Collection of Maths Problems

Matrices 3 x 3

Task number: 2658

• Variant 1

  \( \begin{pmatrix} 2 & -1 & 2 \\ 5 & -3 & 3 \\ -1 & 0 & -2 \\ \end{pmatrix} \)

• Hint

  A matrix is diagonalizable if and only if the spaces of eigenvectors have dimension equal to the algebraic multiplicity of the associated eigenvalue.

  This is satisfied e.g. when the matrix is of order \(n\) and has \(n\) distinct eigenvalues.

• Result

  \(\lambda_1=\lambda_2=\lambda_3=-1\), \(\mathbf x=c\cdot(1, 1, -1)^T\).

  The matrix is not diagonalizable.

• Variant 2

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

• Result

  \(\lambda_1=2\), \(\mathbf x=c\cdot(1, 0, 0)^T\); \(\lambda_2=1\), \(\mathbf x=c\cdot(1, 0, 1)^T\); \(\lambda_3=-1\), \(\mathbf x=c\cdot(0, 1, -1)^T\);

  The matrix is diagonalizable.

• Variant 3

  \( \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -4 \\ -1 & 0 & 4 \\ \end{pmatrix} \)

• Result

  \(\lambda_1=\lambda_2=3\), \(\mathbf x=c\cdot(1, -2, 1)^T\); \(\lambda_3=0\), \(\mathbf x=c\cdot(4, 4, 1)^T\);

  The matrix is not diagonalizable.