Question

Let \( A = \begin{bmatrix} a+1 & b & c \\ a & b+1 & c \\ a & b & c+1 \end{bmatrix} \). If determinant of the matrix \( A \) is zero, then \( (a + b + c)^3 = \_\_\_\_\_\_ \)

• A. 0
• B. –1
• C. abc
• D. 3abc

Hint:

For structured matrices, try decomposing into simpler parts or use trace and determinant relationships.

Answer 1

The Correct Option is B

We are given:
\[ A = \begin{bmatrix} a+1 & b & c \\ a & b+1 & c \\ a & b & c+1 \end{bmatrix} \]

Let us consider \( A \) as a matrix obtained by adding 1 to the diagonal elements of another matrix:
\[ A = M + I \quad \text{where} \quad M = \begin{bmatrix} a & b & c \\ a & b & c \\ a & b & c \end{bmatrix} \]
The matrix \( M \) has rank 1 (all rows are linearly dependent). Hence, \( \det(M) = 0 \). Now, the matrix \( A = M + I \), so we find the determinant of this type of matrix.

Since all rows differ only in diagonal by 1, perform row operations or apply properties to simplify:
\[ \det(A) = 0 \Rightarrow \text{Characteristic determinant gives:} (a + b + c + 1)^3 - 1 = 0 \]
\[ \Rightarrow (a + b + c)^3 = -1 \]

Final Answer: (2) –1