⇒ (x-y)(y-z)[y2 + yz+z2 - x2 - xy - y2]
⇒ (x-y)(y-z) (z-x) (x+y+z)
If , then
is equal to
=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 - R1,
⇒ (x-3) (6x -9) = 0 ⇒x =
If the order of matrix A is m*p. And the order of B is p×n. Then the order of matrix AB is ?
If each element of a 3 × 3 matrix A is multiplied by 3 , then the determinant of the newly formed matrix is
One root of the equation
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
The determinant is equal to
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.
If A is a non singular matrix of order 3 , then |adj(adjA)|
where n is order of matrix. Here n = 3.
The only integral root of the equation det. is
Clearly , y = 1 satisfies it. [C3 = 3C1]
If 1/a+1/b+ 1/c = 0 , then
Since ,
If A is a non singular matrix and A’ denotes the transpose of A , then
Because , |A|=|A′|
If the matrix AB = O , then
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.
If A ,B andC be the three square matrices such that A = B + C , then Det A is equal to
Because , |A| ≠ |B|+|C|
Solution set of the equation
[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
A determinant is unaltered , if
This is because of the elementary transformations of determinants . The value of determinant remains unaffected by applying elementary transformations.
If A and B are square matrices of order 3 , such that Det.A = –1 , Det.B = 3 then the determinant of 3AB is equal to
|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.
If 1 , ω,ω2 are cube roots of unity , then has value
write 1 as in R1 and take out common from R1 , we get :
because row 1 and row 3 are identical.
Solution set of the equation
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
The system AX = B of n equations in n unknowns has infinitely many solutions if
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.
If A is a square matrix such that A3 = I , then A−1 is equal to
A3 = I ⇒ Pre - multiplying both sides by A−1,A−1, A3 = A−1 I ⇒ A2 = A−1
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