If `x=-2` and `Delta=|(x+y,x,x),(5x+4y,4x,2x),(10x+8y,8x,3x)|` then numerical value of `Delta` is

Options

A. 8
B. `-8`
C. `4`
D. `-4`

Answer

To find the numerical value of \(\Delta\) given that \(x = -2\) and

\[ \Delta = \begin{vmatrix} x+y & x & x \\ 5x + 4y & 4x & 2x \\ 10x + 8y & 8x & 3x \end{vmatrix} \]

we will simplify the determinant using row operations and then substitute the value of \(x\).

### Step 1: Substitute \(x = -2\)

First, we substitute \(x = -2\) into the determinant:

\[ \Delta = \begin{vmatrix} -2 + y & -2 & -2 \\ 5(-2) + 4y & 4(-2) & 2(-2) \\ 10(-2) + 8y & 8(-2) & 3(-2) \end{vmatrix} \]

This simplifies to:

\[ \Delta = \begin{vmatrix} -2 + y & -2 & -2 \\ -10 + 4y & -8 & -4 \\ -20 + 8y & -16 & -6 \end{vmatrix} \]

### Step 2: Apply Row Operations

Next, we will perform row operations to simplify the determinant. We can replace row 2 with \(R_2 - 2R_1\) and row 3 with \(R_3 - 3R_1\):

- For \(R_2\):
\[ R_2 = (-10 + 4y) - 2(-2 + y) = -10 + 4y + 4 - 2y = -6 + 2y \]
So, the new row 2 becomes:
\[ (-6 + 2y, -8 + 4, -4 + 4) = (-6 + 2y, -4, 0) \]

- For \(R_3\):
\[ R_3 = (-20 + 8y) - 3(-2 + y) = -20 + 8y + 6 - 3y = -14 + 5y \]
So, the new row 3 becomes:
\[ (-14 + 5y, -16 + 6, -6 + 6) = (-14 + 5y, -10, 0) \]

Now, the determinant looks like:

\[ \Delta = \begin{vmatrix} -2 + y & -2 & -2 \\ -6 + 2y & -4 & 0 \\ -14 + 5y & -10 & 0 \end{vmatrix} \]

### Step 3: Calculate the Determinant

Since the last column has two zeros, we can expand the determinant along the last column:

\[ \Delta = (-2) \begin{vmatrix} -6 + 2y & -4 \\ -14 + 5y & -10 \end{vmatrix} \]

Calculating the 2x2 determinant:

\[ \Delta = -2 \left[ (-6 + 2y)(-10) - (-4)(-14 + 5y) \right] \]

Calculating the terms:

\[ = -2 \left[ 60 - 20y - 56 + 20y \right] \]
\[ = -2 \left[ 4 \right] = -8 \]

### Step 4: Conclusion

Thus, the numerical value of \(\Delta\) is:

\[ \Delta = -8 \]