Question

Question:Find the inverse of the given matrix.

15.$\left(\begin{array}{ccc}-1& 2& 3\\ 2& 0& -4\\ -1& -1& 1\end{array}\right)$

Short answer

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

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

Given parameters

The given matrix is$\left(\begin{array}{ccc}-1& 2& 3\\ 2& 0& -4\\ -1& -1& 1\end{array}\right)$.

Find the inverse of the given matrix.

Find the inverse of the matrix.

Find the determinant of the given matrix.

$\begin{array}{rcl}\text{det(}M\text{)}& =& \left|\begin{array}{ccc}-1& 2& 3\\ 2& 0& 4\\ -1& -1& 1\end{array}\right|\\ & =& 2C_{12}+1C_{32}\\ & =& (2\times (-6\left)\right)+(1\times 10)\\ & =& -2\end{array}$

Find the minors of the given matrix.

localid="1664185095839" $\begin{array}{rcl}{\text{M}}_{\text{11}}& =& \left|\begin{array}{cc}0& 4\\ -1& 1\end{array}\right|\\ & =& 4\\ {\text{M}}_{\text{12}}& =& -\left|\begin{array}{cc}2& 4\\ -1& 1\end{array}\right|\\ & =& -6\end{array}$

Evaluate further minors.

$\begin{array}{rcl}{\text{M}}_{\text{13}}& =& \left|\begin{array}{cc}2& 0\\ -1& -1\end{array}\right|\\ & =& -2\\ {\text{M}}_{\text{21}}& =& -\left|\begin{array}{cc}2& 3\\ -1& 1\end{array}\right|\\ & =& -5\end{array}$

Evaluate further minors.

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

Evaluate further minors.

$\begin{array}{rcl}{\text{M}}_{\text{31}}& =& \left|\begin{array}{cc}2& 3\\ 0& 4\end{array}\right|\\ & =& 8\\ {\text{M}}_{\text{32}}& =& -\left|\begin{array}{cc}-1& 3\\ 2& 4\end{array}\right|\\ & =& 10\end{array}$

Evaluate further minors.

$\begin{array}{rcl}{\text{M}}_{33}& =& \left|\begin{array}{cc}-1& 2\\ 2& 0\end{array}\right|\\ & =& -4\end{array}$

Cofactors:$\left(\begin{array}{ccc}4& -6& -2\\ -5& 2& -3\\ 8& 10& -4\end{array}\right)$

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

role="math" localid="1664185558251" ${\left(\begin{array}{ccc}4& -6& -2\\ -5& 2& -3\\ 8& 10& -4\end{array}\right)}^T=\left(\begin{array}{ccc}4& -5& 8\\ -6& 2& 10\\ -2& -3& -4\end{array}\right)$

Find the inverse of the given matrix.

role="math" localid="1664185638259" $$\begin{gathered} {\text{M}}^{-1}=\text{adj(M)}.\frac{1}{\left|\text{M}\right|} \\ {\text{M}}^{-1}=\frac{1}{6}\left(\begin{array}{ccc}-4& 5& -8\\ 6& -2& -10\\ 2& 3& 4\end{array}\right) \end{gathered}$$

Therefore, the inverse of the given matrix is $\frac{1}{2}\left(\begin{array}{ccc}-4& 5& -8\\ 6& -2& -10\\ 2& 3& 4\end{array}\right)$.