giving \(- \lambda^{3} + 14 \lambda^{2} - 35 \lambda + 22\).

\(\det(A - \lambda I) = - \lambda^{3} + 14 \lambda^{2} - 35 \lambda + 22\)

Step 5

Step 5 - Solve the polynomial for the eigenvalues.

In this problem: Setting that polynomial to zero and solving gives \(\lambda = 1\), \(\lambda = 2\), \(\lambda = 11\).

\(\lambda = 1,\; \lambda = 2,\; \lambda = 11\)

Final Answer

\(\lambda = 1,\; \lambda = 2,\; \lambda = 11\)
See Example 2 Hide Example 2

Problem

\(A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & 2 \end{pmatrix}\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Subtract lambda from the main diagonal.
  4. Expand the determinant into a polynomial.
  5. Solve the polynomial for the eigenvalues.

Step 1

Step 1 - Set up the problem.

In this problem: We are finding the values of \(\lambda\) that make \(A - \lambda I\) collapse, for the matrix \(\begin{pmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & 2 \end{pmatrix}\).

\(A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & 2 \end{pmatrix}\)

Step 2

Step 2 - Write the formula.

In this problem: An eigenvalue is a number that leaves \(A - \lambda I\) with no inverse, and a matrix has no inverse exactly when its determinant is zero.

\(\det(A - \lambda I) = 0\)

Step 3

Step 3 - Subtract lambda from the main diagonal.

In this problem: Subtracting \(\lambda\) from each entry on the main diagonal gives \(\begin{pmatrix} 1 - \lambda & 2 & 0 \\ 0 & 3 - \lambda & 0 \\ 2 & -4 & 2 - \lambda \end{pmatrix}\).

\(A - \lambda I = \begin{pmatrix} 1 - \lambda & 2 & 0 \\ 0 & 3 - \lambda & 0 \\ 2 & -4 & 2 - \lambda \end{pmatrix}\)

Step 4

Step 4 - Expand the determinant into a polynomial.

In this problem: Taking the determinant of that matrix turns it into a polynomial in \(\lambda\), giving \(- \lambda^{3} + 6 \lambda^{2} - 11 \lambda + 6\).

\(\det(A - \lambda I) = - \lambda^{3} + 6 \lambda^{2} - 11 \lambda + 6\)

Step 5

Step 5 - Solve the polynomial for the eigenvalues.

In this problem: Setting that polynomial to zero and solving gives \(\lambda = 1\), \(\lambda = 2\), \(\lambda = 3\).

\(\lambda = 1,\; \lambda = 2,\; \lambda = 3\)

Final Answer

\(\lambda = 1,\; \lambda = 2,\; \lambda = 3\)
See Example 3 Hide Example 3

Problem

\(A = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 1 \end{pmatrix}\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Subtract lambda from the main diagonal.
  4. Expand the determinant into a polynomial.
  5. Solve the polynomial for the eigenvalues.

Step 1

Step 1 - Set up the problem.

In this problem: We are finding the values of \(\lambda\) that make \(A - \lambda I\) collapse, for the matrix \(\begin{pmatrix} 3 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 1 \end{pmatrix}\).

\(A = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 1 \end{pmatrix}\)

Step 2

Step 2 - Write the formula.

In this problem: An eigenvalue is a number that leaves \(A - \lambda I\) with no inverse, and a matrix has no inverse exactly when its determinant is zero.

\(\det(A - \lambda I) = 0\)

Step 3

Step 3 - Subtract lambda from the main diagonal.

In this problem: Subtracting \(\lambda\) from each entry on the main diagonal gives \(\begin{pmatrix} 3 - \lambda & 0 & 0 \\ 0 & 1 - \lambda & 2 \\ 0 & 2 & 1 - \lambda \end{pmatrix}\).

\(A - \lambda I = \begin{pmatrix} 3 - \lambda & 0 & 0 \\ 0 & 1 - \lambda & 2 \\ 0 & 2 & 1 - \lambda \end{pmatrix}\)

Step 4

Step 4 - Expand the determinant into a polynomial.

In this problem: Taking the determinant of that matrix turns it into a polynomial in \(\lambda\), giving \(- \lambda^{3} + 5 \lambda^{2} - 3 \lambda - 9\).

\(\det(A - \lambda I) = - \lambda^{3} + 5 \lambda^{2} - 3 \lambda - 9\)

Step 5

Step 5 - Solve the polynomial for the eigenvalues.

In this problem: Setting that polynomial to zero and solving gives \(\lambda = -1\), \(\lambda = 3\), \(\lambda = 3\).

\(\lambda = -1,\; \lambda = 3,\; \lambda = 3\)

Final Answer

\(\lambda = -1,\; \lambda = 3,\; \lambda = 3\)
See Example 4 Hide Example 4

Problem

\(A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 5 \end{pmatrix}\)

Approach

Step-by-step method

  1. Set up the problem.
  2. Write the formula.
  3. Subtract lambda from the main diagonal.
  4. Expand the determinant into a polynomial.
  5. Solve the polynomial for the eigenvalues.

Step 1

Step 1 - Set up the problem.

In this problem: We are finding the values of \(\lambda\) that make \(A - \lambda I\) collapse, for the matrix \(\begin{pmatrix} 2 & 1 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 5 \end{pmatrix}\).

\(A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 5 \end{pmatrix}\)

Step 2

Step 2 - Write the formula.

In this problem: An eigenvalue is a number that leaves \(A - \lambda I\) with no inverse, and a matrix has no inverse exactly when its determinant is zero.

\(\det(A - \lambda I) = 0\)

Step 3

Step 3 - Subtract lambda from the main diagonal.

In this problem: Subtracting \(\lambda\) from each entry on the main diagonal gives \(\begin{pmatrix} 2 - \lambda & 1 & 0 \\ 0 & 4 - \lambda & 0 \\ 0 & 0 & 5 - \lambda \end{pmatrix}\).

\(A - \lambda I = \begin{pmatrix} 2 - \lambda & 1 & 0 \\ 0 & 4 - \lambda & 0 \\ 0 & 0 & 5 - \lambda \end{pmatrix}\)

Step 4

Step 4 - Expand the determinant into a polynomial.

In this problem: Taking the determinant of that matrix turns it into a polynomial in \(\lambda\), giving \(- \lambda^{3} + 11 \lambda^{2} - 38 \lambda + 40\).

\(\det(A - \lambda I) = - \lambda^{3} + 11 \lambda^{2} - 38 \lambda + 40\)

Step 5

Step 5 - Solve the polynomial for the eigenvalues.

In this problem: Setting that polynomial to zero and solving gives \(\lambda = 2\), \(\lambda = 4\), \(\lambda = 5\).

\(\lambda = 2,\; \lambda = 4,\; \lambda = 5\)

Final Answer

\(\lambda = 2,\; \lambda = 4,\; \lambda = 5\)