Test: Minors And Cofactors

Only non-singular matrices have inverses.
Suppose that A is invertible. This means that we have the inverse matrix A⁻¹ of A.
Consider the equation Ax=0. We show that this equation has only zero solution.
Multiplying it by　A⁻¹ on the left, we obtain
A⁻¹ Ax = A^-1 0 
⇒ x = 0

|kA| = kⁿ|A|
|3A| = (3)ⁿ |A|
= 27|A| = k|A|
k = 27

Correct Answer :- d

Explanation:- 3(6-6) -2(6-9) +3(4-6)

= 3(0) + 6 - 6

= 0

Co factor of element 9 will be {(6 * 8) -(-5 * 0)}
= 48

A minor, Mij, of the element aij is the determinant of the matrix obtained by deleting the ith row and jth column.

A21 = {(0, -1) (3,4)}
⇒ (0 - (-3)) = 3
A21 = (-1)⁽²⁺¹⁾ (3)
= -3

½{(0,0,1) (1,3,1) (x,y,1)} = 0
{(0,0,1) (1,3,1) (x,y,1)} = 0/(½)
{(0,0,1) (1,3,1) (x,y,1)} = 0
0(3-y) -0(1-x) +1(y-3x) = 0
=> y - 3x = 0
=> y = 3x

Δ = Sum of products of element of row(or column)  with their corresponding co-factors.
Δ = a11 A11 + a21 A21 + a31 A31