Target Exam

CUET

Subject

Maths. Section B1

Chapter

Determinants

Question:

Find minors and cofactors of all the elements of the determinant $\begin{vmatrix} 1 & -2 \\ 4 & 3 \end{vmatrix}$

Options:

Minors: $M_{11}=3, M_{12}=4, M_{21}=-2, M_{22}=1$
Cofactors: $A_{11}=3, A_{12}=-4, A_{21}=2, A_{22}=1$

Minors: $M_{11}=3, M_{12}=-2, M_{21}=4, M_{22}=1$
Cofactors: $A_{11}=3, A_{12}=2, A_{21}=-4, A_{22}=1$

Minors: $M_{11}=1, M_{12}=-2, M_{21}=4, M_{22}=3$
Cofactors: $A_{11}=1, A_{12}=2, A_{21}=-4, A_{22}=3$

Minors: $M_{11}=3, M_{12}=4, M_{21}=-2, M_{22}=1$
Cofactors: $A_{11}=-3, A_{12}=4, A_{21}=-2, A_{22}=-1$

Correct Answer:

Minors: $M_{11}=3, M_{12}=4, M_{21}=-2, M_{22}=1$
Cofactors: $A_{11}=3, A_{12}=-4, A_{21}=2, A_{22}=1$

Explanation:

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

$A_{11} = (-1)^{1+1} M_{11} = (-1)^2 (3) = 3$

$A_{12} = (-1)^{1+2} M_{12} = (-1)^3 (4) = -4$

$A_{21} = (-1)^{2+1} M_{21} = (-1)^3 (-2) = 2$

$A_{22} = (-1)^{2+2} M_{22} = (-1)^4 (1) = 1$