Question

Let \(a,b\in\mathbb{R}\). If the system of linear equations

Hint:

For infinitely many solutions in a linear system, one equation must become a linear combination of the others when the first two equations are independent.

Answer 1

Correct Answer: 6

Step 1: Write the first two equations.
\[ x+2y+2z=1 \] \[ 2x+3y+z=2 \]

Step 2: For infinitely many solutions, the third equation must be dependent on the first two equations.
So, there exist constants \(\lambda\) and \(\mu\) such that
\[ (ax+5y+bz=b)=\lambda(x+2y+2z=1)+\mu(2x+3y+z=2) \]

Step 3: Compare coefficients of \(x,y,z\) and constant terms.
\[ a=\lambda+2\mu \] \[ 5=2\lambda+3\mu \] \[ b=2\lambda+\mu \] \[ b=\lambda+2\mu \]

Step 4: Compare the two expressions for \(b\).
\[ 2\lambda+\mu=\lambda+2\mu \]
\[ \lambda=\mu \]

Step 5: Substitute \(\lambda=\mu\) in the \(y\)-coefficient equation.
\[ 5=2\lambda+3\lambda \]
\[ 5=5\lambda \]
\[ \lambda=1 \]
Hence,
\[ \mu=1 \]

Step 6: Find \(a\) and \(b\).
\[ a=\lambda+2\mu=1+2=3 \]
\[ b=\lambda+2\mu=1+2=3 \]

Step 7: Find \(a+b\).
\[ a+b=3+3=6 \]
\[ \boxed{6.0} \]