Question

If we have a determinant $f\left( x \right)=\left|\begin{matrix} x \\\ 2\lambda \\\ \end{matrix} \right.\text{ }\left. \begin{matrix} & \lambda \\\ & x \\\ \end{matrix} \right|$, then $f\left( \lambda x \right)-f\left( x \right)$ is equal to:
(a) $x\left( {{\lambda }^{2}}-1 \right)$
(b) $2\lambda \left( {{x}^{2}}-1 \right)$
(c) ${{\lambda }^{2}}\left( {{x}^{2}}-1 \right)$
(d) $\lambda \left( {{x}^{2}}-1 \right)$
(e) ${{x}^{2}}\left( {{\lambda }^{2}}-1 \right)$

Answer 1

Here, we can apply the property of determinant that $\det \left( \lambda A \right)=\lambda \times \text{det}\left( A \right)$, where A is any given matrix.

Complete step-by-step answer:
Determinant is a scalar value that can be computed from the elements of a square matrix and encodes certain properties of the linear transformations described by the matrix. The determinant of the matrix A is denoted by det (A), det A or |A|.
The determinant tells us the things about the matrix that are useful in systems of linear equations, helps us to find the inverse of a matrix and it is useful in calculus and more.
In case of $2\times 2$ matrix, the determinant may be defined as:
$|A|=\left| \begin{matrix} a \\\ c \\\ \end{matrix} \right.\left. \begin{matrix} & b \\\ & d \\\ \end{matrix} \right|=ad-bc$
Determinants possess many algebraic properties.
Determinant is defined only for square matrices.
Here, we are given a square matrix f(x) as:
$f\left( x \right)=\left| \begin{matrix} x \\\ 2\lambda \\\ \end{matrix} \right.\text{ }\left. \begin{matrix} & \lambda \\\ & x \\\ \end{matrix} \right|$
This is a square matrix of order 2 $\times$ 2.
Now, according to the question, to get the value of a given expression that is $f\left( \lambda x \right)-f\left( x \right)$, we need to find the values of both the terms in this expression. So, we have;

\lambda x \\\ 2\lambda \\\ \end{matrix} \right.\text{ }\left. \begin{matrix} & \lambda \\\ & \lambda x \\\ \end{matrix} \right|$$ Since, $|\lambda A|=|{{\lambda }^{n}}|A|$, where n is the order of determinant. Therefore, $$f\left( \lambda x \right)={{\lambda }^{2}}\left| \begin{matrix} x \\\ 2 \\\ \end{matrix} \right.\text{ }\left. \begin{matrix} & 1 \\\ & x \\\ \end{matrix} \right|={{\lambda }^{2}}\left( {{x}^{2}}-2 \right)$$ And, the value of the other term is $f\left( x \right)={{x}^{2}}-2{{\lambda }^{2}}$. So, the value of the given expression will be as: $ f\left( \lambda x \right)-f\left( x \right)={{\lambda }^{2}}\left( {{x}^{2}}-2 \right)-\left( {{x}^{2}}-2{{\lambda }^{2}} \right) \\\ \Rightarrow f\left( \lambda x \right)-f\left( x \right)={{\lambda }^{2}}{{x}^{2}}-2{{\lambda }^{2}}-{{x}^{2}}+2{{\lambda }^{2}} \\\ \Rightarrow f\left( \lambda x \right)-f\left( x \right)={{x}^{2}}\left( {{\lambda }^{2}}-1 \right) \\\ $ **Hence, option (e) is the correct answer.** **Note:** Students should note the property of the determinants that we used here $\det \left( \lambda A \right)=\lambda \times \text{det}\left( A \right)$. The calculations must be done properly to avoid the unnecessary mistakes.