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 $3 \times 3$:

{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.