Eigenvalue of a 3x3 Matrix Calculator

This Eigenvalue Calculator finds the eigenvalues of a 3×3 matrix. It forms the characteristic equation det(A − λI) = 0, expands it into a cubic in λ, and solves for the eigenvalues.

Step-by-step method

1. Write the 3x3 matrix A.
2. Write the characteristic equation det(A - λ I) = 0.
3. Expand the determinant into a cubic polynomial in λ.
4. Solve the polynomial for the eigenvalues.

Formula:

Example 1:

Step 1 - Write the 3x3 matrix A.

In this problem: We find the eigenvalues of the matrix \(A = \begin{pmatrix} 2 & 0 & 0 \\ 1 & 2 & 1 \\ -1 & 0 & 1 \end{pmatrix}\).

Step 2 - Write the characteristic equation det(A - λ I) = 0.

In this problem: Eigenvalues solve the characteristic equation det(A - \(\lambda\) I) = 0.

Step 3 - Expand the determinant into a cubic polynomial in λ.

In this problem: Expand the determinant of A minus lambda times the identity.

Step 4 - Solve the polynomial for the eigenvalues.

In this problem: Solving \(- \lambda^{3} + 5 \lambda^{2} - 8 \lambda + 4 = 0\) gives the eigenvalues \(\lambda = 1,\; \lambda = 2,\; \lambda = 2\).

Final answer:

Example 2:

Step 1 - Write the 3x3 matrix A.

In this problem: We find the eigenvalues of the matrix \(A = \begin{pmatrix} 2 & -3 & 1 \\ 1 & -2 & 1 \\ 1 & -3 & 2 \end{pmatrix}\).

Step 2 - Write the characteristic equation det(A - λ I) = 0.

In this problem: Eigenvalues solve the characteristic equation det(A - \(\lambda\) I) = 0.

Step 3 - Expand the determinant into a cubic polynomial in λ.

In this problem: Expand the determinant of A minus lambda times the identity.

Step 4 - Solve the polynomial for the eigenvalues.

In this problem: Solving \(- \lambda^{3} + 2 \lambda^{2} - \lambda = 0\) gives the eigenvalues \(\lambda = 0,\; \lambda = 1,\; \lambda = 1\).

Final answer:

See Example 1 Hide Example 1

Problem

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

Approach

Step-by-step method

1. Write the 3x3 matrix A.
2. Write the characteristic equation det(A - λ I) = 0.
3. Expand the determinant into a cubic polynomial in λ.
4. Solve the polynomial for the eigenvalues.

Step 1

Step 1 - Write the 3x3 matrix A.

In this problem: We find the eigenvalues of the matrix \(A = \begin{pmatrix} 2 & 0 & 0 \\ 1 & 2 & 1 \\ -1 & 0 & 1 \end{pmatrix}\).

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

Step 2

Step 2 - Write the characteristic equation det(A - λ I) = 0.

In this problem: Eigenvalues solve the characteristic equation det(A - \(\lambda\) I) = 0.

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

Step 3

Step 3 - Expand the determinant into a cubic polynomial in λ.

In this problem: Expand the determinant of A minus lambda times the identity.

\(\det(A - \lambda I) = - \lambda^{3} + 5 \lambda^{2} - 8 \lambda + 4\)

Step 4

Step 4 - Solve the polynomial for the eigenvalues.

In this problem: Solving \(- \lambda^{3} + 5 \lambda^{2} - 8 \lambda + 4 = 0\) gives the eigenvalues \(\lambda = 1,\; \lambda = 2,\; \lambda = 2\).

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

Final Answer

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See Example 2 Hide Example 2

Problem

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

Approach

Step-by-step method

1. Write the 3x3 matrix A.
2. Write the characteristic equation det(A - λ I) = 0.
3. Expand the determinant into a cubic polynomial in λ.
4. Solve the polynomial for the eigenvalues.

Step 1

Step 1 - Write the 3x3 matrix A.

In this problem: We find the eigenvalues of the matrix \(A = \begin{pmatrix} 2 & -3 & 1 \\ 1 & -2 & 1 \\ 1 & -3 & 2 \end{pmatrix}\).

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

Step 2

Step 2 - Write the characteristic equation det(A - λ I) = 0.

In this problem: Eigenvalues solve the characteristic equation det(A - \(\lambda\) I) = 0.

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

Step 3

Step 3 - Expand the determinant into a cubic polynomial in λ.

In this problem: Expand the determinant of A minus lambda times the identity.

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

Step 4

Step 4 - Solve the polynomial for the eigenvalues.

In this problem: Solving \(- \lambda^{3} + 2 \lambda^{2} - \lambda = 0\) gives the eigenvalues \(\lambda = 0,\; \lambda = 1,\; \lambda = 1\).

\(\lambda = 0,\; \lambda = 1,\; \lambda = 1\)

Final Answer

\(\)