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
5
For infinitely many solutions, the third equation's coefficients must be a linear combination of the first two: Row3 = m·Row1 + n·Row2, and the same combination must hold for the constants.
Matching coefficients: for x: \(m+2n=1\); for y: \(-2m+n=-7\); for z: \(3m+n=a\).
Solving the first two: from \(m=1-2n\), substitute into \(-2m+n=-7\): \(-2(1-2n)+n=-7\Rightarrow -2+4n+n=-7\Rightarrow5n=-5\Rightarrow n=-1\), so \(m=1-2(-1)=3\).
Then \(a=3m+n=3(3)+(-1)=9-1=8\).
For the constant term: \(24=9m+bn=9(3)+b(-1)=27-b\Rightarrow b=27-24=3\).
Verification: with \(a=8\), the coefficient determinant \(\begin{vmatrix}1&-2&3\\2&1&1\\1&-7&8\end{vmatrix}=1(8+7)+2(16-1)+3(-14-1)=15+30-45=0\), confirming rank deficiency (consistent with infinite solutions).
Therefore \(a-b=8-3=5\).
Similar Questions
If the system of equations \(x+y+z=2\), \(2x+4y-z=6\), \(3x+2y+\lambda z=r\) has infinitely many solutions then:
Important Questions from Application of Determinants
The system of linear equation kx + y + z = 1, x + ky + z = 1 and x + y + kz = 1 has a unique solution under which one of the following conditions?
Which of the following are correct in respect of the system of equation
x + y + z = 8,
x – y + 2z = 6 and
3x – y + 5z = k?
1. They have no solution if k = 15
2. They have infinitely many solutions, if k = 20
3. They have a unique solution if k = 25
Select the correct answer using the code given below:Under what condition does the above system of equations have unique solutions?
- The number of values of $k$, for which the system of equations: $(k^2 - 4)x + (k - 2)y = k^2 - 2k$ and $(k + 2)x + y = k$ have infinitely many solutions, is -
For what values of k is the system of equations 2k 2x + 3y - 1 = 0, 7x - 2y + 3 = 0, 6kx + y + 1 = 0 consistent?
