Let `plamda^(4)+qlamda^(3)+rlamda^(2)+slamda+t`
`=|(lamda^(2)+3lamda,lamda-1,lamda-3),(lamda-1,-2lamda,lamda-4),(lamda-3,lamda+4,3lamda)|`
where p,q,r,s, and t are constant. Then value of t is

Options

A. 0
B. `-1`
C. `2`
D. `3`

Answer

To solve the equation

\[ p \lambda^4 + q \lambda^3 + r \lambda^2 + s \lambda + t = \begin{vmatrix} \lambda^2 + 3\lambda & \lambda - 1 & \lambda - 3 \\ \lambda - 1 & -2\lambda & \lambda - 4 \\ \lambda - 3 & \lambda + 4 & 3\lambda \end{vmatrix} \]

we need to find the value of \( t \).

### Step 1: Substitute \( \lambda = 0 \)

We start by substituting \( \lambda = 0 \) into both sides of the equation.

**LHS:**
\[ p(0)^4 + q(0)^3 + r(0)^2 + s(0) + t = t \]

**RHS:**
\[ \begin{vmatrix} 0^2 + 3(0) & 0 - 1 & 0 - 3 \\ 0 - 1 & -2(0) & 0 - 4 \\ 0 - 3 & 0 + 4 & 3(0) \end{vmatrix} = \begin{vmatrix} 0 & -1 & -3 \\ -1 & 0 & -4 \\ -3 & 4 & 0 \end{vmatrix} \]

### Step 2: Calculate the Determinant

Now we need to calculate the determinant:

\[ D = \begin{vmatrix} 0 & -1 & -3 \\ -1 & 0 & -4 \\ -3 & 4 & 0 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the second and third rows respectively.

Substituting the values:

\[ D = 0 \cdot (0 \cdot 0 - (-4) \cdot 4) - (-1) \cdot (-1 \cdot 0 - (-4)(-3)) + (-3) \cdot (-1 \cdot 4 - 0 \cdot (-3)) \]

Calculating each term:

1. The first term is \( 0 \).
2. The second term is \( -1 \cdot (0 - 12) = 12 \).
3. The third term is \( -3 \cdot (-4) = 12 \).

Thus,

\[ D = 0 + 12 + 12 = 24 \]

### Step 3: Set the Equation

Now we set the left-hand side equal to the right-hand side:

\[ t = 24 \]

### Conclusion

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

\[ \boxed{24} \]