Maths Determinants
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New answer posted
a month agoIf Δ =
|x-2 2x-3 3x-4|
|2x-3 3x-4 4x-5|
|3x-5 5x-8 10x-17|
= Ax³ + Bx² + Cx + D, then B + C is equal to:
Contributor-Level 10
Δ = | x-2 2x-3 3x-4 |
| 2x-3 3x-4 4x-5 |
| 3x-5 5x-8 10x-17|
= Ax³ + Bx² + Cx + D.
R? → R? - R? , R? → R? - R?
Δ = | x-2 2x-3 3x-4 |
| x-1 |
| x-2 (x-2) 6 (x-2) |
= (x-1) (x-2) | 1 2x-3 3x-4 |
| 1 |
| 1 2 6 |
= -3 (x - 1)² (x - 2) = -3x³ + 12x² - 15x + 6
∴ B + C = 12 - 15 = -3
New answer posted
a month agoContributor-Level 10
Find the area of a triangle with vertices A(?2,?3), B(4,0), and C(1,5).
New answer posted
a month agoContributor-Level 10
This is a mnemonic method that computes the determinant of 3*3 times. It involves drawing diagonal lines for remembring how to multiply and sum elements of the matrix. Say there is a matrix:
| a b c |
| d e f |
| g h i |
We will first multiply the elements that are connected by three diagonals that run from top-left to bottom right:
a*e*i
b*f*g
c*d*h
Let us now sum these products: (aei)+ (bfg)+ (cdh)
Now, we will multiply the elements connected by three diagonals from top-right to bottom left:
c*e*g
b*d*i
a*f*h
Let us now sum these products:
(ceg)+ (bdi)+ (afh)
After this, let us subtract the sum of negative terms from the sum of positive terms:
Deter
New answer posted
a month agoContributor-Level 9
M = {A = [a b; c d] | a, b, c, d ∈ {±3, ±2, ±1,0}
f (A) = det (A) = ad-bc=15
Case 1: ad=9 (2 ways: 3*3, -3*-3), bc=-6 (4 ways: 3*-2, -3*2, 2*-3, -2*3). Total = 2*4=8.
Case 2: ad=6 (4 ways), bc=-9 (2 ways). Total = 4*2=8.
Total no. of possible such cases = 8+8=16.
New question posted
a month agoNew answer posted
a month agoContributor-Level 9
Note D = |3, 4, 5; 1, 2, 3; 4, 4| = 0 (R? → R? - 2R? + 3R? )
Now let P? = 4x+4y+4z-δ=0. If the system has solutions it will have infinite solution, so P? = αP? + βP?
Hence 3α+β=4 and 4α+2β=4 ⇒ α=2 and β=-2
So for infinite solution 2µ-2=δ ⇒ for 2µ ≠ δ+2 System inconsistent
New answer posted
a month agoContributor-Level 10
So D = 0 → |, [1,3, k²], | = 0 ⇒ k² = 9
x + y + 3z = 0
x + 3y + 9z = 0
3x + y + 3z = 0
(1)- (3)
x = 0 ⇒ y + 3z = 0
y/z = -3
So x + (y/z) = -3
New answer posted
a month agoContributor-Level 10
f (x) = |sin²x, -2+cos²x, cos2x; 2+sin²x, cos²x, cos2x; sin²x, cos²x, 1+cos2x|.
R? →R? -R? , R? →R? -R?
f (x) = |sin²x, -2+cos²x, cos2x; 2, 2-2cos²x, 0; 0, 2-2cos²x, 1|.
f (x) = sin²x (2-2cos²x) - (-2+cos²x) (2) + cos2x (2 (2-2cos²x).
This seems tedious. From the solution, f (x)=4+2cos2x.
Max value when cos2x=1, f (x)=6.
New answer posted
a month agoContributor-Level 10
R? → R? -R? , R? → R? -R?
|sinx-cosx, cosx-sinx, 0; 0, sinx-cosx, cosx-sinx; cosx, sinx| = 0
(sinx-cosx)² |1, -1, 0; 0, 1, -1; cosx, sinx| = 0
(sinx-cosx)² (1 (sinx+cosx) + 1 (cosx) = 0
(sinx-cosx)² (sinx + 2cosx) = 0
sin x = cos x
tan x = 1 ⇒ x = π/4
or
sin x = -2cos x
tan x = -2
Not within given range.
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