Practical Calculation of Determinants¶
Elementary Operations¶
Row elementary operations convert a square matrix to the echelon, i.e. upper triangular, form. The determinant of such a transformed matrix is simply the product of its diagonal elements. Application of elementary operations may be therefore a useful method of calculating determinants. However, one must take into account that elementary operations upon rows (columns) of a matrix can modify its determinant. Namely:
Transposition of any two rows (columns) changes the sign of the determinant.
Multiplying a row (column) by a non-zero scalar multiplies the determinant by that scalar.
Adding to a row a multiple of another row, as well as adding to a column a multiple of another column, does not change the determinant of the matrix.
Example 1. \(\qquad\left|\begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 1 & 1 & 1 \\ 1 & 1 & 3 & 1 & 1 \\ 1 & 1 & 1 & 4 & 1 \\ 1 & 1 & 1 & 1 & 5 \end{array} \right|\ \ = \ \ \left|\begin{array}{ccccc} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 4 \end{array} \right|\ \ =\ \ 24\,.\)
Operations performed on rows \(\ \boldsymbol{R}_1,\,\boldsymbol{R}_2,\,\boldsymbol{R}_3,\, \boldsymbol{R}_4,\,\boldsymbol{R}_5:\quad \boldsymbol{R}_i = \boldsymbol{R}_i - \boldsymbol{R}_1\,,\quad i = 2,3,4,5.\)
Example 2.
Operations performed on rows \(\ \boldsymbol{R}_1,\,\boldsymbol{R}_2,\, \boldsymbol{R}_3,\,\boldsymbol{R}_4:\)
\(\ \boldsymbol{R}_2 = \boldsymbol{R}_2 - 2\,\boldsymbol{R}_1,\ \ \boldsymbol{R}_3 = \boldsymbol{R}_3 + \boldsymbol{R}_1,\ \ \boldsymbol{R}_4 = \boldsymbol{R}_4 + 2\,\boldsymbol{R}_1\,;\)
\(\ \boldsymbol{R}_3 = \boldsymbol{R}_3 + 2\,\boldsymbol{R}_2,\ \ \boldsymbol{R}_4 = \boldsymbol{R}_4 + 3\,\boldsymbol{R}_2\,;\)
\(\ \boldsymbol{R}_4 = \boldsymbol{R}_4 - 3\,\boldsymbol{R}_3\,.\)
Example 3. \(\ \ \) Making use of the equalities \(\quad\left\{\ \, \begin{array}{l} 1798\ =\ 31\,\cdot\,58 \\ 2139\ =\ 31\,\cdot\,69 \\ 3255\ =\ 31\,\cdot\,105 \\ 4867\ =\ 31\,\cdot\,157 \end{array}\right.\)
without evaluating the determinant, validate that \(\quad\left|\begin{array}{llll} 1 & 7 & 9 & 8 \\ 2 & 1 & 3 & 9 \\ 3 & 2 & 5 & 5 \\ 4 & 8 & 6 & 7 \end{array} \right|\quad\) is also divisible by 31.
Solution. \(\,\)
The operation \(\ \,\boldsymbol{C}_4\ =\ 1000\,\cdot\,\boldsymbol{C}_1\,+\,100\,\cdot\,\boldsymbol{C}_2\,+\, 10\,\cdot\,\boldsymbol{C}_3\,+\,\boldsymbol{C}_4\ \,\) on columns \(\ \boldsymbol{C}_1,\,\boldsymbol{C}_2,\, \boldsymbol{C}_3,\,\boldsymbol{C}_4\ \) does not change the value of the determinant. As a result we obtain
The Laplace Expansion¶
Definition.
A minor of size \(\,k\ \) of a matrix \(\,\boldsymbol{A}\in M_{m\times n}(K)\ \) is the determinant of the square matrix obtained from \(\,\boldsymbol{A}\ \) by removing some selected \(\ m-k\ \) rows \(\,\) and \(\ n-k\ \) columns \(\ (1\leq k \leq m,n).\)
\(\,\)
Example 4. \(\ \) Let \(\ \ \boldsymbol{A}\ \,=\ \, \left[\begin{array}{rrrrr} -2 & 0 & 3 & 4 & 1 \\ 4 & 7 & 6 & -2 & 0 \\ 3 & 3 & 5 & 1 & 1 \\ -1 & 2 & 3 & 0 & 4 \end{array}\right]\in M_{4\times 5}(Q)\,.\)
Removing the third row and the second and fourth column we get a minor of size three:
\(\,\)
Definition.
