Question

Consider the system of equations
\[ x+2y-z=a \]
\[ x+y+3z=b \]
\[ 2x+3y+2z=c \]

Consider the following statements:
(I) For every \((a,b,c)\in\mathbb{R}^3\), the above system has a solution.
(II) For \((a,b,c)=(0,0,0)\), the solution set is given by
\[ \{(-7t,\ 4t,\ t):t\in\mathbb{R}\}. \]
Which of the following statements is correct?

• A. Only statement (I) is correct
• B. Only statement (II) is correct
• C. Both statements (I) and (II) are correct
• D. Neither statement (I) nor statement (II) is correct

Hint:

The coefficient matrix has determinant \(0\), so the system is not solvable for every \((a,b,c)\); solving the homogeneous system \(a=b=c=0\) by elimination gives \((x,y,z)=(-7t,4t,t)\).

Answer 1

The Correct Option is B

Step 1: Write the coefficient matrix and test statement (I) via its determinant.
The coefficient matrix of the system is
\[ M=\begin{pmatrix}1 & 2 & -1\\ 1 & 1 & 3\\ 2 & 3 & 2\end{pmatrix}. \]
A system with \(3\) equations and \(3\) unknowns has a solution for every right-hand side \((a,b,c)\) only if \(M\) is invertible, that is \(\det M\neq0\). Compute the determinant by expanding along the first row:
\[ \det M=1(1\cdot2-3\cdot3)-2(1\cdot2-3\cdot2)+(-1)(1\cdot3-1\cdot2) \]
\[ =1(2-9)-2(2-6)-1(3-2)=(-7)-2(-4)-1=-7+8-1=0. \]

Step 2: Interpret the zero determinant.
Since \(\det M=0\), the matrix \(M\) is singular, so the system cannot have a (unique) solution for an arbitrary right-hand side; a solution exists only when \((a,b,c)\) satisfies a compatibility condition. So statement (I), which claims a solution for every \((a,b,c)\), is FALSE.

Step 3: Row reduce to find the compatibility condition.
\[ \left[\begin{array}{ccc|c}1 & 2 & -1 & a\\ 1 & 1 & 3 & b\\ 2 & 3 & 2 & c\end{array}\right] \]
Replace \(R_2\to R_2-R_1\) and \(R_3\to R_3-2R_1\):
\[ \left[\begin{array}{ccc|c}1 & 2 & -1 & a\\ 0 & -1 & 4 & b-a\\ 0 & -1 & 4 & c-2a\end{array}\right] \]
Now replace \(R_3\to R_3-R_2\): the last row simplifies to \(0=c-a-b\), so the system is consistent only when \(c=a+b\). For \((a,b,c)\) not satisfying this, the system has no solution at all, confirming statement (I) is FALSE.

Step 4: Check \((a,b,c)=(0,0,0)\) is consistent.
Here \(c-a-b=0-0-0=0\), so the condition \(c=a+b\) holds, and the homogeneous system is consistent.

Step 5: Solve the homogeneous system for statement (II).
From the reduced second row, \(-y+4z=0\), so \(y=4z\). Substitute into the first equation \(x+2y-z=0\):
\[ x=-2y+z=-2(4z)+z=-8z+z=-7z. \]
Setting \(z=t\), the solution set is
\[ (x,y,z)=(-7t,\ 4t,\ t),\quad t\in\mathbb{R}, \]
which is exactly the set given in statement (II). A quick check with the third original equation confirms this: \(2(-7t)+3(4t)+2(t)=-14t+12t+2t=0\), so it is satisfied automatically. Statement (II) is TRUE.

Final Answer:
Only statement (II) is correct. \[ \boxed{\text{Only statement (II) is correct}} \]