For what values of p, the equations : x + y + z = 1; x + 2y + 4z = p & x + 4y + 10z = p2 have a solution? Solve them completely in each case. Share with your friends Share 4 Vijay Kumar Gupta answered this Dear Student, Consider the following system of equations. x+y+z=1 x+2y+4z=p x+4y+10z=p2The above system of equation can be written in the form AX=B, where A=1111241410, X=xyz, B=1pp2First find the determinanat of matrix A. A=1111241410=124410-114110+11214 =20-16-10-4+4-2 =4-6+2 =6-6 =0This means that the given system of equations does not have a unique solution.Either it has infinitely many solutions or no solutions.To check the same, we will consider adj A. In that case,If adj A B≠O Zero matrix, then the system of equations has no solutions.If adj A B=O Zero matrix, then the system of equations has infinitely many solutions.Find the cofactors of matrix A to get, A11=4, A12=-6, A13=2 A21=-6, A22=9, A23=-3 A31=2, A32=-3, A33=1Thus, adj A is given by adj A=A11A12A13A21A22A23A31A32A33T =4-62-69-32-31 Now adj A B=O implies that 4-62-69-32-31 1pp2=000⇒ 4-6p+2p2-6+9p-3p22-3p+p2=000⇒ 4-6p+2p2=0 ; -6+9p-3p2=0, 2-3p+p2=0⇒2 p2-3p+2=0 ; -3p2-3p+2=0, p2-3p+2=0Thus, we will get only two solutions of p from the equation p2-3p+2=0 p2-3p+2 =0⇒p2-2p-p+2=0⇒pp-2-1p-2=0⇒p-2p-1=0⇒p=1, 2Thus for p=1, 2, the given system of equations has infinitely many solutionsFor p=1, it becomes x+y+z=1 x+2y+4z=1 x+4y+10z=1Thw augumented matrix is, A:B=111 1124 11410 1Apply row operations: R2→R2-R1, R3→R3-R1 A:B=111 1013 0039 0 Apply row operations: R3→R3-3R2 A:B=111 1013 0000 0 This gives, y+3z=0 and x+y+z=1let z=k, which gives y=-3z=-3kAlso, x=1-y-z=1+3k-k=1+2kThus, the solutions can be represented by 1+2k, -3k, k ,where k∈ZSimilary we can found solutions when p=2. Regards 6 View Full Answer
