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.

Let A=[a?]?. tr(AA?)=3.
Σa?² = 3.
Possible cases for non-zero elements are (1,1,1) or (-1,-1,-1) or (1,1,-1) etc.
Total combinations = ?C? * 2³ = 84 * 8 = 672.C? * 2³ = 84 * 8 = 672.

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

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

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