Question

Question: Find the inverse of the given matrix.

16.$\left(\begin{array}{ccc}-2& 0& 1\\ 1& -1& 2\\ 3& 1& 0\end{array}\right)$

Short answer

The inverse of the given matrix$\left(\begin{array}{ccc}-2& 0& 1\\ 1& -1& 2\\ 3& 1& 0\end{array}\right)$is $\frac{1}{8}\left(\begin{array}{ccc}-2& 1& 1\\ 6& -3& 5\\ 4& 2& 2\end{array}\right)$.

Step-by-step solution

Definition of Inverse Matrix

The inverse of a matrix is another matrix that produces the multiplicative identity when multiplied by the given matrix. ${\text{AA}}^{\text{- 1}}={\text{A}}^{\text{- 1}}\text{A}=\text{I}$.

where I is the identity matrix, ${\text{A}}^{\text{- 1}}$is the inverse of a matrix A.

For matrix A, the inverse matrix formula is ; $A^{-1}=\frac{adj\left(A\right)}{\left|A\right|};\left|\text{A}\right|\ne \text{0}$, where A is a square matrix.

Given parameters

The given matrix is$\left(\begin{array}{ccc}-2& 0& 1\\ 1& -1& 2\\ 3& 1& 0\end{array}\right)$.

Find the inverse of the given matrix.

Find the inverse of the matrix

Find the determinant of the given matrix

$\begin{array}{rcl}\text{det(M)}& =& \left|\begin{array}{ccc}-2& 0& 1\\ 1& -1& 2\\ 3& 1& 0\end{array}\right|\\ & =& -\text{2[}-\text{1(0)}-\text{2(1)]}-\text{0[1(0)}-\text{2(3)]}+\text{1[1(1)}-\text{(}-\text{1)(3)]}\\ & =& \text{4}+\text{4}\\ & =& \text{8}\end{array}$

Find the minors of the given matrix

role="math" localid="1664188418143" $\begin{array}{rcl}{\text{M}}_{\text{11}}& =& \left|\begin{array}{cc}-1& 2\\ 1& 0\end{array}\right|\\ & =& -2\\ {\text{M}}_{12}& =& -\left|\begin{array}{cc}1& 2\\ 3& 0\end{array}\right|\\ & =& 6\end{array}$

Evaluate further minors

$\begin{array}{rcl}{\text{M}}_{\text{13}}& =& \left|\begin{array}{cc}1& -1\\ 3& 1\end{array}\right|\\ & =& 4\\ {\text{M}}_{21}& =& -\left|\begin{array}{cc}0& 1\\ 1& 0\end{array}\right|\\ & =& 1\end{array}$

Evaluate further minors

$\begin{array}{rcl}{\text{M}}_{22}& =& \left|\begin{array}{cc}-2& 1\\ 3& 0\end{array}\right|\\ & =& -3\\ {\text{M}}_{23}& =& -\left|\begin{array}{cc}-2& 0\\ 3& 1\end{array}\right|\\ & =& 2\end{array}$

Evaluate further minors

$\begin{array}{rcl}{\text{M}}_{31}& =& \left|\begin{array}{cc}0& 1\\ -1& 2\end{array}\right|\\ & =& 1\\ {\text{M}}_{32}& =& -\left|\begin{array}{cc}-2& 1\\ 1& 2\end{array}\right|\\ & =& 5\end{array}$

Evaluate further minors

$\begin{array}{rcl}{\text{M}}_{33}& =& \left|\begin{array}{cc}-2& 0\\ 1& -1\end{array}\right|\\ & =& 2\end{array}$

Cofactors:$\left(\begin{array}{ccc}-2& 6& 4\\ 1& -3& 2\\ 1& 5& 2\end{array}\right)$

Take the transpose of the cofactors of the given matrix to find the adjoint of the matrix.

role="math" localid="1664188708014" ${\left(\begin{array}{ccc}-2& 6& 4\\ 1& -3& 2\\ 1& 5& 2\end{array}\right)}^T=\left(\begin{array}{ccc}-2& 1& 1\\ 6& -3& 5\\ 4& 2& 2\end{array}\right)$

Find the inverse of the given matrix

role="math" localid="1664188723389" $$\begin{gathered} {\text{M}}^{-1}=\text{adj(M)}.\frac{1}{\left|\text{M}\right|} \\ {\text{M}}^{-\text{1}}=\frac{\text{1}}{\text{8}}\left(\begin{array}{ccc}-2& 1& 1\\ 6& -3& 5\\ 4& 2& 2\end{array}\right) \end{gathered}$$

Therefore, the inverse of the given matrix is $\frac{\text{1}}{\text{8}}\left(\begin{array}{ccc}-2& 1& 1\\ 6& -3& 5\\ 4& 2& 2\end{array}\right)$.