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

Hint: 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;
\[f\left( \lambda x \right)=\left| \begin{matrix}
 \lambda x \\
 2\lambda \\
\end{matrix} \right.\text{ }\left. \begin{matrix}
  & \lambda \\
 & \lambda x \\
\end{matrix} \right|\]
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.