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: Recall when a square system is solvable for every right-hand side.
A system $Mx=v$ has exactly one solution for every choice of $v$ precisely when the rows of $M$ are linearly independent. If one row can be written as a combination of the others, then most choices of $(a,b,c)$ will fail.

Step 2: Test statement (I) by trying to eliminate directly.
Subtract the first equation from the second: $(x+y+3z)-(x+2y-z)=b-a$, giving $-y+4z=b-a$.

Step 3: Build the third equation from the first two.
$2(x+2y-z)+(-y+4z)=2x+4y-2z-y+4z=2x+3y+2z$, exactly the left side of equation 3. So equation 3 is not independent. Tracking the right-hand side: $2a+(b-a)=a+b$. So consistency needs $c=a+b$; this cannot hold for an arbitrary $(a,b,c)$, e.g. $(0,0,1)$ has no solution. So statement (I) is FALSE.

Step 4: Confirm $(0,0,0)$ passes the consistency test.
With $a=b=c=0$, $c=a+b$ reads $0=0$, true.

Step 5: Solve using the relation found in Step 2.
With $a=b=0$, $-y+4z=0$ gives $y=4z$. Put into $x+2y-z=0$: $x+8z-z=0$, so $x=-7z$.

Step 6: Write and verify the solution set.
Letting $z=t$ gives $(x,y,z)=(-7t,4t,t)$. Check the third equation: $2(-7t)+3(4t)+2(t)=0$ for every $t$. This matches statement (II), so it is TRUE.

Step 7: Sanity check with $t=1$.
$(x,y,z)=(-7,4,1)$: eq1: $-7+8-1=0$; eq2: $-7+4+3=0$; eq3: $-14+12+2=0$. All hold.

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