Test: Determinants - 2 - JEE MCQ

⇒ (x-y)(y-z)[y² + yz+z² - x² - xy - y²]

⇒ (x-y)(y-z) (z-x) (x+y+z)

=1(-5) -2(10) + 3(11)
=-5-20+33 = 8

=1(-30) - 6(20) + 3(66)

= -30-120+198= 48

D = 8 ⇒ 6D = 48

Apply, R₂ →R₂ - R₁,

Apply, R₃ →R₃ - 4R₁,

⇒ (x-3) (6x -9)  = 0 ⇒x = 

Apply , R₁→R₁ +R₂+R₃,
 

⇒either (3x -2) = 0

Apply , C₁ → C₁+C₂+C₃+C₄

Apply, R1 → R₁ - R₂,

Apply, R₁→R₁ - R₂

⇒ (10+ x) x³

Apply, C₁ → C₁ + C₂ + C₃,

The determinant of a lower triangular (or an upper triangular matrix is equal to the product of the diagonal elements.

where n is order of matrix. Here n = 3.

Clearly , y = 1 satisfies it. [C₃ = 3C₁]

Since , 

Because , |A|=|A′|

If the matrix AB = O , then , marix A can be a non zero matrix as well as matrix B can be a non zero matrix. Which means det.A = 0 and det.B = 0.

Because , |A| ≠ |B|+|C|

[x(-3x(x+2) - 2x(x-3)]+6[2(x+2)+3(x-3)]-1(4x-9x) = 0

⇒ - 5x³ + 35x - 30 = 0
⇒ (x-1)(x-2)(x+3) = 0 ⇒ x=1,2,-3

This is because of the elementary transformations of determinants . The value of determinant remains unaffected by applying elementary transformations.

|3AB| = 27|A||B| = 27(-1)(3) = -81

Because , the determinant of a skew symmetric matrix of odd order is always zero and of even order is a non zero perfect square.

write 1 as in R₁ and take out  common from R₁ , we get : 

because row 1 and row 3 are identical.

x(x² - 12) - 3(2x - 14)+7(12 - 7x) = 0

⇒ x³ - 67x + 126 = 0

⇒(x-2)(x-7)(x+9) = 0 ⇒x = 2,7,-9

Explanation here if det. A = 0 , (adj A) B = O ⇒ The system AX = B of n equations in n unknowns may be consistent with infinitely many solutions or it may be inconsistent.

A³ = I ⇒  Pre - multiplying both sides by A⁻¹,A⁻¹, A³ = A⁻¹ I ⇒ A² = A⁻¹