[9-1,21+3y=2],[62+(1-2b)y=6]

(a - 1) x + 3y = 2 , 6x + (1 - 2b) y =6 .

If [(x+3,z+4,2y-7),(-6,a-1,0),(b-3,-21,0)] =[(0,6,3y-2),(-6,-3,2c+2),(2b+4,-21,0)] Find the values of a, b, c,x,y and z

If x != y != z and |[[x,x^2,1+x^3],[y,y^2,1+y^3],[z,z^2,1+z^3]]|=0 then using properties of determinants, show that xyz= -1.

If A=[(1,-2,0),(2,1,3),(0,-2,1)] and B=[(7,2,-6),(-2,1,-3),(-4,2,5)] , find AB Hence , solve the system of equation x-2y=10, 2x+y+3z=8 and -2y+z=7.

If A=[{:(1,-2,0),(2,1,3),(0,-2,1):}],B=[{:(7,2,-6),(-2,1,-3),(-4,2,5):}] , find AB Also solve x-2 y=10, 2x+y+3z=8, -2y+z=7

The x-coordinates of the vertices of a square of unit area are the roots of the equation x^2-3|x|+2=0 . The y-coordinates of the vertices are the roots of the equation y^2-3y+2=0. Then the possible vertices of the square is/are (a)(1,1),(2,1),(2,2),(1,2) (b)(-1,1),(-2,1),(-2,2),(-1,2) (c)(2,1),(1,-1),(1,2),(2,2) (d)(-2,1),(-1,-1),(-1,2),(-2,2)

The x-coordinates of the vertices of a square of unit area are the roots of the equation x^2-3|x|+2=0 . The y-coordinates of the vertices are the roots of the equation y^2-3y+2=0. Then the possible vertices of the square is/are (a)(1,1),(2,1),(2,2),(1,2) (b)(-1,1),(-2,1),(-2,2),(-1,2) (c)(2,1),(1,-1),(1,2),(2,2) (d)(-2,1),(-1,-1),(-1,2),(-2,2)