\(\,\)
Suppose we have a square matrix \(\ \boldsymbol{A}\,=\,[a_{ij}]_{n\times n}\in M_{n}(K).\) \(\\\) Select an element \(\ a_{ij}\ \) and delete the \(\,i\)-th row and the \(\,j\)-th column of \(\ \boldsymbol{A}.\) \(\\\) The obtained minor \(\,M_{ij}\,\) multiplied by the sign factor \(\ (-1)^{i+j}\ \) is termed \(\\\) the cofactor of the element \(\ a_{ij}\ \) in the matrix \(\ \boldsymbol{A}\,\) or shortly the \(\ i,j\ \) cofactor of \(\,\boldsymbol{A},\) \(\\\) and is denoted by \(\ A_{ij}:\)
\(\,\)
Example 5. \(\ \) Let \(\ \ \boldsymbol{A}\ \,=\ \, \left[\begin{array}{rrrr} 2 & 1 & 0 & 6 \\ 3 & -1 & 7 & 4 \\ 1 & 0 & 5 & 2 \\ -1 & -2 & 1 & 5 \end{array}\right]\in M_4(Q)\,.\)
The cofactors of the elements \(\ a_{12} = 1\ \) and \(\ a_{33} = 5\ \) are:
It is worth noting that the cofactor \(\ A_{ij}\,\) of an element \(\,a_{ij}\,\) does not depend on that element nor on any element in the \(\,i\)-th row or in the \(\,j\)-th column of the matrix. \(\\\)
Theorem 4. \(\,\) The Laplace Expansion. \(\\\)
The determinant of a matrix \(\ \boldsymbol{A}\,=\,[a_{ij}]_{n\times n}\in M_{n}(K)\ \) is given by: \(\\\)
the sum of products of consecutive elements of a selected \(\,i\)-th row by the cofactors of these elements (expansion of the determinant along the \(\,i\)-th row):
(1)¶\[\det\boldsymbol{A}\ =\ a_{i1}\,A_{i1}\,+\,a_{i2}\,A_{i2}\,+\,\dots\,+\,a_{in}\,A_{in}\,, \quad i=1,2,\ldots,n.\]the sum of products of consecutive elements of a selected \(\,j\)-th column by the cofactors of these elements (expansion of the determinant along the \(\,j\)-th column):
(2)¶\[\det\boldsymbol{A}\ =\ a_{1j}\,A_{1j}\,+\,a_{2j}\,A_{2j}\,+\,\dots\,+\,a_{nj}\,A_{nj}\,, \quad j=1,2,\ldots,n.\]
Interestingly enough, the Laplace expansion may be performed along any row or any column: the result will be always the same.
To prove the Laplace expansion it suffices to check that the r-h-sides of Equations (1) and (2) fulfill the requirements of the axiomatic definition of the determinant.
As a method of computing determinants, the Laplace expansion is the most efficient when in a row or in a column there is only one non-zero element (this form may be achieved by elementary operations). Thus a practical calculation of a determinant is performed in two stages:
use of elementary operations to convert the matrix into the form in which in a row \(\\\) (or in a column) there is only one non-zero element;
application of the Laplace expansion along that row (or that column).
Example 6. \(\,\) The applied operations are described beneath the calculations.
Operations on rows \(\ \boldsymbol{R}_1,\,\boldsymbol{R}_2,\, \boldsymbol{R}_3,\,\boldsymbol{R}_4:\) \(\boldsymbol{R}_2 = \boldsymbol{R}_2 + \boldsymbol{R}_1,\ \ \boldsymbol{R}_3 = \boldsymbol{R}_3 - 2\,\boldsymbol{R}_1,\ \ \boldsymbol{R}_4 = \boldsymbol{R}_4 - \boldsymbol{R}_1.\) Laplace expansion along the 3rd column. Number \(\ 3\ \) pulled out of the 3rd column.
Operations performed on columns \(\ \boldsymbol{C}_1,\,\boldsymbol{C}_2,\,\boldsymbol{C}_3:\ \) \(\boldsymbol{C}_2 = \boldsymbol{C}_2 - \boldsymbol{C}_1,\ \ \boldsymbol{C}_3 = \boldsymbol{C}_3 - \boldsymbol{C}_1.\ \) \(\\\) Number \(\ 3\ \) pulled out of the 2nd column. Laplace expansion along the 2nd row. \(\\\)
Finally: \(\qquad\left|\begin{array}{rrrr} 2 & -5 & 1 & 2 \\ -3 & 7 & -1 & 4 \\ 5 & -9 & 2 & 7 \\ 4 & -6 & 1 & 2 \end{array}\right|\ \ =\ \ 3\ \ \left|\begin{array}{rrr} -1 & 2 & 2 \\ 1 & 1 & 1 \\ 2 & -1 & 0 \end{array}\right|\ \ =\ \ -\ 9\,. \\\)
In Sage a determinant of a given square matrix is computed by the function
(method) determinant(), in short det(). We shall call it to calculate
the determinant from the above example and to confirm the theorem on
determinant of a transpose matrix:
sage: A = matrix(QQ,[[ 2,-5, 1, 2],
[-3, 7,-1, 4],
[ 5,-9, 2, 7],
[ 4,-6, 1, 2]])
sage: det_A = A.determinant()
# Shorthand notation for a determinant of a transposed matrix:
sage: det_At = A.T.det()
sage: print "det A =", det_A; det_A==det_At
det A = -9
True
