Maths Matrices
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New answer posted
3 weeks agoContributor-Level 10
all i = 1, 2, 3
Case 1 7 one's and two zeroes which can occur in
Case 2 One 2 three 1's five zeroes =
total such matrices = 504 + 36 = 540
New answer posted
3 weeks agoContributor-Level 10
(I + M)² = (I + M) (I + M) = I + 2M + M² = (I + 2M)
(I + M)³ = (I + 2M) (I + M) = I + 3M + 2M² = (I + 3M)
(I + M)? = I + 50M
det (I + M)? - 50M) = det (I) = 1
New answer posted
4 weeks agoContributor-Level 10
A = [1]
[0 1]
Now A² = [1] [1] = [1 2]
[0 1] [0 1] [0 1]
A³ = A².A = [1 2] [1] = [1 3]
[0 1] [0 1] [0 1]
Similarly A²? ¹¹ = [1 2011]
[0 1]
New answer posted
4 weeks agoContributor-Level 10
A² = [1 2 3; 0 1 2; 0 1]
A³=A².A= [1 3 6; 0 1 3; 0 1]
A²? = [1 20 1+2+3.20; 0 1 20; 0 1] = [1 20 210; 0 1 20; 0 1]
M= [20 210 520; 0 20 210; 0 20]
M (a? )=T? =n (n+1)/2
S? = 1/2 [ n (n+1) (2n+1)/6 + n (n+1)/2 ]
⇒S? =1540
⇒M=2020
New answer posted
4 weeks agoContributor-Level 10
A? =B? (i)
A³B²=A²B³. (ii)
Subtract (i) & (ii)
⇒A³ (A²−B²)=B³ (B²−A²)
⇒ (A²−B²) (A³+B³)=0
A²−B² is invertible matrix
∴A²−B²≠0
⇒A³+B³=0
∴? A³+B³? =0
New answer posted
4 weeks agoContributor-Level 10
x + 2y + z = 2
αx + 3y – z = α
–αx + y + 2z = –α
Δ = | (1, 2, 1), (α, 3, -1), (-α, 1, 2) | = 1 (6+1) – 2 (2α–α) + 1 (α+3α) = 7+2α
α = –7/2
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