Collection of Maths Problems

Inverses

Task number: 2425

• Variant 1

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

• Resolution

  We build \((\mathbf A|\mathbf I)\) and by elementary transformations we change it into the form \((\mathbf I|\mathbf A^{-1})\).

  In our case

  \( \begin{pmatrix} \begin{array}{cccc|cccc} 1 & 0 & 1 & 1 & 1 & 0 & 0 & 0 \\ 2 & 0 & 1 & 1 & 0 & 1 & 0 & 0 \\ 2 & 1 & 0 & 0 & 0 & 0 & 1 & 0 \\ 1 & 2 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \end{pmatrix}\sim \begin{pmatrix} \begin{array}{cccc|cccc} 1 & 0 & 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 &-1 &-1 &-2 & 1 & 0 & 0 \\ 0 & 1 &-2 &-2 &-2 & 0 & 1 & 0 \\ 0 & 2 & 0 &-1 &-1 & 0 & 0 & 1 \\ \end{array} \end{pmatrix}\sim\\ \)
  \(\sim \begin{pmatrix} \begin{array}{cccc|cccc} 1 & 0 & 0 & 0 &-1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 2 &-2 & 1 & 0 \\ 0 & 0 & 1 & 1 & 2 &-1 & 0 & 0 \\ 0 & 2 & 0 &-1 &-1 & 0 & 0 & 1 \\ \end{array} \end{pmatrix}\sim \begin{pmatrix} \begin{array}{cccc|cccc} 1 & 0 & 0 & 0 &-1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 2 &-2 & 1 & 0 \\ 0 & 0 & 1 & 0 &-3 & 3 &-2 & 1 \\ 0 & 0 & 0 & 1 & 5 &-4 & 2 &-1 \\ \end{array} \end{pmatrix}\)

• Result

  \(\mathbf A^{-1} = \begin{pmatrix} -1 & 1 & 0 & 0 \\ 2 & -2 & 1 & 0 \\ -3 & 3 & -2 & 1 \\ 5 & -4 & 2 & -1 \\ \end{pmatrix}\).

• Variant 2

  \( \mathbf B= \begin{pmatrix} 0 & 2 & 2 & 1 \\ 1 & 0 & 2 & 0 \\ 2 & 1 & 0 & 2 \\ 2 & 2 & 1 & 1 \\ \end{pmatrix} \).

• Result

  \(\mathbf B^{-1} = \begin{pmatrix} -2/5 & 1/5 & 0 & 2/5 \\ 2/15 & -2/5 & -1/3 & 8/15 \\ 1/5 & 2/5 & 0 & -1/5 \\ 1/3 & 0 & 2/3 & -2/3 \\ \end{pmatrix}\).

• Variant 3

  \( \mathbf C= \begin{pmatrix} 2 & 0 & 1 & 0 \\ 1 & 2 & 0 & 1 \\ 2 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 \end{pmatrix} \).

• Result

  \(\mathbf C^{-1} = \begin{pmatrix} 3/2 & 1/2 & -1 & -1/2 \\ -1 & 0 & 1 & 0 \\ -2 & -1 & 2 & 1 \\ 1/2 & 1/2 & -1 & 1/2 \\ \end{pmatrix}\).

• Variant 4

  \( \mathbf D= \begin{pmatrix} 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 1 \\ 1 & 1 & 2 & 0 \\ 0 & 1 & 2 & 1 \\ \end{pmatrix} \).

• Result

  \(\mathbf D^{-1}\) does not exists, the matrix is singular.

• Variant 5

  \( \mathbf E= \begin{pmatrix} 1 & 1 & 2 & 0 \\ 1 & 2 & 1 & 1 \\ 0 & 1 & 2 & 1 \\ 1 & 2 & 0 & 0 \\ \end{pmatrix} \).

• Result

  \(\mathbf E^{-1} = \begin{pmatrix} 2/3 & 4/3 & -4/3 & -1 \\ -1/3 & -2/3 & 2/3 & 1 \\ 1/3 & -1/3 & 1/3 & 0 \\ -1/3 & 4/3 & -1/3 & -1 \\ \end{pmatrix}\).