Evaluating Determinants
Let A be an n×n matrix with ajk denoting the entry in the j’th row and k’th column.
For any j, k we let Mjk be the (n − 1) × (n − 1) matrix obtained by deleting the j’th row
and k’th column, known as the j, k minor matrix. For example if


1 2 4
A = 5 2 0
3 8 1
then
M11
2 0
=
8 1
M32
1 4
=
5 0
The determinant det A may be computed in the following way.
P
det A = nk=1 (−1)j+k ajk det Mjk
fixed j
Pn
j+k
= j=1 (−1) ajk det Mjk
fixed k
• According the above definition you can choose any row or any column of the matrix
(fixed j or fixed k) and expand the determinant by means of that row or column –
multiply each entry of the row or column by the determinant of the corresponding
minor matrix and an alternating factor of ±1, and then sum these values. For
example if A is the matrix given above, and we expand the determinant using the
first row, we get
det A = a11 det M11 − a12 det M12 + a13 det M13 = (1)(2) − (2)(5) + (4)(34) = 128
• An n×n determinant is therefore defined as a sum of (n−1)×(n−1) determinants,
each of which is then defined as a sum of (n − 2) × (n − 2) determinants, etc.
Eventually it gets down to a combination of 2 × 2 determinants which we have a
direct formula for.
• The above definition incorporates the non-obvious fact that using any row or any
column to expand the determinant leads to exactly the same answer. You should
try an example to see this.
• To make evaluating the determinant as quick and easy as possible, look for a row
or column which contains lots of zeros, and expand using that row or column.
• The matrix A is said to be upper triangular if it contains only zeros below the
diagonal, i.e. ajk = 0 if j > k. In this case you should check that det A =
a11 a22 . . . ann . In other words it is just the product of the diagonal entries. The
same is true for a lower triangular matrix, one which contains only zeros above the
diagonal.