[[[x+y,2],[5+x,0]]=[[6,2],[5,1]],([4,3],[x,5]]=[[y,z],[1,5]]],[,[[2a+b,a-2b],[5c-d,4c-14]]=[[4,-3],[11,24]]]

The solutions of the linear system of [[2,2,2],[7,1,1],[0,6,5]][[x],[y],[z]]=[[3y+11],[6z-1],[5y+11]]+[[x],[x],[4z]]+[[z],[3x],[4y]] is

If A=[[1,-2],[3,-5][-6,0]],B=[-1,-2],[4,2],[1,5]] and C=[[2,4],[-1,-4],[-3,6]] ,then find the matrix X such that 3A-4B+5X=C

Find the value of x 'if : [[2x-y,5],[3,y]]=[[6,5],[3,-2]] a.1 b.2 c.6 d.5

If [(2a+b,a-2b),(5c-d,4c+3d)]=[(4,-3),(11,24)] , then value of a+b-c +2d is :

If [(2a+b, a-2b),(5c-d, 4c+3d)]=[(4,-3),(11,24)] , then value of a + b – c + 2d is:

If [[1,2],[3,1]][[x],[y]] = [[5],[5]] , then value of x+y is (A) 4 (B) 2 (C) 0 (D) 3

If [[1,2],[3,1]][[x],[y]] = [[5],[5]] , then value of x+y is (A) 4 (B) 2 (C) 0 (D) 3

If X= [(2,3,0),(1,-1,5)] ,Y= [(-1,-2,3),(-1,0,2)] ,Z= [(0,4,-1),(5,6,-5)] find 3X+4Y-Z

If A=[(9,1),(4,3)],B=[(1,5),(6,11)] and 3A+5B+2X=0 then X =