Question

If \(\begin{vmatrix} a & 1 & 1 1 & b & 1 1 & 1 & c \end{vmatrix}=2\), where \(a,b\) and \(c\) are positive integers, then \(a+b+c\) is equal to

• A. \(6\)
• B. \(8\)
• C. \(12\)
• D. \(18\)
• E. \(28\)

Hint:

In determinant questions with integer conditions, first simplify the determinant completely into an algebraic equation. After that, use number properties like positivity and factorization to find the integer solution.

Answer 1

The Correct Option is A

Step 1: Write the determinant clearly.
We are given:
\[ \begin{vmatrix} a & 1 & 1 1 & b & 1 1 & 1 & c \end{vmatrix}=2 \] We need to find the value of \(a+b+c\), where \(a,b,c\) are positive integers.

Step 2: Expand the determinant.
Expanding along the first row:
\[ \begin{vmatrix} a & 1 & 1 1 & b & 1 1 & 1 & c \end{vmatrix} = a\begin{vmatrix} b & 1 1 & c \end{vmatrix} -1\begin{vmatrix} 1 & 1 1 & c \end{vmatrix} +1\begin{vmatrix} 1 & b 1 & 1 \end{vmatrix} \] Now simplify each \(2\times 2\) determinant:
\[ \begin{vmatrix} b & 1 1 & c \end{vmatrix}=bc-1 \] \[ \begin{vmatrix} 1 & 1 1 & c \end{vmatrix}=c-1 \] \[ \begin{vmatrix} 1 & b 1 & 1 \end{vmatrix}=1-b \]

Step 3: Simplify the determinant expression.
So, \[ a(bc-1)-(c-1)+(1-b)=2 \] \[ abc-a-c+1+1-b=2 \] \[ abc-a-b-c+2=2 \] Therefore, \[ abc-a-b-c=0 \] or \[ abc=a+b+c \]

Step 4: Use the fact that \(a,b,c\) are positive integers.
Now we must solve:
\[ abc=a+b+c \] for positive integers \(a,b,c\). Since the variables are positive integers, let us test the smallest possibilities.

Step 5: Check whether any variable can be greater than \(1\) for all three.
If all of \(a,b,c\geq 2\), then:
\[ abc \geq 2\cdot 2\cdot 2=8 \] but the smallest possible value of \(a+b+c\) in that case is only \(2+2+2=6\). Since product grows faster than sum, we try small values first.

Step 6: Try one variable equal to \(1\).
Let \(a=1\). Then the equation becomes:
\[ bc=1+b+c \] \[ bc-b-c=1 \] Add \(1\) on both sides:
\[ bc-b-c+1=2 \] \[ (b-1)(c-1)=2 \] Since \(b\) and \(c\) are positive integers, the factor pairs of \(2\) are \((1,2)\) and \((2,1)\). Thus:
\[ (b-1,c-1)=(1,2) \quad \text{or} \quad (2,1) \] which gives:
\[ (b,c)=(2,3) \quad \text{or} \quad (3,2) \] Hence one solution is \((a,b,c)=(1,2,3)\), up to order.

Step 7: Find the required sum.
Therefore, \[ a+b+c=1+2+3=6 \] So the required value is:
\[ \boxed{6} \] which matches option \((1)\).