Question

Question:Find the inverse of the given matrix.

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

Short answer

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

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. ${\text{A.A}}^{-1}={\text{A}}^{-1}\text{.A}=\text{I}$, 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 $\left(\begin{array}{cc}2& 1\\ 0& -3\end{array}\right)$.

Find the inverse of the given matrix.

Find the inverse of the matrix

Find the determinant of the given matrix.

$\begin{array}{rcl}\text{det(}M\text{)}& =& \left|\begin{array}{cc}2& 1\\ 0& -3\end{array}\right|\\ & =& [\text{2}\times (-\text{3)]}-0\\ & =& -6\end{array}$

Find the minors of the given matrix.

$$\begin{gathered} {\text{M}}_{\text{11}}=-3 \\ {\text{M}}_{\text{12}}=0 \\ {\text{M}}_{\text{21}}=-1 \\ {\text{M}}_{\text{22}}=2 \end{gathered}$$

Solve further.

Cofactors:role="math" localid="1664182212314" $\left(\begin{array}{cc}-3& 0\\ -1& -2\end{array}\right)$

Find the adjoint of the matrix.

role="math" localid="1664182281131" ${\left(\begin{array}{cc}-3& 0\\ -1& -2\end{array}\right)}^T=\left(\begin{array}{cc}-3& -1\\ 0& 2\end{array}\right)$

Find the inverse of the given matrix.

role="math" localid="1664182337943" $$\begin{gathered} M^{-1}=\text{adj(}M\text{)}.\frac{1}{\left|M\right|} \\ {M}^{-1}=\frac{1}{6}\left(\begin{array}{cc}-3& -1\\ 0& 2\end{array}\right) \\ =\frac{1}{6}\left(\begin{array}{cc}3& 1\\ 0& -2\end{array}\right) \end{gathered}$$

Therefore, the inverse of the given matrix is $\frac{1}{6}\left(\begin{array}{cc}3& 1\\ 0& -2\end{array}\right)$.