MHT CET · Maths · Determinants
If \(\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{3 \times 3}=\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 1 & 5 \\ 2 & 4 & 7\end{array}\right]\) and \(\mathrm{Aij}\) is a cofactor of \(\mathrm{a}_{\mathrm{ij}}\), then \(a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23}\) is equal to
- A \(-1\)
- B \(2\)
- C \(0\)
- D \(1\)
Answer & Solution
Correct Answer
(C) \(0\)
Step-by-step Solution
Detailed explanation
If we multiply elements of one row to the corresponding cofactors of another row and add them then we get zero hence,
\(a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23}=0\)
\(a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23}=0\)
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