If the matrix A = `[{:(y+a,b,c),(a,y+b,c),(a,b,y+c):}]` has rank 3, then

Options

A. `y ne ` (a+ b + c )
B. `y ne ` 1
C. y = 0
D. y `ne ` - (a + b + c) and y`ne ` 0

Answer

To determine the conditions under which the matrix

\[ A = \begin{pmatrix} y + a & b & c \\ a & y + b & c \\ a & b & y + c \end{pmatrix} \]

has a rank of 3, we need to ensure that the determinant of the matrix is not equal to zero.

### Step 1: Calculate the Determinant of Matrix A

The determinant of a 3x3 matrix

\[ \begin{pmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ z_1 & z_2 & z_3 \end{pmatrix} \]

is given by:

\[ \text{det}(A) = x_1(y_2z_3 - y_3z_2) - x_2(y_1z_3 - y_3z_1) + x_3(y_1z_2 - y_2z_1) \]

Applying this to our matrix A:

\[ \text{det}(A) = (y + a) \left( (y + b)(y + c) - c^2 \right) - b \left( a(y + c) - c a \right) + c \left( a b - b a \right) \]

### Step 2: Simplify the Determinant

Now we simplify the determinant expression:

1. Calculate \( (y + b)(y + c) - c^2 \):
\[ = y^2 + (b+c)y + bc - c^2 = y^2 + (b+c)y + (b-c)c \]

2. Calculate \( a(y + c) - c a = ay + ac - ac = ay \).

3. The last term \( ab - ba = 0 \).

Thus, we have:

\[ \text{det}(A) = (y + a) \left( y^2 + (b+c)y + (b-c)c \right) - b(ay) \]

### Step 3: Set the Determinant Not Equal to Zero

For the matrix A to have a rank of 3, we need:

\[ \text{det}(A) \neq 0 \]

This means we need to ensure that:

\[ (y + a)(y^2 + (b+c)y + (b-c)c) - aby \neq 0 \]

### Step 4: Analyze the Conditions

From the determinant expression, we can derive two conditions:

1. \( y + a \neq 0 \) implies \( y \neq -a \).
2. The quadratic \( y^2 + (b+c)y + (b-c)c \neq 0 \) must also hold.

### Conclusion

Thus, the conditions for the matrix A to have a rank of 3 are:

1. \( y \neq -a \)
2. The quadratic \( y^2 + (b+c)y + (b-c)c \neq 0 \)