Practicing Success
For $A=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$ if $M_{i j}$ and $A_{i j}$ are the minor and the cofactor of $a_{i j}$ respectively, then |A| equals to : (A) $a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13}$ (B) $a_{21} A_{12}+a_{22} A_{22}+a_{23} A_{32}$ (C) $a_{11} M_{11}-a_{12} M_{12}+a_{13} M_{13}$ (D) $a_{11} M_{11}+a_{12} M_{12}+a_{13} M_{13}$ Choose the correct answer from the options given below : |
(A), (B) Only (A), (C) Only (C), (D) Only (A), (C), (D) Only |
(A), (C) Only |
$A=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$ $M_{ij}$ = determinant of matrix formed by eliminating ith row and jth column $A_{ij} = (-1)^{i+j} M_{ij}$ → Relation between minor and cofactor sum of products of corresponding aij and Aij = |A| ⇒ |A| = a11 A11 + a12 A12 + a13 A13 = a11(-1)1+1 M11 + a12(-1)1+2 M12 + a13(-1)1+3 M13 |A| = a11 M11 - a12 M12 + a13 M13 Option → (2) (A), (C) |