# How the inverse of this matrix be found?

How can the inverse of matrix $A = \left( \begin{smallmatrix} 6&5\\5&4 \end{smallmatrix} \right)$ be $A^{-1} = \left( \begin{smallmatrix} -4&5\\ 5&-6 \end{smallmatrix} \right)$ where $\frac{1}{ad-bc} = \frac{1}{24-25} = \frac{1}{-1}$?

I thought that an inverse to this matrix was not possible, but my textbox and Wolfram Alpha says otherwise. Can someone tell me how this is possible, or if I have misunderstood the formula?

• why did you think it was not possible? May 18, 2013 at 11:15
• If you multiply the two matrices, do you not get the identity matrix? What's the problem? May 18, 2013 at 11:16
• I'm an idiot. I didn't calculate 1/(-1) correctly
– Sam
May 18, 2013 at 11:18

You can see that this matrix is invertible in couple of ways.

For example:

1. $|A|=6\cdot4-5\cdot5=-1\neq0$

2. The rows of $A$ are linearly independent .

3. The columns of $A$ are linearly independent .

4. $0$ is not an eigenvalue of $A$ (you can calculate the eigenvalues by finding the characteristic polynomial of $A$)

5. You have found a matrix $B$ s.t $AB=BA=I$

• thanks for the great answer
– Sam
May 18, 2013 at 11:24
• @Sam you're welcome! I'm glad you liked it May 18, 2013 at 11:25