#### Please solve RD Sharma class 12 chapter 5 Determinants exercise Fill in the blanks question 13 maths textbook solution

Hint: Here we use basic concept of determinant of matrix

Given: A is $3\times 3$ matrix. So, n = 3

\begin{aligned} &|A|=5\\ &C_{i y} \text { is cofactor of } C_{11 j} \end{aligned}

Solution:

\begin{aligned} &A_{i j}=\left(\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right) \\ &\text { and cofactor }\left(A_{i j}\right)=(-1)^{i+j}|\operatorname{mij}| \end{aligned}

So, here

\begin{aligned} &a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23} \\ &A_{21}=(-1)^{2+1}\left[a_{12} a_{33}-a_{13} a_{32}\right] \\ &A_{22}=(-1)^{2+2}\left[a_{11} a_{33}-a_{13} a_{31}\right] \\ &A_{23}=(-1)^{2+3}\left[a_{11} a_{32}-a_{12} a_{31}\right] \end{aligned}

\begin{aligned} &=a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23} \\ &=\left(-a_{11} a_{12} a_{33}\right)+a_{11} a_{13} a_{32}+a_{11} a_{12} a_{33}-a_{12} a_{13} a_{31}-a_{12} a_{13} a_{32}+a_{12} a_{13} a_{31} \\ &=\left[a_{11} a_{12} a_{33}-a_{11} a_{12} a_{33}\right]+\left[a_{11} a_{13} a_{32}-a_{11} a_{13} a_{32}\right]+\left[a_{12} a_{13} a_{31}-a_{12} a_{13} a_{31}\right] \\ &\text { So } 0+0+0 \\ &=0 \end{aligned}