Notation, complementary minors and adjoint matrix

Notation

It is known that a matrix $3 \times 3$ is written as follows:

$(\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix})$

where the subscripts indicate the row and the column respectively.

If we write:

$| \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} |$

it means that the we want to calculate the determinant of this matrix.

Obviously this has been written for a $3 \times 3$ matrix, –because this one is the most common–, but the determinants of matrices $2 \times 2$ , $4 \times 4$ or $N \times N$ can also be computed. It only makes sense to speak of determinants of square matrices.

Complementary minors

Let's consider the $3 \times 3$ matrix:

$(\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix})$

The complementary minor of the element $a_{11}$ is the determinant of order 2 that survives when row 1 and column 1 are eliminated.

In other words, the complementary minor we are looking for is:

$M_{11} = | \begin{matrix} / 1 & / 2 & / 3 \\ / 4 & 5 & 6 \\ / 7 & 8 & 9 \end{matrix} | = 5 \cdot 9 - 8 \cdot 6 = - 3$

Now let's calculate the complementary minor of the element $a_{23}$ , in other words, the determinant of order 2 that survives when we eliminate the second row and the third column.

$M_{23} = | \begin{matrix} 1 & 2 & / 3 \\ / 4 & / 5 & / 6 \\ 7 & 8 & / 9 \end{matrix} | = 1 \cdot 8 - 7 \cdot 2 = - 6$

Generally, the complementary minor of an element $a_{i j}$ is written as $M_{i j}$ and it is the determinant of lower order that survives when row $i$ and column $j$ are eliminated.

Cofactors

We call the cofactor of an element of a matrix, its complementary minor but placing before it:

The sign $+$ when $i + j$ is even
The sign $-$ when $i + j$ is odd

Following the previous examples, the cofactor of the element $a_{11}$ is written as $C_{11}$ and must have the sign $+$ ( $1 + 1 = 2$ , which is even), while the cofactor of the element $a_{23}$ is written as $C_{23}$ and must have the sign $-$ ( $2 + 3 = 5$ , which is odd).

Using the precise notation, we conclude $C_{11} = + M_{11} = - 3$ and $C_{23} = - M_{23} = 6$ .

Let's see another example:

Example

Consider the matrix:

$(\begin{matrix} - 1 & 0 & 2 \\ 1 & 1 & - 3 \\ 0 & 2 & 4 \end{matrix})$

We want to find the cofactor of the element $a_{11}$ .

First we calculate the complementary minor:

$M_{11} = | \begin{matrix} - / 1 & / 0 & / 2 \\ / 1 & 1 & - 3 \\ / 0 & 2 & 4 \end{matrix} | = 1 \cdot 4 - 2 \cdot (- 3) = 10$

We check which sign corresponds: $1 + 1 = 2$ , pair, and therefore the sign is positive. Then the cofactor of $a_{11}$ is $C_{11} = 10$ .

Now let's find the cofactor of the element $a_{22}$ :

$M_{22} = | \begin{matrix} - 1 & / 0 & 2 \\ - / 1 & / 1 & - / 3 \\ 0 & / 2 & 4 \end{matrix} | \to | \begin{matrix} - 1 & 2 \\ 0 & 4 \end{matrix} | = (- 1) \cdot 4 - 2 \cdot 0 = - 4$

We verify the sign: $2 + 2 = 4$ , pair, and therefore the sign does not change, so $C_{22} = M_{22}$ .

And this way, we can successively find the cofactors of all the elements $a_{i j}$ of the matrix.

Adjoint matrix

If we replace every element of the matrix $A$ by its cofactor we obtain the adjoint matrix, which is written as $A d j (A)$ .

Let's calculate it using the previous example, starting with the complementary minors:

$\begin{matrix} M_{11} = | \begin{matrix} 1 & - 3 \\ 2 & 4 \end{matrix} | = 10 & M_{12} = | \begin{matrix} 1 & - 3 \\ 0 & 4 \end{matrix} | = 4 & M_{13} = | \begin{matrix} 1 & 1 \\ 0 & 2 \end{matrix} | = 3 \\ M_{21} = | \begin{matrix} 0 & 2 \\ 2 & 4 \end{matrix} | = - 4 & M_{22} = | \begin{matrix} - 1 & 2 \\ 0 & 4 \end{matrix} | = - 4 & M_{23} = | \begin{matrix} - 1 & 0 \\ 0 & 2 \end{matrix} | = - 2 \\ M_{31} = | \begin{matrix} 0 & 2 \\ 1 & - 3 \end{matrix} | = - 2 & M_{32} = | \begin{matrix} - 1 & 2 \\ 0 & 4 \end{matrix} | = - 4 & M_{33} = | \begin{matrix} - 1 & 0 \\ 1 & 1 \end{matrix} | = - 1 \end{matrix}$

Nine complementary minors have been found, but the signs of each one must be added depending on the sum $i + j$ being even or odd. Adding up, the signs will be as follows:

$(\begin{matrix} + & - & + \\ - & + & - \\ + & - & + \end{matrix})$

and therefore the adjoint matrix will be:

$A d j (\begin{matrix} - 1 & 0 & 2 \\ 1 & 1 & - 3 \\ 0 & 2 & 4 \end{matrix}) = (\begin{matrix} 10 & - 4 & 3 \\ 4 & - 4 & 2 \\ - 2 & 4 & - 1 \end{matrix})$