If \[a{x^4} + b{x^3} + c{x^2} + dx + e = \left| {\begin{array}{*{20}{c}}
  {{x^3} + 3x}&{x - 1}&{x + 3} \\
  {x + 1}&{ - 2x}&{x - 4} \\
  {x - 3}&{x + 4}&{3x}
\end{array}} \right|\] then, \[e = \]
A) \[1\]
B) \[0\]
C) \[2\]
D) \[ - 1\]

Hint: In linear algebra, the determinant of a matrix is a scalar value which can be calculated from the elements of a square matrix. It is denoted by \[\det A\]or\[\left| A \right|\].
For \[2 \times 2\] matrix, \[\left| A \right|\] is nothing but the cross multiplication on the diagonal elements.
\[\left| A \right| = \left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right| = ad - bc\]
For \[3 \times 3\]matrix, \[\left| A \right|\]is found using the following formulae,
\[\left| A \right| = \left( {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right) = a(ei - hf) - b(di - gf) + c(dh - eg)\]
For \[4 \times 4\] and higher order matrices, it is quite difficult to find the determinant by applying the formula. For these types of sums, we will do rows or column operation as per the requirement and transform the matrix into simpler form.
We can find the determinant for any square matrices.
Here by finding the determinant value of \[3 \times 3\]matrix given we will find a polynomial equation with that found polynomial equation we will compare the given polynomial equation and find the value of\[e\].

Complete step-by-step answer:
To find the value of \[e\] we will find the value of the determinant and compare the polynomial on both sides of the equation.
It is given that,
\[a{x^4} + b{x^3} + c{x^2} + dx + e = \left| {\begin{array}{*{20}{c}}
  {{x^3} + 3x}&{x - 1}&{x + 3} \\
  {x + 1}&{ - 2x}&{x - 4} \\
  {x - 3}&{x + 4}&{3x}
\end{array}} \right|\]
By the formula of determinant of \[3 \times 3\]matrix, we get,
\[ = ({x^3} + 3x)[ - 6{x^2} - (x + 4)(x - 4)] - (x - 1)[3x(x - 1) - (x - 4)(x - 3)] + (x + 3)[(x + 1)(x + 4) + 2x(x - 3)]\]
We should now simplify the above equation, we get,
\[ = ({x^3} + 3x)[ - 6{x^2} - {x^2} + 16] - (x - 1)[3{x^2} - 3x - {x^2} + 7x - 12] + (x + 3)[{x^2} + 5x + 4 + 2{x^2} - 6x]\]
Now again we should simplify the addition and subtractions in above equation, we get,
\[ = ({x^3} + 3x)[ - 7{x^2} + 16] - (x - 1)[2{x^2} + 4x - 12] + (x + 3)[3{x^2} - x + 4]\]
Now we should again multiply the terms in the bracket, hence we get,
\[ = - 7{x^5} + 16{x^3} - 21{x^3} + 48x - 2{x^3} - 4{x^2} + 12x + 2{x^2} + 4x - 12 + 3{x^3} - {x^2} + 4x + 9{x^2} - 3x + 12\]
Now on further simplification we will arrive at a polynomial of degree five,
\[ = - 7{x^5} - 4{x^3} + 6{x^2} + 65x + 0\]
Now we should compare the above equation with \[a{x^4} + b{x^3} + c{x^2} + dx + e\] we get,
\[e = 0\]

We can come to an conclusion that the correct option is (B) \[e = 0\]

Note: There is another way to solve this question.
We put \[x = 0\] in the given condition.
So, substitute \[x = 0\]in \[a{x^4} + b{x^3} + c{x^2} + dx + e = \left| {\begin{array}{*{20}{c}}
  {{x^3} + 3x}&{x - 1}&{x + 3} \\
  {x + 1}&{ - 2x}&{x - 4} \\
  {x - 3}&{x + 4}&{3x}
\end{array}} \right|\] we get,
\[a(0) + b(0) + c(0) + d(0) + e = \left| {\begin{array}{*{20}{c}}
  0&{ - 1}&3 \\
  1&0&{ - 4} \\
  { - 3}&4&0
\end{array}} \right|\]
By simplifying the determinant, we get,
\[a(0) + b(0) + c(0) + d(0) + e = 0\][Since the diagonal of the above determinant zero the value of the determinant will be zero]
Hence, the value of \[e = 0\]