Question

The value of the determinant \left| {\begin{array}{*{20}{c}} {x + 1}&{x + 2}&{x + 4} \\\ {x + 3}&{x + 5}&{x + 8} \\\ {x + 7}&{x + 10}&{x + 14} \end{array}} \right| is
A) -2
B) ${x^2} + 2$
C) 2
D) None of these

Answer 1

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 determinant further by using a basic arithmetical operation to get the required solution. Here the terms are in the form of algebraic expressions.

Complete step by step solution:
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 $3 \times 3$:

{x + 1}&{x + 2}&{x + 4} \\\ {x + 3}&{x + 5}&{x + 8} \\\ {x + 7}&{x + 10}&{x + 14} \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 \,\,x + 1 \cdot \left| {\begin{array}{*{20}{c}} {x + 5}&{x + 8} \\\ {x + 10}&{x + 14} \end{array}} \right| - (x + 2) \cdot \left| {\begin{array}{*{20}{c}} {x + 3}&{x + 8} \\\ {x + 7}&{x + 14} \end{array}} \right| + (x + 4) \cdot \left| {\begin{array}{*{20}{c}} {x + 3}&{x + 5} \\\ {x + 7}&{x + 10} \end{array}} \right|$$ On simplifying the determinant, we have

\Rightarrow ,,(x + 1) \cdot \left( {(x + 5)(x + 14) - (x + 10)(x + 8)} \right) - (x + 2) \cdot \left( {(x + 3)(x + 14) - (x + 7)(x + 8)} \right) + (x + 4) \cdot \left( {(x + 3)(x + 10) - (x + 7)(x + 5)} \right) \\

$$On multiplying the algebraic terms we have$$

\Rightarrow ,,(x + 1) \cdot \left( {{x^2} + 19x + 70 - ({x^2} + 18x + 80)} \right) - (x + 2) \cdot \left( {({x^2} + 17x + 42) - ({x^2} + 15x + 56)} \right) + (x + 4) \cdot \left( {({x^2} + 13x + 30) - ({x^2} + 12x + 35)} \right) \\

$$On simplifying the terms we have$$

\Rightarrow ,,(x + 1) \cdot \left( {{x^2} + 19x + 70 - {x^2} - 18x - 80} \right) - (x + 2) \cdot \left( {{x^2} + 17x + 42 - {x^2} - 15x - 56} \right) + (x + 4) \cdot \left( {{x^2} + 13x + 30 - {x^2} - 12x - 35} \right) \\

Cancelling the terms we are inverse to each other $$ \Rightarrow \,\,(x + 1) \cdot \left( {x - 10} \right) - (x + 2) \cdot \left( {2x - 14} \right) + (x + 4) \cdot \left( {x - 5} \right)$$ On multiplying the terms $$ \Rightarrow \left( {{x^2} - 10x + x - 10} \right) - \left( {2{x^2} - 14x + 4x - 28} \right) + \left( {{x^2} - 5x + 4x - 20} \right)$$ On simplifying the above terms $$ \Rightarrow \left( {{x^2} - 9x - 10} \right) - \left( {2{x^2} - 10x - 28} \right) + \left( {{x^2} - x - 20} \right)$$ $$ \Rightarrow {x^2} - 9x - 10 - 2{x^2} + 10x + 28 + {x^2} - x - 20$$ On further simplification we have $$ \Rightarrow - 2$$ Hence, the value of the determinant $$\left| {\begin{array}{*{20}{c}} {x + 1}&{x + 2}&{x + 4} \\\ {x + 3}&{x + 5}&{x + 8} \\\ {x + 7}&{x + 10}&{x + 14} \end{array}} \right|$$ is -2. **Therefore, the determinant of the given matrix is $-2. $ So, option (A) is correct.** **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.