Collection of Maths Problems

Adjugate matrix and inverse

Task number: 2606

• Variant 1

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

• Hint

  Adjugate matrix: \((\mbox{adj}(\mathbf A))_{i,j}=(-1)^{i+j}|\mathbf A^{j,i}|\), where \(\mathbf A^{j,i}\) is the matrix obtained by the deletion of the \(j\)-th row and the \(i\)-th column from the matrix \(\mathbf A\) (note the exchange of the indices).

  The inverse is then given by: \(\mathbf A^{-1}=\frac{1}{|\mathbf A|}\mbox{adj}(\mathbf A)\).

• Answer

  Over \(\mathbb R\)

  \(|\mathbf A|=-1\), \( \mbox{adj}(\mathbf A)= \begin{pmatrix} 1 & 1 & -1 \\ 2 & 1 & -2 \\ -2 &-1 & 1 \\ \end{pmatrix} \), \(\mathbf A^{-1}= \begin{pmatrix} -1 & -1 & 1 \\ -2 & -1 & 2 \\ 2 & 1 & -1 \\ \end{pmatrix} \)

  Over \(\mathbb Z_5\)

  \(|\mathbf A|=-1 = 4\), \( \mbox{adj}(\mathbf A)= \begin{pmatrix} 1 & 1 & 4 \\ 2 & 1 & 3 \\ 3 & 4 & 1 \\ \end{pmatrix} \), \(\mathbf A^{-1}= \begin{pmatrix} 4 & 4 & 1 \\ 3 & 4 & 2 \\ 2 & 1 & 4 \\ \end{pmatrix} \)

• Variant 2

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

• Answer

  Over \(\mathbb R\)

  \(|\mathbf A|=4\), \( \mbox{adj}(\mathbf A)= \begin{pmatrix} 2 & 2 & -4 \\ -1 & 1 & 2 \\ 1 & -1 & 2 \\ \end{pmatrix} \), \(\mathbf A^{-1}= \begin{pmatrix} 1/2 & 1/2 & -1 \\ -1/4 & 1/4 & 1/2 \\ 1/4 & -1/4 & 1/2 \\ \end{pmatrix} \)

  Over \(\mathbb Z_5\)

  \(|\mathbf A|=4\) , \( \mbox{adj}(\mathbf A)= \begin{pmatrix} 2 & 2 & 1 \\ 4 & 1 & 2 \\ 1 & 4 & 2 \\ \end{pmatrix} \), \(\mathbf A^{-1}= \begin{pmatrix} 3 & 3 & 4 \\ 1 & 4 & 3 \\ 4 & 1 & 3 \\ \end{pmatrix} \)

• Variant 3

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

• Answer

  Over \(\mathbb R\)

  \(|\mathbf A|=-5\), \( \mbox{adj}(\mathbf A)= \begin{pmatrix} 1 & -2 & -2 \\ -1 & -3 & 7 \\ -1 & 2 & -3 \\ \end{pmatrix} \), \(\mathbf A^{-1}= \begin{pmatrix} -1/5 & 2/5 & 2/5 \\ 1/5 & 3/5 &-7/5 \\ 1/5 &-2/5 & 3/5 \\ \end{pmatrix} \)

  Over \(\mathbb Z_5\)

  \(|\mathbf A|= 0\), hence \(\mathbf A\) is singular over \(\mathbb Z_5\) and \(\mathbf A^{-1}\) does not exist, \( \mbox{adj}(\mathbf A)= \begin{pmatrix} 1 & 3 & 3 \\ 4 & 2 & 2 \\ 4 & 2 & 2 \\ \end{pmatrix} \).