Find minors and cofactors of all the elements of the determinant $\begin{vmatrix} 1 & -2 \\ 4 & 3 \end{vmatrix}$ |
Minors: $M_{11}=3, M_{12}=4, M_{21}=-2, M_{22}=1$ Minors: $M_{11}=3, M_{12}=-2, M_{21}=4, M_{22}=1$ Minors: $M_{11}=1, M_{12}=-2, M_{21}=4, M_{22}=3$ Minors: $M_{11}=3, M_{12}=4, M_{21}=-2, M_{22}=1$ |
Minors: $M_{11}=3, M_{12}=4, M_{21}=-2, M_{22}=1$ |
The correct answer is Option (4) → Minors: $M_{11}=3, M_{12}=4, M_{21}=-2, M_{22}=1$ Cofactors: $A_{11}=3, A_{12}=-4, A_{21}=2, A_{22}=1$ ## Minor of the element $a_{ij}$ is $M_{ij}$ Here $a_{11} = 1$. So $M_{11} = \text{Minor of } a_{11} = 3$ $M_{12} = \text{Minor of the element } a_{12} = 4$ $M_{21} = \text{Minor of the element } a_{21} = -2$ $M_{22} = \text{Minor of the element } a_{22} = 1$ Now, cofactor of $a_{ij}$ is $A_{ij}$. So $A_{11} = (-1)^{1+1} M_{11} = (-1)^2 (3) = 3$ $A_{12} = (-1)^{1+2} M_{12} = (-1)^3 (4) = -4$ $A_{21} = (-1)^{2+1} M_{21} = (-1)^3 (-2) = 2$ $A_{22} = (-1)^{2+2} M_{22} = (-1)^4 (1) = 1$ |