Determinant

Computing the determinant of a matrix is performed by following a given row and computing the value of the adjugates. For example, suppose we have a $2\times2$ matrix $A$ of the form:

$A = \begin{pmatrix} e_{0} & e_{1} \\ e_{2} & e_{3} \end{pmatrix}$

that would mean that:

$det(A)=e_{3}*e_{0}-e_{2}*e_{1}$

For larger matrices, supposing that $e_{3}$ represents a sub-matrix of $A$ , its determinant has to be computed. This is performed recursively by traversing a matrix row and then calling wasDeterminant on any sub-matrix.

For a larger matrix, such as:

$A = \begin{pmatrix} e_{0} & e_{1} & e_{2} \\ e_{3} & e_{4} & e_{5} \\ e_{6} & e_{7} & e_{8} \\ \end{pmatrix}$

the element must be multiplied with the value of the sub-determinant while minding the alternation of the signs. In this case:

$det(A)=e_{0} * ( e_{8} * e_{4} - e_{7} * e_{5} ) - e_{1} * ( e_{8} * e_{3} - e_{6} * e_{5} ) ...$

Adjugate

The adjugate step consists in finding the cofactor (determinants) of every element in the matrix and replacing the original value with the cofactor of the sub-matrix. For example, for a matrix $B$ :

$B = \begin{pmatrix} e_{0} & e_{1} & e_{2} \\ e_{3} & e_{4} & e_{5} \\ e_{6} & e_{7} & e_{8} \\ \end{pmatrix}$

One would take each element, ignore the row and column it appears in and compute the determinant of the sub-matrix $B$ . For example, in this particular case, the element $e_{0}$ will be replaced with the value of the determinant:

$B_{e_{0}} = \left[ \begin{matrix} e_{4} & e_{5} \\ e_{7} & e_{8} \end{matrix} \right]$

The signs alternate on rows and columns, for example, the cofactor of the second element $e_{1}$ will have a negative sign:

$B_{e_{1}} = - \left[ \begin{matrix} e_{3} & e_{5} \\ e_{6} & e_{8} \\ \end{matrix} \right]$

The same applies to the cofactor of element $e_{3}$ , that will have a negative sign as well:

$B_{e_{3}} = - \left[ \begin{matrix} e_{1} & e_{2} \\ e_{7} & e_{8} \end{matrix} \right]$

By going through every element of the matrix and replacing each element with value of the determinants of the sub-matrix (also called the co-factor), we obtain a new matrix that has to finally be transposed in order to finally obtain the adjugate of the matrix.

One can note that at the time of writing the cofactor function does not bother with applying signs to the values - this is performed later on in by the adjugate or determinant functions.

Scalar Product

For a multiplication between a scalar $p$ and a matrix $A$ so that $C=p*A$ :

$C = p * \begin{pmatrix} a_{1,1} & a_{1,2} & a_{...} & a_{1,j} \\ a_{2,1} & a_{2,2} & a_{...} & a_{2,j} \\ \cdots & \cdots & \cdots & \cdots \\ a_{i,1} & a_{i,2} & a_{...} & a_{i,j} \\ \end{pmatrix}_{i,j} = \begin{pmatrix} p*a_{1,1} & p*a_{1,2} & p*a_{...} & p*a_{1,j} \\ p*a_{2,1} & p*a_{2,2} & p*a_{...} & p*a_{2,j} \\ \cdots & \cdots & \cdots & \cdots \\ p*a_{i,1} & p*a_{i,2} & p*a_{...} & p*a_{i,j} \\ \end{pmatrix}_{i,j}$

we multiply the scalar $p$ with every element in $A$ in order to obtain the resulting matrix $C$ .

Dot Product

In order to multiply two matrixes, the number of columns of the first matrix $A$ has to be equal to the number of rows of the second matrix $B$ .

$\begin{pmatrix} a_{1,1} & a_{1,2} & a_{...} & a_{1,j} \\ a_{2,1} & a_{2,2} & a_{...} & a_{2,j} \\ \cdots & \cdots & \cdots & \cdots \\ a_{i,1} & a_{i,2} & a_{...} & a_{i,j} \\ \end{pmatrix}_{i,j} \times \begin{pmatrix} b_{1,1} & b_{1,2} & b_{...} & b_{1,k} \\ b_{2,1} & b_{2,2} & b_{...} & b_{2,k} \\ \cdots & \cdots & \cdots & \cdots \\ b_{j,1} & b_{j,2} & b_{...} & b_{j,k} \\ \end{pmatrix}_{j,k} = \begin{pmatrix} c_{1,1} & c_{1,2} & c_{...} & c_{1,k} \\ c_{2,1} & c_{2,2} & c_{...} & c_{2,k} \\ \cdots & \cdots & \cdots & \cdots \\ c_{i,1} & c_{i,2} & c_{...} & c_{i,k} \\ \end{pmatrix}_{i,k}$

in this case, we have $A_{i,j}$ and $B_{j,k}$ with $j=j$ . For every element in $C_{i,k}$ we multiply elements from the rows of $A$ with elements from the column of $B$ and add them up to obtain the elements in $C$ .

$\begin{eqnarray*} c_{1,1} &=& a_{1,1}*b_{1,1} + a_{1,2}*b_{2,1} + \cdots + a_{1,j}*b_{j,1} \\ c_{1,2} &=& a_{1,1}*b_{1,2} + a_{1,2}*b_{2,2} + \cdots + a_{1,j}*b_{j,2} \\ &\cdots& \\ c_{i,1} &=& a_{i,1}*b_{1,1} + a_{i,2}*b_{2,1} + \cdots + a_{i,j}*b_{j,1} \\ &\cdots& \\ c_{i,k} &=& a_{i,1}*b_{1,k} + a_{i,2}*b_{2,k} + \cdots + a_{i,j}*b{j,k} \end{eqnarray*}$

Table of Contents

Determinant

Adjugate

Scalar Product

Dot Product