[" The solution of "(x,y,z)" the equation "],[[[1,0,1],[-1,1,0],[0,-1,1]][[x],[y],[z]]=[[1],[1],[2]]" is "(x,y,z)]

The solutiion of (x,y,z) the equation [(-1,0,1),(-1,1,0),(0,-1,1)][(x),(y),(z)]=[(1),(1),(2)] is (x,y,z)

The solutiion of (x,y,z) the equation [(-1,0,1),(-1,1,0),(0,-1,1)][(x),(y),(z)]=[(1),(1),(2)] is (x,y,z)

If [[1,0,0],[0,1,0],[0,0,1]][[x],[y],[z]]=[[1],[-1],[0]] then value of (x+y+z)is

If [[1,0,0],[0,y,0],[0,0,1]][[x],[-1],[z]]=[[1],[2],[1]], then the value of x+y+z is

If [[1 1 1],[0 1 1],[0 0 1]] [[x],[y],[z]] = [[6],[3],[2]] ,then 2x+y-z=?

.If [[x-1,2,y-5],[z,0,2],[1,-1,1+a]]=[[1-x,2,-y],[2,0,2],[1,-1,1]] then find x,y,z,a

Write the value of x+y+z , if [(1,0,0),(0,1,0),(0,0,1)][(x),(y),(z)]=[(1),(-1),(0)]

If x ,\ y ,\ z are non-zero real numbers, then the inverse of the matrix A=[[x,0, 0],[ 0,y,0],[ 0, 0,z]] , is (a) [[x^(-1),0 ,0 ],[0,y^(-1),0],[ 0, 0,z^(-1)]] (b) x y z[[x^(-1),0 ,0],[ 0,y^(-1),0],[ 0, 0,z^(-1)]] (c) 1/(x y z)[[x,0, 0],[ 0,y,0],[ 0, 0,z]] (d) 1/(x y z)[[1, 0, 0],[ 0 ,1, 0],[ 0, 0, 1]]

If [(1,0,0),(0, 0, 1),(0,1,0)][(x),(y),(z)]=[(2),(-1),(3)] , find x ,\ y ,\ z .

If [(1,2,-3),(0,4,5),(0,0,1)][(x),(y),(z)]=[(1),(1),(1)] , then (x, y, z) is equal to