Question

How to find all the minors and cofactors of the matrix A = \left( {\begin{array}{*{20}{c}}1&{ - 2}&3\\\6&7&{ - 1}\\\\{ - 3}&1&4\end{array}} \right)?

Answer 1

For the matrix $A = {\left[ {{a_{ij}}} \right]_{n \times n}}$ of order $n$, with elements represented by term ${a_{ij}}$ such that $1 \le i,j \le n$; the minor for any element ${a_{ij}}$ is the determinant of the matrix formed by deleting the ${i^{th}}$ row and ${j^{th}}$ column.

The cofactor of any element ${a_{ij}}$ for the matrix $A = {\left[ {{a_{ij}}} \right]_{n \times n}}$ is equal to ${\left( { - 1} \right)^{i + j}}{M_{ij}}$. Here ${M_{ij}}$is the minor of the matrix $A$corresponding to element ${a_{ij}}$.

Complete step by step solution:
The given square matrix $A$ can be written as shown below.
A = \left( {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right)
= \left( {\begin{array}{*{20}{c}}1&{ - 2}&3\\\6&7&{ - 1}\\\\{ - 3}&1&4\end{array}} \right)

Let’s start to find minor ${M_{11}}$ corresponding to element ${a_{11}} = 1$ as follows:

Delete the first row and first column of the matrix $A$ where the element ${a_{11}} = 1$ is situated and obtains the determinant of the matrix formed in this process.
{M_{11}} = \left| {\begin{array}{*{20}{c}}7&{ - 1}\\\1&4\end{array}} \right|

Evaluate the determinant as shown below.
${M_{11}} = \left( 7 \right)\left( 4 \right) - \left( 1 \right)\left( { - 1} \right)$
$= 28 + 1$
$= 29$
Therefore, the minor of the element ${a_{11}} = 1$ for the matrix $A$ is ${M_{11}} = 29$.

now, use the relation between cofactor and minor which is ${C_{ij}} = {\left( { - 1} \right)^{i + j}}{M_{ij}}$ to obtain the cofactor ${C_{11}}$ corresponding to element ${a_{11}} = 1$ for the matrix $A$ is shown below.
${C_{11}} = {\left( { - 1} \right)^{1 + 1}}{M_{11}}$
$= {\left( { - 1} \right)^2}\left( {29} \right)$
$= 29$
Therefore, corresponding to ${a_{11}} = 1$ for the matrix $A$ the minor is ${M_{11}} = 29$ and the cofactor is ${C_{11}} = 29$.

Similarly, obtain other minors and cofactors corresponding to each element of matrix $A$.

Delete the first row and second column of the matrix $A$ and obtain the determinant of the matrix formed in this process.
{M_{12}} = \left| {\begin{array}{*{20}{c}}6&{ - 1}\\\\{ - 3}&4\end{array}} \right|

Evaluate the determinant as shown below.
${M_{12}} = \left( 6 \right)\left( 4 \right) - \left( { - 3} \right)\left( { - 1} \right)$
$= 24 - 3$
$= 21$
Therefore, the minor of the element ${a_{12}} = - 2$ for the matrix $A$ is ${M_{12}} = 21$.

The cofactor ${C_{12}}$ corresponding to element ${a_{12}} = - 2$ for the matrix $A$ is shown below.
${C_{12}} = {\left( { - 1} \right)^{1 + 2}}{M_{12}}$
$= {\left( { - 1} \right)^3}\left( {21} \right)$
$= - 21$

Therefore, corresponding to ${a_{12}} = - 2$ for the matrix $A$ the minor is ${M_{12}} = 21$ and the cofactor is ${C_{12}} = - 21$.

The complete table for each minor and cofactor corresponding to each element from matrix $A$ is shown below.

${a_{ij}}$	${M_{ij}}$	${C_{ij}}$
${a_{11}} = 1$	$29$	$29$
${a_{12}} = - 2$	$21$	$- 21$
${a_{13}} = 3$	$27$	$27$
${a_{21}} = 6$	$- 11$	$11$
${a_{22}} = 7$	$13$	$13$
${a_{23}} = - 1$	$- 5$	$5$
${a_{31}} = - 3$	$- 19$	$- 19$
${a_{32}} = 1$	$- 19$	$19$
${a_{32}} = 4$	$19$	$19$

Therefore, the minor matrix and cofactor matrix for matrix $A$ is written as shown below.
\Rightarrow M = \left( {\begin{array}{*{20}{c}}{29}&{21}&{27}\\\\{ - 11}&{13}&{ - 5}\\\\{ - 19}&{ - 19}&{19}\end{array}} \right)
\Rightarrow C = \left( {\begin{array}{*{20}{c}}{29}&{ - 21}&{27}\\\\{11}&{13}&5\\\\{ - 19}&{19}&{19}\end{array}} \right)

Note: The cofactor and minor for an element at even position like ${a_{11}}$, ${a_{13}}$, ${a_{22}}$ etc. are the same and at odd positions negative to each other. In other words, ${C_{ij}} = {M_{ij}}$ if $\left( {i + j} \right)$ is even and ${C_{ij}} = - {M_{ij}}$ if $\left( {i + j} \right)$ is odd where ${a_{ij}}$ is the element.