Question

If $\vartriangle $ $ = $ $\left| {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right|$ and ${A_{ij}}$ is Cofactors of ${a_{ij}}$ , then the value of $\vartriangle $ is given by
A. ${a_{11}}{A_{31}} + {a_{12}}{A_{32}} + {a_{13}}{A_{31}}$
B. ${a_{11}}{A_{11}} + {a_{12}}{A_2}1 + {a_{13}}{A_{31}}$
C. ${a_{21}}{A_{11}} + {a_{22}}{A_{12}} + {a_{23}}{A_{13}}$
D. ${a_{11}}{A_{11}} + {a_{21}}{A_{21}} + {a_{31}}{A_{31}}$

Answer 1

Hint:Use the determinant of a matrix method convert the $3 \times 3$ determinant matrix into $2 \times 2$ determinant matrix by taking first column as constant and convert the $2 \times 2$ determinant matrix into cofactor and use the formula ${A_{ij}} = {( - 1)^{i + j}}\det {M_{ij}}$.

Complete step-by-step answer:
Cofactor $({A_{ij}})$ : Cofactor is a number which is obtained by eliminating the elements of designated ${i^{th}}$ row and ${j^{th}}$ column of a matrix , cofactor can have negative or positive sign depending upon the designated element
Minor $({M_{ij}})$ : Minor is a matrix of order $(n - 1) \times (n - 1)$ which is obtained by deleting ${i^{th}}$ row and ${j^{th}}$column of a matrix of order $n \times n$
Formula $ \Rightarrow $ ${A_{ij}} = {( - 1)^{i + j}}\det {M_{ij}}$
Transpose of a matrix : A transpose matrix is a matrix obtained by interchanging rows and columns of a matrix . The new elements in the transpose matrix will be of the form
${A^T} = \left[ {{a_{ji}}} \right]$
Given determinant matrix $\vartriangle $ $ = $ $\left| {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right|$
since we know that determinant of a matrix is equal to determinant of its transpose matrix
$\vartriangle = {\vartriangle ^T}$
By the definition of determinant converting $3 \times 3$ determinant matrix into $2 \times 2$ determinant matrix by taking first row as constant
${\vartriangle ^T}$ = ${a_{11}}\left( {\begin{array}{*{20}{c}}
  {{a_{22}}}&{{a_{23}}} \\
  {{a_{32}}}&{{a_{33}}}
\end{array}} \right) - {a_{21}}\left( {\begin{array}{*{20}{c}}
  {{a_{12}}}&{{a_{13}}} \\
  {{a_{32}}}&{{a_{33}}}
\end{array}} \right) + {a_{31}}\left( {\begin{array}{*{20}{c}}
  {{a_{12}}}&{{a_{13}}} \\
  {{a_{22}}}&{{a_{23}}}
\end{array}} \right)$
And by the definition of minor matrix we can obtain the terms of ${M_{11}},{M_{21}},{M_{31}}$
${M_{11}} = $ $\left( {\begin{array}{*{20}{c}}
  {{a_{22}}}&{{a_{23}}} \\
  {{a_{32}}}&{{a_{33}}}
\end{array}} \right)$ ${M_{21}} = $ $\left( {\begin{array}{*{20}{c}}
  {{a_{12}}}&{{a_{13}}} \\
  {{a_{32}}}&{{a_{33}}}
\end{array}} \right)$ ${M_{31}} = $ $\left( {\begin{array}{*{20}{c}}
  {{a_{12}}}&{{a_{13}}} \\
  {{a_{22}}}&{{a_{23}}}
\end{array}} \right)$
By the definition of cofactor and by using the formula of co factor ${A_{ij}} = {( - 1)^{i + j}}\det {M_{ij}}$ we get
${A_{11}} = $ ${( - 1)^{1 + 1}}$$\left( {\begin{array}{*{20}{c}}
  {{a_{22}}}&{{a_{23}}} \\
  {{a_{32}}}&{{a_{33}}}
\end{array}} \right)$ ${A_{21}} = $ ${( - 1)^{2 + 1}}\left( {\begin{array}{*{20}{c}}
  {{a_{12}}}&{{a_{13}}} \\
  {{a_{32}}}&{{a_{33}}}
\end{array}} \right)$ ${A_{31}} = $ ${( - 1)^{3 + 1}}\left( {\begin{array}{*{20}{c}}
  {{a_{12}}}&{{a_{13}}} \\
  {{a_{22}}}&{{a_{23}}}
\end{array}} \right)$
${A_{11}} = $ $\left( {\begin{array}{*{20}{c}}
  {{a_{22}}}&{{a_{23}}} \\
  {{a_{32}}}&{{a_{33}}}
\end{array}} \right)$ ${A_{21}} = $ $\left( {\begin{array}{*{20}{c}}
  {{a_{12}}}&{{a_{13}}} \\
  {{a_{32}}}&{{a_{33}}}
\end{array}} \right)$ ${A_{31}} = $ $\left( {\begin{array}{*{20}{c}}
  {{a_{12}}}&{{a_{13}}} \\
  {{a_{22}}}&{{a_{23}}}
\end{array}} \right)$
Substituting the cofactors ${A_{11}},{A_{21}},{A_{31}}$
$\vartriangle $ = ${a_{11}}\left( {\begin{array}{*{20}{c}}
  {{a_{22}}}&{{a_{23}}} \\
  {{a_{32}}}&{{a_{33}}}
\end{array}} \right) + {a_{21}}( - 1)\left( {\begin{array}{*{20}{c}}
  {{a_{12}}}&{{a_{13}}} \\
  {{a_{32}}}&{{a_{33}}}
\end{array}} \right) + {a_{31}}\left( {\begin{array}{*{20}{c}}
  {{a_{12}}}&{{a_{13}}} \\
  {{a_{22}}}&{{a_{23}}}
\end{array}} \right)$
After substituting the values of ${A_{11}},{A_{21}},{A_{31}}$
Therefore the determinant of the above question is equivalent to
 $\vartriangle = $ ${a_{11}}{A_{11}} + {a_{21}}{A_{21}} + {a_{31}}{A_{31}}$

So, the correct answer is “Option A”.

Note:Determinant of a matrix AB is equivalent to the determinant of the matrix A multiplied with determinant of matrix B.If each element of a row or a column of a square matrix is multiplied by a number K , then the determinant of the matrix obtained is K times the determinant of the given matrix.If two rows or columns of matrix are interchanged, then the sign of matrix changes.Be cautious with the positive and negative sign before cofactor with a small mistake in positive and negative sign the entire question can be wrong.