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: Understanding the Concept:
This problem requires evaluating a 3x3 determinant and then solving a Diophantine equation (an equation where we seek integer solutions) for the sum of the variables. A careful analysis of the problem statement and the provided answer key suggests there might be a typo in the determinant's value. We will proceed by assuming the intended value makes the problem solvable.
Step 2: Key Formula or Approach:
The determinant of a 3x3 matrix \( \begin{vmatrix} a & b & c
d & e & f
g & h & i \end{vmatrix} \) is \(a(ei-fh) - b(di-fg) + c(dh-eg)\).
Let's calculate the determinant of the given matrix:
\[ \Delta = a(bc-1) - 1(c-1) + 1(1-b) \] \[ \Delta = abc - a - c + 1 + 1 - b \] \[ \Delta = abc - (a+b+c) + 2 \] Step 3: Detailed Explanation:
We are given that the determinant equals -2.
\[ abc - (a+b+c) + 2 = -2 \] \[ abc - (a+b+c) = -4 \] Let's test the options. We are looking for positive integers a, b, c.
If we take Option (A), a+b+c = 6.
Substituting this into our equation:
\[ abc - 6 = -4 \implies abc = 2 \] We need to find three positive integers a, b, c such that their sum is 6 and their product is 2. The only set of positive integers whose product is 2 is \{1, 1, 2\}.
Let's check the sum for this set: \(1+1+2 = 4\).
This does not equal 6. So there is a contradiction. This means that for a+b+c=6, it's impossible for the determinant to be -2. There appears to be a typo in the question's value for the determinant.
Let's assume the determinant should have been +2.
If \(\Delta = 2\), the equation becomes:
\[ abc - (a+b+c) + 2 = 2 \] \[ abc - (a+b+c) = 0 \implies abc = a+b+c \] Let's test the options again with this new condition.
If we take Option (A), a+b+c = 6.
Then we need \(abc = 6\). We are looking for three positive integers whose sum and product are both 6. The set of integers \{1, 2, 3\} satisfies this:
Sum: \(1+2+3 = 6\).
Product: \(1 \times 2 \times 3 = 6\).
This set provides a consistent solution. Let's verify the determinant with a=1, b=2, c=3:
\[ \Delta = (1)(2)(3) - (1+2+3) + 2 = 6 - 6 + 2 = 2 \] This confirms our hypothesis that the determinant value was intended to be +2.
Step 4: Final Answer:
Based on the analysis that the determinant value in the question is likely a typo and should be +2, the set of positive integers {1, 2, 3} is a valid solution. For this set, the sum a+b+c is 6. Therefore, option (A) is the correct answer.