Q: Suppose p, q, r 0 and system of equation gathered (p+a) x+b y+c z=0 \\ a x+(q+b) y+c z=0 gathered a x+b y+(r+c) z=0, has a non-trivial solution, then the value of a p + b q + c r is

Let |A|= | array ccc p+a & b & c \\ a & q+b & c \\ a & b & r+c array | Also, given that equations has non trivial solution |A|=0 aligned & | array ccc p+a & b & c \\ a & q+b & c \\ a & b & r+c array |=0 \\ & (p+a)[(q+b)(r+c)-b c]-b[a(r+c)-c a] \\ & +c[a b-a(q+b)]=0 \\ & (p+a)[q r+q c+b r]-b[a r]+c[-a q]=0 \\ & p q r+p q c+p b r+a q r=0 aligned Dividing whole equation by p q r aligned & p q r p q r + p q c p q r + p r b p q r + q r a p q r & =0 \\ & 1+ c r + b q + a p & =0 \\ & a p + b q + c r &

Q: Given 2 x-y+2 z=2, x-2 y+z=-4, x+y+ z=4, then the value of such that the given system of equation has no solution is

Since, the system has no solution. | array ccc 2 & -1 & 2 \\ 1 & -2 & -1 \\ 1 & 1 & array |=0 2(-2 +1)+1( +1)+2(3)=0 -4 +2+ +1+6=0 3 =9 =3

Question 1

Suppose p, q, r 0 and system of equation gathered (p+a) x+b y+c z=0 \ a x+(q+b) y+c z=0 gathered a x+b y+(r+c) z=0, has a non-trivial solution, then the value of a p + b q + c r is

Accepted Answer

Let |A|= | array ccc p+a & b & c \ a & q+b & c \ a & b & r+c array | Also, given that equations has non trivial solution |A|=0 aligned & | array ccc p+a & b & c \ a & q+b & c \ a & b & r+c array |=0 \ & (p+a)[(q+b)(r+c)-b c]-b[a(r+c)-c a] \ & +c[a b-a(q+b)]=0 \ & (p+a)[q r+q c+b r]-b[a r]+c[-a q]=0 \ & p q r+p q c+p b r+a q r=0 aligned Dividing whole equation by p q r aligned & p q r p q r + p q c p q r + p r b p q r + q r a p q r & =0 \ & 1+ c r + b q + a p & =0 \ & a p + b q + c r &

Question 2

If x is a complex root of the equation | array ccc 1 & x & x \ x & 1 & x \ x & x & 1 array |+ | array ccc 1-x & 1 & 1 \ 1 & 1-x & 1 \ 1 & 1 & 1-x array |=0, then x^ 2007 +x^ -2007 =

Accepted Answer

Expanding the two determinants, we get (1-3 x^2+2 x^3 )+ (3 x^2-x^3 )=0 x^3+1=0 x=- ,- ^2,-1 x^ 2007 +x^ -2007 =-1-1=-2

Question 3

The equations x-y+2 z=4 3 x+y+4 z=6 x+y+z=1 have

Accepted Answer

Given equations, x-y+2 z=4 3 x+y+4 z=6 x+y+z=1 Let = | array ccc 1 & -1 & 2 \ 3 & 1 & 4 \ 1 & 1 & 1 array | =1(1-4)+1(3-4)+2(3-1) =-3-1+4=0 and _1= | array ccc 4 & -1 & 2 \ 6 & 1 & 4 \ 1 & 1 & 1 array | =4(1-4)+1(6-4)+2(6-1) =-12+2+10=0 Now, =0 and _1=0 These equations have infinitely many solutions.

Question 4

If the system of linear equations 2 x+y-z=7 x-3 y+2 z=1 ; x+4 y+ z=k where , k R has infinitely many solutions, then +k is equal to:

Accepted Answer

For getting infinite solutions D =0, D _1= D _2= D =0 then check all the three equations Let = | array ccc 2 & 1 & -1 \ 1 & -3 & 2 \ 1 & 4 & array |=0 =-3 And _1= | array ccc 7 & 1 & -1 \ 1 & -3 & 2 \ k & 4 & -3 array |=0 k=6 +k=3

Question 5

Given 2 x-y+2 z=2, x-2 y+z=-4, x+y+ z=4, then the value of such that the given system of equation has no solution is

Accepted Answer

Since, the system has no solution. | array ccc 2 & -1 & 2 \ 1 & -2 & -1 \ 1 & 1 & array |=0 2(-2 +1)+1( +1)+2(3)=0 -4 +2+ +1+6=0 3 =9 =3

Question 6

If p a, q b, r c and the system of equations aligned & p x+a y+a z=0 \ & b x+q y+b z=0 \ & c x+c y+r z=0 aligned has a non-trivial solution, then the value of p p-a + q q-b + r r-c is

Accepted Answer

As the given system of equations has a non-trivial solution. = | array lll p & a & a \ b & q & b \ c & c & r array |=0 Applying C_2 C_2-C_1 and C_3 C_3-C_1 = | array ccc p & a-p & a-p \ b & q-b & 0 \ c & 0 & r-c array |=0 Expanding along C_3, we get (a-p) | array cc b & q-b \ c & 0 array |+(r-c) | array cc p & a-p \ b & q-b array |=0 aligned & (a-p)(-c)(q-b)+(r-c) \ & \ p(q-b)-b(a-p)\ =0 \ & \ & (p-a)(q-b) c+p(r-c)(q-b)+b(r-c)(p-a)=0 aligned Dividing by (p-a)(q-b)(r-c), we get c r-c + p

Question 7

Suppose p , q , r 0 and system of equation ( p + a ) x + by + cz =0 a x+(q+b) y+c z=0 a x+b y+(r+c) z=0 has a non-trivial solution, then value of a p + b q + c r is

Accepted Answer

Since the system is homogeneous and has a non-trivial solution 0= | array ccc p+a & b & c \ a & q+b & c \ a & b & r+c array |= | array ccc p & -q & 0 \ 0 & q & -r \ a & b & r+c array | using R _ 1 R _ 1 - R _ 2 and R _ 2 R _ 2 - R _ 3 Expanding along C , we get p [ q ( r + c )+ br ]+ aqr =0 pqr + pqc + prb + qra =0 a p + b q + c r =-1

Determinants — Previous Year Questions & Solutions