Question

If a determinant is given in terms of x as $f\left( x \right)=\left| \left( \begin{matrix} 1 & x & x+1 \\\ 2x & x\left( x-1 \right) & \left( x+1 \right)x \\\ 3x\left( x-1 \right) & x\left( x-1 \right)\left( x-2 \right) & \left( x+1 \right)x\left( x-1 \right) \\\ \end{matrix} \right) \right|$
then f(100) is equal to:
(a) 0
(b) 1
(c) 100
(d) -100

Answer 1

Hint: In this question, we first need to look into the definitions of matrices and determinants. Then by using the formula of third order determinant we need to find the value of f(x) and substitute 100 in the respective function obtained.

Complete step-by-step solution -
Let us first look into some of the basic definitions of matrices and determinants.
MATRIX: A matrix is a rectangular arrangement of numbers (real or complex) which may be represented as

{{a}_{11}} & \ldots & {{a}_{1n}} \\\ \vdots & \ddots & \vdots \\\ {{a}_{m1}} & \cdots & {{a}_{mn}} \\\ \end{matrix} \right)$$ Matrix is enclosed by ( ) or [ ] Compact from the above matrix is represented by$${{\left[ {{a}_{ij}} \right]}_{m\times n}}$$ or $$A=\left[ {{a}_{ij}} \right]$$. ELEMENT OF A MATRIX: The numbers a11, a12, etc., in the above matrix are known as the elements of the matrix, generally represented as$${{a}_{ij}}$$, which denotes element in the $$i\text{th row and }j\text{th column}\text{.}$$ ORDER OF A MATRIX: In above matrix has m rows and n columns, then A is of order $$m\times n$$. TYPES OF MATRICES: Row Matrix Column Matrix Rectangular Matrix Horizontal Matrix Vertical Matrix Null / Zero Matrix Square Matrix Diagonal Matrix Scalar Matrix Unit / Identity Matrix Upper Triangular Matrix Lower Triangular Matrix Sub Matrix Equal Matrices Principal Diagonal of a Matrix Singular Matrix EQUAL MATRICES: Two matrices A and B are said to be equal, if both are having the same order and corresponding elements of both the matrices are equal. DETERMINANT: Every square matrix A is associated with a number, called its determinant and it is denoted by $$\det \left( A \right)$$ or $$\left| A \right|$$. Only square matrices have determinants. The matrices which are not square do not have determinants. THIRD ORDER DETERMINANT: If $$A=\left( \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right)$$, then $$\left| A \right|={{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{23}}{{a}_{32}} \right)-{{a}_{12}}\left( {{a}_{21}}{{a}_{33}}-{{a}_{23}}{{a}_{31}} \right)+{{a}_{13}}\left( {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \right)$$ Now, by comparing the given function with the above formula we get, $$\Rightarrow f\left( x \right)=\left| \left( \begin{matrix} 1 & x & x+1 \\\ 2x & x\left( x-1 \right) & \left( x+1 \right)x \\\ 3x\left( x-1 \right) & x\left( x-1 \right)\left( x-2 \right) & \left( x+1 \right)x\left( x-1 \right) \\\ \end{matrix} \right) \right|$$ $$\begin{aligned} & \Rightarrow f\left( x \right)=1\left( x\left( x-1 \right)\left( x+1 \right)x\left( x-1 \right)-\left( x+1 \right)xx\left( x-1 \right)\left( x-2 \right) \right) \\\ & \text{ }-x\left( 2x\left( x+1 \right)x\left( x-1 \right)-\left( x+1 \right)x3x\left( x-1 \right) \right) \\\ & \text{ }+\left( x+1 \right)\left( 2xx\left( x-1 \right)\left( x-2 \right)-x\left( x-1 \right)3x\left( x-1 \right) \right) \\\ \end{aligned}$$ Now, considering the each part of the above equation by assuming them as A, B, C and then simplify them separately we get, $$\Rightarrow f\left( x \right)=A+B+C$$ $$\Rightarrow A=1\left( x\left( x-1 \right)\left( x+1 \right)x\left( x-1 \right)-\left( x+1 \right)xx\left( x-1 \right)\left( x-2 \right) \right)$$ Let us take out the common terms then we get, $$\begin{aligned} & \Rightarrow A=x\left( x-1 \right)\left( x+1 \right)x\left( x-1-x+2 \right) \\\ & \Rightarrow A=x\left( x-1 \right)\left( x+1 \right) \\\ & \therefore A={{x}^{2}}\left( {{x}^{2}}-1 \right) \\\ \end{aligned}$$ $$\Rightarrow B=-x\left( 2x\left( x+1 \right)x\left( x-1 \right)-\left( x+1 \right)x3x\left( x-1 \right) \right)$$ $$\begin{aligned} & \Rightarrow B=-x\cdot x\left( x+1 \right)x\left( x-1 \right)\left( 2-3 \right) \\\ & \Rightarrow B=x\cdot x\left( x+1 \right)x\left( x-1 \right) \\\ & \therefore B=x\cdot {{x}^{2}}\left( {{x}^{2}}-1 \right) \\\ \end{aligned}$$ $$\Rightarrow C\text{=}\left( x+1 \right)\left( 2xx\left( x-1 \right)\left( x-2 \right)-x\left( x-1 \right)3x\left( x-1 \right) \right)$$ $$\begin{aligned} & \Rightarrow C\text{=}\left( x+1 \right)xx\left( x-1 \right)\left( 2\left( x-2 \right)-3\left( x-1 \right) \right) \\\ & \Rightarrow C\text{=}\left( x+1 \right)xx\left( x-1 \right)\left( 2x-4-3x+3 \right) \\\ & \Rightarrow C\text{=}\left( x+1 \right)xx\left( x-1 \right)\left( -x-1 \right) \\\ & \therefore C=-\left( x+1 \right)\cdot {{x}^{2}}\left( {{x}^{2}}-1 \right) \\\ \end{aligned}$$ $$\Rightarrow f\left( x \right)={{x}^{2}}\left( {{x}^{2}}-1 \right)+x\cdot {{x}^{2}}\left( {{x}^{2}}-1 \right)-\left( x+1 \right)\cdot {{x}^{2}}\left( {{x}^{2}}-1 \right)$$ $$\begin{aligned} & \Rightarrow f\left( x \right)={{x}^{2}}\left( {{x}^{2}}-1 \right)\left[ 1+x-\left( x+1 \right) \right] \\\ & \Rightarrow f\left( x \right)={{x}^{2}}\left( {{x}^{2}}-1 \right)\cdot 0 \\\ & \therefore f\left( x \right)=0 \\\ \end{aligned}$$ $$\therefore f\left( 100 \right)=0$$ Hence, the correct option is (a). Note: Instead of expanding every term and then multiplying to find the third order determinant we can also find it by using the second order determinant while expanding and then find its value there itself which also gives the same result. $$\left| A \right|={{a}_{11}}\left| \left( \begin{matrix} {{a}_{22}} & {{a}_{23}} \\\ {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right) \right|-{{a}_{12}}\left| \left( \begin{matrix} {{a}_{21}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{33}} \\\ \end{matrix} \right) \right|+{{a}_{13}}\left| \left( \begin{matrix} {{a}_{21}} & {{a}_{22}} \\\ {{a}_{31}} & {{a}_{32}} \\\ \end{matrix} \right) \right|$$ While expanding the term we need to be careful when we take out the common terms and the addition or subtraction of the terms inside. Because neglecting any of the terms or any of the signs changes the function completely and then the result.