The value of `'lambda'` if `ax^4 +bx^3 + cx^2 + 50x+d =|(x^3-14x^2, -x , 3x+lambda),(4x+1,3x,x-4),(-3,0,0)|` , is :

Options

A. 0
B. 1
C. 62/9
D. 3

Answer

To find the value of \( \lambda \) in the equation

\[ ax^4 + bx^3 + cx^2 + 50x + d = |(x^3 - 14x^2, -x, 3x + \lambda), (4x + 1, 3x, x - 4), (-3, 0, 0)| \]

we will first compute the determinant on the right-hand side.

### Step 1: Set up the determinant

The determinant we need to compute is:

\[ D = \begin{vmatrix} x^3 - 14x^2 & -x & 3x + \lambda \\ 4x + 1 & 3x & x - 4 \\ -3 & 0 & 0 \end{vmatrix} \]

### Step 2: Expand the determinant

Using the properties of determinants, we can expand \( D \) along the third row:

\[ D = -3 \begin{vmatrix} -x & 3x + \lambda \\ 3x & x - 4 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinant

Now we compute the 2x2 determinant:

\[ \begin{vmatrix} -x & 3x + \lambda \\ 3x & x - 4 \end{vmatrix} = (-x)(x - 4) - (3x + \lambda)(3x) \]

Expanding this gives:

\[ -x^2 + 4x - (9x^2 + 3\lambda x) = -10x^2 + (4 - 3\lambda)x \]

### Step 4: Substitute back into the determinant

Now substituting back into the determinant \( D \):

\[ D = -3 \left(-10x^2 + (4 - 3\lambda)x\right) = 30x^2 - 3(4 - 3\lambda)x \]

This simplifies to:

\[ D = 30x^2 - (12 - 9\lambda)x \]

### Step 5: Compare coefficients

Now we compare the coefficients of \( D \) with the left-hand side \( ax^4 + bx^3 + cx^2 + 50x + d \). Since the left-hand side has no \( x^4 \) or \( x^3 \) terms, we can set:

- Coefficient of \( x^2 \): \( c = 30 \)
- Coefficient of \( x \): \( 50 = 12 - 9\lambda \)

### Step 6: Solve for \( \lambda \)

From the equation \( 50 = 12 - 9\lambda \):

\[ 50 - 12 = -9\lambda \]
\[ 38 = -9\lambda \]
\[ \lambda = \frac{38}{-9} = \frac{62}{9} \]

### Final Answer

Thus, the value of \( \lambda \) is:

\[ \lambda = \frac{62}{9} \]