Question

The value of determinant \[\Delta = \left| {\begin{array}{*{20}{c}}
  {1!}&{2!}&{3!} \\
  {2!}&{3!}&{4!} \\
  {3!}&{4!}&{5!}
\end{array}} \right|\] is

Answer 1

Hint: Here in this question, we have to find the value of determinant of order \[3 \times 3\]. To solve this first we have to expand the factorial numbers which their in the determinant and later by determinant expansion expand the determinant further simplify using a basic arithmetical operation to get the required solution.

Complete step-by-step answer:
Determinants are mathematical objects that are very useful in the analysis and solution of systems of linear equations. As shown by Cramer’s rule, a nonhomogeneous system of linear equations has a unique solution if and only if the determinant of the system's Matrix is non zero (i.e., the matrix is non-singular).
Now consider the given determinant of order \[3 \times 3\]:
\[ \Rightarrow \,\,\Delta = \left| {\begin{array}{*{20}{c}}
  {1!}&{2!}&{3!} \\
  {2!}&{3!}&{4!} \\
  {3!}&{4!}&{5!}
\end{array}} \right|\]
First, expand the factorial numbers which their in the determinant, it means to multiply together all the numbers descending from the factorial number, then
\[ \Rightarrow \,\,\Delta = \left| {\begin{array}{*{20}{c}}
  1&{2 \times 1}&{3 \times 2 \times 1} \\
  {2 \times 1}&{3 \times 2 \times 1}&{4 \times 3 \times 2 \times 1} \\
  {3 \times 2 \times 1}&{4 \times 3 \times 2 \times 1}&{5 \times 4 \times 3 \times 2 \times 1}
\end{array}} \right|\]
On simplification, we have
\[ \Rightarrow \,\,\Delta = \left| {\begin{array}{*{20}{c}}
  1&2&6 \\
  2&6&{24} \\
  6&{24}&{120}
\end{array}} \right|\]
Now, expand the Determinant of a above \[3 \times 3\] matrix by cofactor expansion theorem:
The cofactor expansion theorem states that any determinant can be computed by adding the products of the elements of a column or row by their respective cofactors.
\[ \Rightarrow \,\,\Delta = 1 \cdot \left| {\begin{array}{*{20}{c}}
  6&{24} \\
  {24}&{120}
\end{array}} \right| - 2 \cdot \left| {\begin{array}{*{20}{c}}
  2&{24} \\
  6&{120}
\end{array}} \right| + 6 \cdot \left| {\begin{array}{*{20}{c}}
  2&6 \\
  6&{24}
\end{array}} \right|\]
On simplifying the determinant, we have
\[ \Rightarrow \,\,\Delta = 1 \cdot \left( {720 - 576} \right) - 2 \cdot \left( {240 - 144} \right) + 6 \cdot \left( {48 - 36} \right)\]
On simplification, we get
\[ \Rightarrow \,\,\Delta = 1 \cdot \left( {144} \right) - 2 \cdot \left( {96} \right) + 6 \cdot \left( {12} \right)\]
\[ \Rightarrow \,\,\Delta = 144 - 192 + 72\]
\[ \Rightarrow \,\,\Delta = 216 - 192\]
\[ \Rightarrow \,\,\Delta = 24\]
Hence, the value of the determinant \[\left| {\begin{array}{*{20}{c}}
  {1!}&{2!}&{3!} \\
  {2!}&{3!}&{4!} \\
  {3!}&{4!}&{5!}
\end{array}} \right|\] is 24.
So, the correct answer is “24”.

Note: When considering the determinant that should be in the square matrix it means the determinant should have a equal number of rows and column otherwise it can be solve by using a determinant method and remember that to find the determinant of a\[2 \times 2\] matrix, you have to multiply the elements on the main diagonal and subtract the product of the elements on the secondary diagonal.