If the system of equations \(x+y+z=2\), \(2x+4y-z=6\), \(3x+2y+\lambda z=r\) has infinitely many solutions then:
\(2\lambda+r=14\)
For infinitely many solutions, Row3 = m·Row1 + n·Row2 for the coefficients, and the same combination must hold for the constants.
Matching coefficients: for x: \(m+2n=3\); for y: \(m+4n=2\); for z: \(m-n=\lambda\).
Subtracting the x-equation from the y-equation: \((m+4n)-(m+2n)=2-3\Rightarrow 2n=-1\Rightarrow n=-\dfrac12\).
Then \(m+2\left(-\dfrac12\right)=3\Rightarrow m-1=3\Rightarrow m=4\). (Check: \(m+4n=4-2=2\) ✓.)
So \(\lambda=m-n=4-\left(-\dfrac12\right)=\dfrac{9}{2}\).
For the constant term: \(r=2m+6n=2(4)+6\left(-\dfrac12\right)=8-3=5\).
Now check the options with \(\lambda=\dfrac92,\ r=5\): \(2\lambda+r=2\left(\dfrac92\right)+5=9+5=14\), which matches option (B). (The other options do not hold: \(\lambda-2r=4.5-10=-5.5\ne-5\); \(\lambda+2r=4.5+10=14.5\ne14\); \(2\lambda-r=9-5=4\ne5\).)
Therefore the correct relation is \(2\lambda+r=14\).
Similar Questions
If the system of equations \(x-2y+3z=9\), \(2x+y+z=b\), \(x-7y+az=24\) has infinitely many solutions then \(a-b\) is equal to
Important Questions from Application of Determinants
The equations 3x - 4y = 5 and 12x - 16y = 20 have:
The system of equations
2x + y - 3z = 5
3x - 2y + 2z = 5 and
5x - 3y - z = 16The system of equations kx + y + z = 1, x + ky + z = k and x + y + kz = k 2has no solution if k equals
The equations 2x - ky + 7 = 0 and 6x - 12y + 15 = 0 have no solution for
If (a, b), (x1, y1) and (x2, y2) are the vertices of a triangle such that the x-coordinates a, x1, x2 are in geometric progression with common ratio r and the y-coordinates b, y1, y2 are also in geometric progression with common ratio s, then the area of the triangle is:
