`[(2x,3)][(1,2),(-3,0)][(x),(3)]=0`

If [ (2x, 3)] [(1,2),(-3,0)][(x),(8)]=0 , find 'x'

If [2x" "3][(1,2),(-3,0)][(x),(3)]=O , find x.

Solve [(1,2),(0,0)] [(x,2),(3,3)]=[(7,0),(1,0)] , and find the value of x.

If [(1,x,1)][(1,3,2),(0,5,1),(0,3,2)][(x),(1),(-2)]=[0] then x is

The roots of the equation det [(1-x, 2, 3),(0,2-x,0),(0,2,3-x)]=0 are (A) 1 and 2 (B) 1 and 3 (C) 2 and 3 (D) 1,2, and 3

The roots of the equation det [(1-x, 2, 3),(0,2-x,0),(0,2,3-x)]=0 are (A) 1 and 2 (B) 1 and 3 (C) 2 and 3 (D) 1,2, and 3

Let A=[(1,2,3),(2,2,-1),(3,0,k)] and f(x)=x^3-2x^2-alphax+beta=0.If A satisfies f(x)=0, then

If [1" "x" "1] [(1,2,3),(0,5,1),(0,3,2)][(x),(1),(-2)]=0 then find the value of x.