Question

If $\alpha $ is a characteristic root of a non-singular matrix, then the corresponding characteristic root of $adj\left( A \right)$ is.
(a) $\dfrac{\left| A \right|}{\alpha }$
(b) $\left| \dfrac{A}{\alpha } \right|$
(c) $\dfrac{\left| adj\left( A \right) \right|}{\alpha }$
(d) $\left| \dfrac{adj\left( A \right)}{\alpha } \right|$

Answer 1

Hint: For solving this question first we will see how we define the characteristic root for a non-singular matrix, and we will use the relation between a matrix and its adjoint matrix to get the correct answer.

Complete step-by-step answer:
Given:
It is given that If $\alpha $ is a characteristic root of a non-singular matrix $A$ .
Now, we know that the characteristic root of any matrix satisfies its characteristic equation. It is defined by this relation $\left| A-\lambda I \right|=0$ where $\lambda $ is the characteristic root of the matrix $A$ and $I$ is the identity matrix.
Now, we can define it another way like if $\lambda $ is the characteristic root of the matrix $A$ . Then, $AX=\lambda X$ where $X$ is not a null matrix or it is non-singular.
Now, we come back to our question that $\alpha $ is a characteristic root of a non-singular matrix $A$ . Then, we can write that $AX=\alpha X$ where $X$ is not a null matrix or it is non-singular.
Now, the relation between matrix $A$ and $adj\left( A \right)$ is written below:
$A\cdot adj\left( A \right)=\left| A \right|\cdot I................\left( 1 \right)$
Now, let $\beta $ is the characteristic root of $adj\left( A \right)$ . Then, $adj\left( A \right)\cdot X=\beta X$ where $X$ is a not a null matrix or it is non-singular.
Now, we will do some mathematical operations on $adj\left( A \right)\cdot X=\beta X$ to find the value of $\beta $ as follow:
1. Pre-multiply the equation $adj\left( A \right)\cdot X=\beta X$ by the matrix $A$ . Then,
\[\begin{align}
  & adj\left( A \right)\cdot X=\beta X \\
 & \Rightarrow A\cdot adj\left( A \right)\cdot X=A\cdot \left( \beta X \right) \\
\end{align}\]
2. Now, from equation (1) substitute $A\cdot adj\left( A \right)=\left| A \right|\cdot I$ in the above equation. Then,
\[\begin{align}
  & A\cdot adj\left( A \right)\cdot X=A\cdot \left( \beta X \right) \\
 & \Rightarrow \left| A \right|\cdot I\cdot X=\beta \cdot \left( AX \right) \\
 & \Rightarrow \left| A \right|\cdot X=\beta \cdot \left( AX \right) \\
 & \Rightarrow AX=\dfrac{\left| A \right|}{\beta }\cdot X \\
\end{align}\]
3. Now, compare the above equation and $AX=\alpha X$ . Then,
\[\begin{align}
  & AX=\dfrac{\left| A \right|}{\beta }\cdot X=\alpha X \\
 & \Rightarrow \dfrac{\left| A \right|}{\beta }=\alpha \\
 & \Rightarrow \dfrac{\beta }{\left| A \right|}=\dfrac{1}{\alpha } \\
 & \Rightarrow \beta =\dfrac{\left| A \right|}{\alpha } \\
\end{align}\]
4. Now, from our assumption, we can say that $\beta =\dfrac{\left| A \right|}{\alpha }$ is the characteristic root of $adj\left( A \right)$ .
Thus, the characteristic root of $adj\left( A \right)$ is $\dfrac{\left| A \right|}{\alpha }$ .
Hence, (a) is the correct option.

Note: Here, the student should perform mathematical operations on matrices accordingly as per the matrix algebra rules. Moreover, we should not confuse between the option (b) and (a) as in the option (b) $\left| \dfrac{A}{\alpha } \right|$ is written where $A$ is a matrix and $\alpha $ is a scalar and we should know that it is different from what is written in option (a).