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#### explain solution RD Sharma class 12 chapter 5 Determinants exercise Fill in the blanks question 12 maths

Hint: Here, we use basic concept of determinant of matrix

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

$|A|=5 \; \; C_{i j}=\operatorname{cofactor} \text { of } a_{i j} . \text { So, let } C_{i j}=A_{i j}, C=A$

Solution: A is $\left | a_{ij} \right |$ be a square matrix

So, $A=\left(\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right) 1 \leq i \leq 3$

\begin{aligned} &|A|=5 \\ &|A|=a_{11}\left[\begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right]-a_{12}\left[\begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array}\right]+a_{13}\left[\begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array}\right] \\ &|A|=a_{11}\left[\begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right]+a_{12}\left[\begin{array}{cc} a_{23} & a_{21} \\ a_{33} & a_{31} \end{array}\right]+a_{13}\left[\begin{array}{ll} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array}\right] \end{aligned}

[Interchange column of a12]

\begin{aligned} &\Rightarrow|A|=a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13} \\ &5=a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13} \\ &5=a_{11} C_{11}+a_{12} C_{12}+a_{13} C_{13} \quad[A=C] \end{aligned}