Consider the system of the equation `k x+y+z=1,x+k y+z=k ,a n dx+y+k z=k^2dot` Statement 1: System equations has infinite solutions when `k=1.` Statement 2: If the determinant `|1 1 1k k1k^2 1k|=0,` t hen `k=-1.`

Consider the system of equations x-2y+3z=-1, x-3y+4z=1 and -x+y-2z=k Statement 1: The system of equation has no solution for k!=3 and Statement 2: The determinant |[1,3,-1], [-1,-2,k] , [1,4,1]| !=0 for k!=0

Consider the system of equations x-2y+3z=1;-x+y-2z=; x-3y+4z=1. Statement 1: The system of equations has no solution for k!=3. Statement2: The determinant |1 3-1-1-2k1 4 1|!=0,\ for\ k!=3. Statement 1 is True: Statement 2 is True; Statement 2 is a correct explanation for statement 1 Statement 1 is true, Statement 2 is true;2 Statement 2 not a correct explanation for statement 1. Statement 1 is true, statement 2 is false Statement 1 is false, statement 2 is true

Consider the system of equations x-2y+3z=-1 -x+y-2z=k x-3y+4z=1 Statement -1 The system of equation has no solutions for k ne 3 . statement -2 The determinant |{:(1,3,-1),(-1,-2,k),(1,4,1):}| ne0, "for"" " kne3.

The system of equation kx+y+z=1, x +ky+z=k and x+y+kz=k^(2) has no solution if k equals

The system of equation kx + y + z = 1, x + ky + z = k and x + y + kz = k^(2) has no solution if k equals :

The system of equations kx + y + z =1, x + ky + z = k and x + y + zk = k^(2) has no solution if k is equal to :

kx+y+z=1 , x+ky+z=k , x+y+kz=k^2 be the system of equations with no solution , then k=

Consider the system of equations x-2y+3z=-1, -x+y-2z=k, x-3y+4z=1 Assertion: The system of equations has no solution for k!=3 and Reason: The determinant |(1,3,-1),(-1,-2,k),(1,4,1)|!=0, for k!=3 (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true R is not te correct explanation of A (C) A is true but R is false. (D) A is false but R is true.

For how many values of k. will the system of equations (k + 1) x + 8y = 4k and kx + (k + 3) y = 3k - 1 , have an infinite number of solutions