# 2.7 Inverse Matrix

2.7 Inverse Matrix
1. If A is a square matrix, is another square matrix and A × B = B × A = I, then matrix is the inverse matrix of matrix and vice versa. Matrix A is called the inverse matrix of for multiplication and vice versa.

2. The symbol A-1 denotes the inverse matrix of A.

3. Inverse matrices can only exist for square matrices but not all square matrices have inverse matrices.

4. If AB ≠ I or BA ≠ I, then is not the inverse of B and B is not the inverse of A.

Example 1:
Determine whether matrix  $A=\left(\begin{array}{cc}2& 9\\ 1& 5\end{array}\right)$  is an inverse matrix of matrix $B=\left(\begin{array}{cc}5& -9\\ -1& 2\end{array}\right).$

Solution:

5. The inverse of a matrix may also be found using a formula.
If $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ , then the inverse matrix of A, A-1, is given by the formula below.
6. ad – bc is known as the determinant of matrix A.

7. If the determinant, ad – bc = 0, then the inverse matrix of A does not exist.

Example 2:
Find the inverse matrix of $A=\left(\begin{array}{cc}6& 1\\ -9& -1\end{array}\right)$  using the formula.

Solution:

$\begin{array}{l}A=\left(\begin{array}{cc}6& 1\\ -9& -1\end{array}\right)\\ a=6,\text{}b=1,\text{}c=-9,\text{}d=-1\\ {A}^{-1}=\frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)\\ {A}^{-1}=\frac{1}{6×-1-\left(1×-9\right)}\left(\begin{array}{cc}-1& -1\\ 9& 6\end{array}\right)\\ {A}^{-1}=\frac{1}{-6+9}\left(\begin{array}{cc}-1& -1\\ 9& 6\end{array}\right)\\ {A}^{-1}=\frac{1}{3}\left(\begin{array}{cc}-1& -1\\ 9& 6\end{array}\right)=\left(\begin{array}{cc}-\frac{1}{3}& -\frac{1}{3}\\ 3& 2\end{array}\right)\end{array}$

Example 3:
The inverse matrix of $\left(\begin{array}{cc}-7& 2\\ -9& 2\end{array}\right)\text{is}r\left(\begin{array}{cc}2& s\\ 9& t\end{array}\right).$  Find the value of r, of s and of t.

Solution:
$\begin{array}{l}\text{Let}A=\left(\begin{array}{cc}-7& 2\\ -9& 2\end{array}\right)\\ {A}^{-1}=\frac{1}{-7×2-\left(-9\right)×2}\left(\begin{array}{cc}2& -2\\ 9& -7\end{array}\right)\\ {A}^{-1}=\frac{1}{4}\left(\begin{array}{cc}2& -2\\ 9& -7\end{array}\right)\\ \therefore r\left(\begin{array}{cc}2& s\\ 9& t\end{array}\right)=\frac{1}{4}\left(\begin{array}{cc}2& -2\\ 9& -7\end{array}\right)\\ \\ \text{By comparison,}\\ r=\frac{1}{4},\text{}s=-2,\text{}t=-7.\end{array}$