Maths Matrices

Kindly consider the following Image

A? A = I
⇒ a²+b²+c²=1 and ab+bc+ca=0
Now, (a+b+c)²=1 ⇒ a+b+c=±1
So, a³+b³+c³-3abc = (a+b+c) (a²+b²+c²-ab-bc-ca) = (±1) (1-0)=±1
⇒ 3abc = 2±1 = 3,1
⇒ abc = 1, 1/3

2x-y+2z=2
x-2y+λz=-4
x+λy+z=4
For no solution:
D=|2, -1,2; 1, -2, λ 1,1,1|=0
⇒ 2 (-2-λ²)+1 (1-λ)+2 (λ+2)=0
⇒ -2λ²+λ+1=0
⇒ λ=1, -1/2
D? =|2, -1,2; -4, -2, λ 4,1,1|
=2 (-2-λ)+1 (-4-4λ)+2 (-4+8)
=2 (1+λ) which is not equal to zero for λ=1, -1/2

|A|≠0
For (P): A≠I?
So, A = [1 0; 0 1] or [1; 0 1] or [1 0; 1]
or [1; 1 0]
So (P) is false.
A = [1 0; 1 0] or [1; 0 1] or [1 0; 1]
⇒ tr (A)=2
⇒ Q is true

2? -2? =112. m=7, n=4. mn=28.

System of equations can be written as
[2 3 6; 1 2 a; 3 5 9] [x; y; z] = [8; 5; b]
R? →R? -1/2R? , R? →R? -3/2R?
. the system will have no solution if 3-a=0 and b-13≠0.
i.e. for a=3 & b≠13.

$\mathrm{}\left|B\right|=\left|\begin{array}{lll}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right|=\left|\begin{array}{lll}{3}^0{a}_{11}& 3^1{a}_{21}& 3^2{a}_{31}\\ 3^1{a}_{12}& 3^2{a}_{22}& 3^3{a}_{32}\\ 3^2{a}_{13}& 3^3{a}_{23}& 3^4{a}_{33}\end{array}\right|$

$81=3^3\cdot 3^3\cdot 3^2\left|\mathrm{A}\right|$

$\Rightarrow \left|A\right|=\frac{1}{9}$

Δ = |1,1, -1; 1,2, α 2, -1,1| = 1 (2+α)-1 (1-2α)-1 (-1-4) = 2+α+2α-1+5 = 3α+6=0 ⇒ α=-2.
Δ? = |2,1, -1; 1,2, α β, -1,1| = 2 (2+α)-1 (1-αβ)-1 (-1-2β) = 4+2α-1+αβ+1+2β = 4+2α+αβ+2β=0.
4-4-2β+2β=0. This holds.
Δ? = |1,2, -1; 1,1, α 2, β,1| = 1 (1-αβ)-2 (1-2α)-1 (β-2) = 1-αβ-2+4α-β+2 = 1+4α-αβ-β=0.
1-8+2β-β=0 ⇒ -7+β=0 ⇒ β=7.
α+β = -2+7 = 5.

A = [, [-1, 4]. |A| = 2 - 1 = 1.
Characteristic equation: λ² - tr (A)λ + |A| = 0 ⇒ λ² - 3λ + 1 = 0.
By Cayley-Hamilton, A² - 3A + I = 0. A? ¹ (A² - 3A + I) = A - 3I + A? ¹ = 0.
A? ¹ = 3I - A.
Comparing with A? ¹ = αI + βA, we get α=3, β=-1.
4 (α - β) = 4 (3 - (-1) = 16.