Collection of Maths Problems

Matrices 2 x 2

Task number: 2657

• Variant 1

  \( \begin{pmatrix} 2 & 6 \\ 6 & -3 \\ \end{pmatrix} \)

• Hint

  Eigenvalues \(\lambda_i\) of the matrix \(A\) are the root of the charakteristic polynomial \(p_A(t)=|A-t I|\).

  Eigenvectors corersponding to \(\lambda_i\) satisfy \(A\mathbf x=\lambda_i\mathbf x\), i.e. the are the solutions of the homogeneous system \((A-\lambda_i I)\mathbf x=\mathbf 0\).

• Resolution

  \(\begin{vmatrix} 2-t & 6 \\ 6 & -3-t \\ \end{vmatrix} =t^2+t-42=(t-6)(t+7)\) hence \(\lambda_1=6\), \(\lambda_2=-7\)

  Eigenvectors are calculated from the associated systems of linear equations \[\begin{eqnarray*} -4x_1 + 6x_2 & = & 0\\ 6x_1 - 9x_2 & = & 0 \end{eqnarray*}\] with a solution \(\mathbf x=(x_1, x_2)^T=c\cdot(3, 2)^T\).

  For the second eigenvalue \(-7\) \[\begin{eqnarray*} 9x_1 + 6x_2 & = & 0\\ 6x_1 + 4x_2 & = & 0 \end{eqnarray*}\] we get \(\mathbf x=c\cdot(2, -3)^T\).

  Matrices of these systems are always singular. Observe that on matrices of order two it means that the second row is a scalar multiple of the first one.

• Variant 2

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

• Result

  \(\lambda_1=1+i\), \(\mathbf x=c\cdot(1, 1+i)^T\);
  \(\lambda_2=1-i\), \(\mathbf x=c\cdot(1, 1-i)^T\);

• Variant 3

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

• Result

  \(\lambda_1=6\), \(\mathbf x=c\cdot(1, 1)^T\);
  \(\lambda_2=-1\), \(\mathbf x=c\cdot(-5, 2)^T\);

• Variant 4

  \( \begin{pmatrix} 5 & 10 \\ 4 & -1 \\ \end{pmatrix} \)

• Result

  \(\lambda_1=9\), \(\mathbf x=c\cdot(5, 2)^T\);
  \(\lambda_2=-5\), \(\mathbf x=c\cdot(-1, 1)^T\);