You can put this solution on YOUR website! A =
| 5 5 |
| 0 -4 |
now A^-1 exists iff
A x A^-1 = A^-1 x A = I where I is the identity matrix
determinant of A is written det A
det A = (5*-4) - (5*0) = -20
A^-1 = (1/-20) *
| -4 -5 |
| 0 5 |
the above matrix is formed using the following algorithm
| a11 a12 |
| a21 a22 |
the resulting matrix is formed as
| a22 -a12 |
| -a21 a11 |
then multiply this matrix by 1 / det A
A^-1 =
| 1/5 1/4 |
| 0 -1/4 |
check the answer
A x A-1 =
|5 5|
|0 -4| x
|1/5 1/4|
| 0 -1/4|
A x A^-1 =
| 1 0 |
| 0 1 |
note matrix multiplication is each element in result is row elements times column elements added
A^-1 x A =
|1/5 1/4|
| 0 -1/4| x
|5 5|
|0 -4| =
| 1 0 |
| 0 1 |
our answer checks since I2 =
| 1 0 |
| 0 1 |
A^-1 =
| 1/5 1/4 |
| 0 -1/4 |
