Question

Question: Use (6.13)to find the inverse of the given matrix.

13.$\left(\begin{array}{cc}6& 9\\ 3& 5\end{array}\right)$

Short answer

The inverse of a given matrix$\left(\begin{array}{cc}6& 9\\ 3& 5\end{array}\right)$ is $\left(\begin{array}{cc}\frac{5}{3}& -3\\ -1& 2\end{array}\right)$.

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

For matrix , the inverse matrix formula is ; , where is a square matrix.

Given parameters

The given matrix is $\left(\begin{array}{cc}6& 9\\ 3& 5\end{array}\right)$.

Find the inverse of the given matrix.

Find the determinant of the matrix

The sum of the products of the elements of any row or column with the corresponding cofactors of the matrix is equal to the value of the determinant.

For example, if$A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ be a matrix of order $2\times 2$, then the determinant of a matrix is defined as $\left|\text{A}\right|=\left(\text{a}\times \text{d}\right)-\left(\text{b}\times \text{c}\right)$.

The matrix is given role="math" localid="1664180786165" $A=\left(\begin{array}{cc}6& 9\\ 3& 5\end{array}\right)$.

Find the determinant of A.

$\begin{array}{rcl}\left|\text{A}\right|& =& \left(\text{6}\times \text{5}\right)-\left(\text{3}\times \text{9}\right)\\ & =& \text{30}-\text{27}\\ & =& \text{3}\end{array}$

Find the Adjoint of a given matrix

The adjoint of a square matrix$\text{A}={\left[{\text{a}}_{\text{ij}}\right]}_{\text{n}\times \text{n}}$ is defined as the transpose of the matrix $\text{A}={\left[{\text{A}}_{\text{ij}}\right]}_{\text{n}\times \text{n}}$, where${\text{A}}_{\text{ij}}$ is the cofactor of the elements ${\text{a}}_{\text{ij}}$.

For example, if role="math" localid="1664180826099" $A=\left(\begin{array}{cc}2& 3\\ 1& 4\end{array}\right)$then ${\text{A}}_{\text{11}}={\text{4,A}}_{\text{12}}=-{\text{1,A}}_{\text{21}}=-{\text{3,A}}_{\text{22}}=\text{2}$

role="math" localid="1664180928744" $\begin{array}{rcl}\text{adj}\left(\text{A}\right)& =& \left(\begin{array}{cc}{\text{A}}_{\text{11}}& {\text{A}}_{\text{21}}\\ {\text{A}}_{\text{12}}& {\text{A}}_{22}\end{array}\right)\\ & =& \left(\begin{array}{cc}4& -3\\ -1& 2\end{array}\right)\end{array}$

The Adjoint of a given matrix is $\begin{array}{rcl}\text{adj}\left(\text{A}\right)& =& \left(\begin{array}{cc}5& -9\\ -3& 6\end{array}\right)\end{array}$.

Find the inverse of a given matrix

The inverse of a matrix A is ${\text{A}}^{\text{- 1}}$such that the product ${\text{AA}}^{\text{- 1}}=\text{1}$.

role="math" localid="1664181111415" $\begin{array}{rcl}{\text{A}}^{\text{- 1}}& =& \frac{\text{1}}{\text{detA}}\text{Adj}\left(\text{A}\right)\\ & & \text{=}\frac{1}{3}\left(\begin{array}{cc}5& -9\\ -3& 6\end{array}\right)\\ & & \text{=}\left(\begin{array}{cc}\frac{5}{3}& -3\\ -1& 2\end{array}\right)\end{array}$

Therefore, the inverse of a given matrix $\begin{array}{rcl}& & \left(\begin{array}{cc}6& 9\\ 3& 5\end{array}\right)\end{array}$is $\begin{array}{rcl}& & \left(\begin{array}{cc}\frac{5}{3}& -3\\ -1& 2\end{array}\right)\end{array}$.