Question

If the matrix A = $\begin{bmatrix} y + a & b & c \\ a & y + b & c \\ a & b & y + c \end{bmatrix}$has rank 3, then –

• A. y¹ (a + b + c)
• B. y ¹ 1
• C. y = 0
• D. y ¹ – (a + b + c) and y ¹ 0

Answer 1

Answer: (D)

y ¹ – (a + b + c) and y ¹ 0

Answer 2

Here the rank of A is 3

Therefore, the minor of order 3 of A ¹ 0.

Ž $\left| \begin{matrix} y + a & b & c \\ a & y + b & c \\ a & b & y + c \end{matrix} \right|$ ¹ 0

Ž (y + a + b + c) $\left| \begin{matrix} 1 & b & c \\ 1 & y + b & c \\ 1 & b & y + c \end{matrix} \right|$ ¹ 0

[Applying C₁ ® C₁ + C₂ + C₃ and taking (y + a + b + c) common from C₁]

Ž (y + a + b + c) $\left| \begin{matrix} 1 & b & c \\ 0 & y & 0 \\ 0 & 0 & y \end{matrix} \right|$¹ 0[Applying R₂ ® R₂ – R₁, R₃ ® R₃ – R₁]

Ž (y + a + b + c) (y²) ¹ 0 [Expanding along C₁]Ž y ¹ 0 and y ¹ –(a + b + c)

Hence (4) is correct answer.