Question

If $px^4+qx^3+rx^2+sx+t = \begin{vmatrix} x^2+3x & x-1 & x+3 \\ x+1 & 2-x & x-3 \\ x-3 & x+4 & 3x \end{vmatrix}$ then t is equal to -

• A. 33
• B. 0
• C. 21
• D. none

Answer 1

Answer: (C)

21

Answer 2

To find the value of 't', we need to understand that 't' is the constant term of the polynomial $px^4+qx^3+rx^2+sx+t$.

The constant term of any polynomial $P(x)$ can be found by substituting $x=0$ into the polynomial.
Let the given polynomial be $P(x) = px^4+qx^3+rx^2+sx+t$.
Then, $P(0) = p(0)^4+q(0)^3+r(0)^2+s(0)+t = t$.

We are given that $P(x) = \begin{vmatrix} x^2+3x & x-1 & x+3 \\ x+1 & 2-x & x-3 \\ x-3 & x+4 & 3x \end{vmatrix}$.
To find 't', we substitute $x=0$ into the determinant: $t = \begin{vmatrix} 0^2+3(0) & 0-1 & 0+3 \\ 0+1 & 2-0 & 0-3 \\ 0-3 & 0+4 & 3(0) \end{vmatrix} = \begin{vmatrix} 0 & -1 & 3 \\ 1 & 2 & -3 \\ -3 & 4 & 0 \end{vmatrix}$

Now, we evaluate this $3 \times 3$ determinant. We can expand it along the first row: $t = 0 \cdot \text{cofactor}(0) - (-1) \cdot \text{cofactor}(-1) + 3 \cdot \text{cofactor}(3)$
$t = 0 \cdot \begin{vmatrix} 2 & -3 \\ 4 & 0 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & -3 \\ -3 & 0 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & 2 \\ -3 & 4 \end{vmatrix}$

Calculate the $2 \times 2$ determinants: $\begin{vmatrix} 2 & -3 \\ 4 & 0 \end{vmatrix} = (2)(0) - (-3)(4) = 0 - (-12) = 12$
$\begin{vmatrix} 1 & -3 \\ -3 & 0 \end{vmatrix} = (1)(0) - (-3)(-3) = 0 - 9 = -9$
$\begin{vmatrix} 1 & 2 \\ -3 & 4 \end{vmatrix} = (1)(4) - (2)(-3) = 4 - (-6) = 4 + 6 = 10$

Substitute these values back into the expression for 't': $t = 0 \cdot (12) + 1 \cdot (-9) + 3 \cdot (10)$
$t = 0 - 9 + 30$
$t = 21$