Calculus of One and Several Variables

19.4.1 A Formula For The Inverse

The definition of the determinant in terms of Laplace expansion along a row or column also provides a way to give a formula for the inverse of a matrix. Recall the definition of the inverse of a matrix in Definition 18.6.2 on Page 1157. Also recall the definition of the cofactor matrix given in Definition 19.2.12 on Page 1241. This cofactor matrix was just the matrix which results from replacing the ij^th entry of the matrix with the ij^th cofactor.

The following theorem says that to find the inverse, take the transpose of the cofactor matrix and divide by the determinant. The transpose of the cofactor matrix is called the adjugate or sometimes the classical adjoint of the matrix A. In other words, A⁻¹ is equal to one divided by the determinant of A times the adjugate matrix of A. This is what the following theorem says with more precision. The proof is presented later in the appendix devoted to the theory of the determinant.

First find the determinant of this matrix. Using Theorems 19.2.23 - 19.2.25 on Page 1247, the determinant of this matrix equals the determinant of the matrix

( ) ( 1 2 3 ) 0 − 6 − 8 0 0 − 2

which equals 12. The cofactor matrix of A is

( ) − 2 − 2 6 ( 4 − 2 0 ) . 2 8 − 6

Each entry of A was replaced by its cofactor. Therefore, from the above theorem, the inverse of A should equal

( − 2 − 2 6 )T ( − 1∕6 1∕3 1∕6 ) 1-( 4 − 2 0 ) = ( − 1∕6 − 1∕6 2∕3 ) . 12 2 8 − 6 1∕2 0 − 1∕2

Does it work? You should check to see if it does. When the matrices are multiplied

( − 1∕6 1∕3 1∕6 ) ( 1 2 3 ) ( 1 0 0 ) ( − 1∕6 − 1∕6 2∕3 ) ( 3 0 1 ) = ( 0 1 0 ) 1∕2 0 − 1∕2 1 2 1 0 0 1

and so it is correct.

First find its determinant. This determinant is

. The inverse is therefore equal to

( | | | | ) ||1∕3 − 1∕2 || || − 1∕6 − 1∕2 || || − 1∕6 1∕3 || T || ||2∕3 − 1∕2 || −|| − 5∕6 − 1∕2 || || || || || | | | | ||− 5∕6 2∕3 || || 6|| −|| 0 1∕2 || || 1∕2 1∕2 || − || 1∕2 0 || || . || || 2∕3 − 1∕2|| ||− 5∕6 − 1∕2 || ||− 5∕6 2∕3|| || ( || 0 1∕2 || || 1∕2 1∕2 || || 1∕2 0 || ) |1∕3 − 1∕2 | −| − 1∕6 − 1∕2 | | − 1∕6 1∕3 |

Expanding all the 2 × 2 determinants this yields

( 1∕6 1∕3 1∕6 ) T ( 1 2 − 1 ) 6( 1∕3 1∕6 − 1∕3 ) = ( 2 1 1 ) − 1∕6 1∕6 1∕6 1 − 2 1

Always check your work.

( ) ( ) ( ) 1 2 − 1 1∕2 0 1∕2 1 0 0 ( 2 1 1 ) ( − 1∕6 1∕3 − 1∕2 ) = ( 0 1 0 ) 1 − 2 1 − 5∕6 2∕3 − 1∕2 0 0 1

and so we got it right. If the result of multiplying these matrices had been something other than the identity matrix, you would know there was an error. When this happens, you need to search for the mistake if you are interested in getting the right answer. A common mistake is to forget to take the transpose of the cofactor matrix.