Notation, complementary minors and adjoint matrix

Notation

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

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

where the subscripts indicate the row and the column respectively.

If we write:

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

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:

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

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:

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

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.

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

Generally, the complementary minor of an element $a_{ij}$ is written as $M_{ij}$ 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:

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:

Consider the matrix:

$$\begin{pmatrix} -1 & 0 & 2 \\ 1 & 1 & -3 \\ 0 & 2 & 4\end{pmatrix}$$

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

First we calculate the complementary minor:

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

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 $a_{22}$:

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

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 $a_{ij}$ 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 $Adj(A)$.

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

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

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:

$$\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & +\end{pmatrix} $$

and therefore the adjoint matrix will be:

$$Adj\begin{pmatrix}-1 & 0 & 2 \\ 1 & 1 & -3 \\ 0 & 2 & 4 \end{pmatrix}= \begin{pmatrix} 10 & -4 & 3 \\ 4 & -4 & 2 \\ -2 & 4 & -1\end{pmatrix}$$

Practice exercises