If `x+a=y+b+1=z+c` then the value of `[[x, (a+y), (a+x)], [y, (b+y), (b+y)], [z, (c+y), (c+z)]]` is

Options

A. `y(a-b)`
B. `y(b-c)`
C. `y(c-a)`
D. 0

Answer

To solve the problem, we need to find the value of the matrix given the condition \( x + a = y + b + 1 = z + c \).

Let's denote this common value as \( k \):
\[ k = x + a = y + b + 1 = z + c \]

From this, we can express \( x \), \( y \), and \( z \) in terms of \( k \):
\[ x = k - a \]
\[ y = k - b - 1 \]
\[ z = k - c \]

Now, we can substitute these values into the matrix:
\[ \begin{bmatrix} x & a + y & a + x \\ y & b + y & b + y \\ z & c + y & c + z \end{bmatrix} \]

Substituting \( x \), \( y \), and \( z \):
\[ \begin{bmatrix} k - a & a + (k - b - 1) & a + (k - a) \\ k - b - 1 & b + (k - b - 1) & b + (k - b - 1) \\ k - c & c + (k - b - 1) & c + (k - c) \end{bmatrix} \]

Now, simplifying each element:
1. First row:
- First element: \( k - a \)
- Second element: \( a + k - b - 1 = k + a - b - 1 \)
- Third element: \( a + k - a = k \)

2. Second row:
- First element: \( k - b - 1 \)
- Second element: \( b + k - b - 1 = k - 1 \)
- Third element: \( k - b - 1 \)

3. Third row:
- First element: \( k - c \)
- Second element: \( c + k - b - 1 = k + c - b - 1 \)
- Third element: \( c + k - c = k \)

Thus, the matrix becomes:
\[ \begin{bmatrix} k - a & k + a - b - 1 & k \\ k - b - 1 & k - 1 & k - b - 1 \\ k - c & k + c - b - 1 & k \end{bmatrix} \]

Next, we will simplify this matrix using column operations. We can perform the operation \( C_3 \to C_3 - C_1 \):
\[ \begin{bmatrix} k - a & k + a - b - 1 & 0 \\ k - b - 1 & k - 1 & 0 \\ k - c & k + c - b - 1 & 0 \end{bmatrix} \]

Now, we can see that the third column is all zeros, which means the determinant of this matrix is zero.

Thus, the value of the matrix is:
\[ \text{Value of the matrix} = 0 \]