Question

Prove that the sum of product of any row or column with corresponding co-factors of some other row or column in a matrix is zero.

Answer 1

Hint: To prove this, definition and knowledge of minors and cofactors is required. The minor of an element is the determinant of the matrix formed by deleting the row and column of that element. If the element is $a_{ij}$, then the cofactor of that element by-
$A_{ij}=(-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor.

Complete step by step answer:
Let X be the a 3x3 square matrix such that-
$X=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}$
We have to prove that the sum of product of any row or column with corresponding co-factors of some other row or column in a matrix is zero, so we will find cofactors of the second row and multiply with the first row. By definition, we can write that-
$M_{21}=det\begin{bmatrix}a_{12}&a_{13}\\a_{32}&a_{33}\end{bmatrix}=a_{12}a_{33}-a_{13}a_{32}\\M_{22}=det\begin{bmatrix}a_{11}&a_{13}\\a_{31}&a_{33}\end{bmatrix}=a_{11}a_{33}-a_{13}a_{31}\\M_{23}=det\begin{bmatrix}a_{11}&a_{12}\\a_{31}&a_{32}\end{bmatrix}=a_{11}a_{32}-a_{12}a_{31}$
The cofactors for the corresponding minors will be-
$A_{21}=\left(-1\right)^{i+j}M_{21}\\=\left(-1\right)^{2+1}\left(a_{12}a_{33}-a_{13}a_{32}\right)\\=-\left(a_{12}a_{33}-a_{13}a_{32}\right)\\A_{22}=\left(-1\right)^{i+j}M_{22}\\=\left(-1\right)^{2+2}\left(a_{11}a_{33}-a_{13}a_{31}\right)\\=\left(a_{11}a_{33}-a_{13}a_{31}\right)\\A_{23}=\left(-1\right)^{i+j}M_{23}\\=\left(-1\right)^{2+3}\left(a_{11}a_{32}-a_{12}a_{31}\right)\\=-\left(a_{11}a_{32}-a_{12}a_{31}\right)$
We have to find the sum of product of the elements and corresponding cofactors which is-
$\begin{align}
  &{a_{11}}{A_{21}} + {a_{12}}{A_{22}} + {a_{13}}{A_{23}} \\
   &= {a_{11}}\left( { - \left( {{a_{12}}{a_{33}} - {a_{13}}{a_{32}}} \right)} \right) + {a_{12}}\left( {{a_{11}}{a_{33}} - {a_{13}}{a_{31}}} \right) + {a_{13}}\left( { - \left( {{a_{11}}{a_{32}} - {a_{12}}{a_{31}}} \right)} \right) \\
   &= - {a_{11}}{a_{12}}{a_{33}} + {a_{11}}{a_{13}}{a_{32}} + {a_{12}}{a_{11}}{a_{33}} - {a_{12}}{a_{13}}{a_{31}} - {a_{13}}{a_{11}}{a_{32}} + {a_{13}}{a_{12}}{a_{31}} \\
  &All\;the\;terms\;cancel\;each\;other, \\
   &= 0 \\
\end{align} $

Hence, proved.

Note: To prove this statement, we can choose any other row or column or any other square matrix. The final result will be the same. Also, one should take care of the sign of the cofactor which has been calculated.