If the value of the determinant , then the value of
is
Determinant = -2(-12 + 12) -2(-6 - 2) +1(-12 - 4)
= - 2(0) - 2(-8) + 1(16)
= 0 + 16 + 16
= 32
The value of x, if = 0 , is
If the value of , then the value of
If you interchange (switch) two rows (or columns) of a matrix A to get B, then det(B) = –det(A). Hence, the determinant is -(-7)
= 7
The operation that will simplify the Determinant
by taking (a+b+c) common from the first Row is
R1 ------> R1 + R2, taking (a+b+c) common from the first row, we get the resultant matrix.
The value of the determinant is
|A|=3((4)-(4))-2((-4)-(2))-1((-8)-(-4))
|A|= -2(-6)-(-6)
|A|= -2(0)
=0
The value of the Δ
If the value of the determinant is equal to 14, then the value of the determinant
is equal to
Taking 3 common from the column I in second determinant, we get the 14 * 3 = 42
The value of is
A = {(5x+2, 2x, 2x) (5x+2, x+2, 2x) (5x+2, 2x, x+2)}
= (5x+2) {(1, 2x, 2x) (1, x+2, 2x) (1, 2x, x+2)}
= (5x+2) [(1(x+2)^2 - 4x^2) - 2x(x+2 - 2x) + 2x(2x - x - 2)]
= (5x+2) [(x+2 - 2x)(x+2+2x)-2x(2 - x) + 2x(x-2)]
= (5x+2) [(2-x)(3x+2) -2x(2-x) -2x(2-x)]
= (5x+2) (2-x)[3x+2-2x-2x]
= (5x+2) (2-x)2
Given
+ Δ. then Δ is
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