Question

The product of the characteristic root of a square matrix A of order x is equal to
$$\begin{array}{rl}& A.{(-1)}^n\left|A\right|\\ & B.\left|A\right|\\ & C.{(-1)}^n|A-I|\\ & D.|A-I|\end{array}$$

Answer 1

Hint: In this question, we need to find the product of the characteristic roots of a square matrix A in order n. For this, we will first suppose the matrix A of the form ${\left(a_{ij}\right)}_{n\times n}$ where i, j = 1, 2, . . . . . n and then draw general matrix. Then, we will suppose ${\lambda }_1,{\lambda }_2\dots \dots {\lambda }_n$ to be the characteristic roots of the A and then $\varnothing \left(\lambda \right)$ will be characteristic equation. Then we will use $\varnothing \left(\lambda \right)=|A-\lambda I|$ to find $|A-\lambda I|$ and simplify to get a general form of a characteristic equation. Finally, putting $\lambda =0$ will give us the final answer.

Complete step-by-step answer:
Here, matrix A is of order n, so let us suppose that, the matrix A is of the form ${\left(a_{ij}\right)}_{n\times n}$ where i, j = 1, 2, . . . . . n and n is the order of the matrix. Hence, our matrix will look like this,
$$A=\left[\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \dots & a_{1n}\\ a_{21}& a_{22}& & & \\ a_{31}& a_{32}& a_{33}& & \\ ⋮& ⋮& & \ddots & \\ a_{n1}& a_{n2}& a_{n3}& \dots & a_{nn}\end{array}\right]$$
Now, since the matrix is of order n, so it will have n characteristic roots. Let us suppose these roots to be ${\lambda }_1,{\lambda }_2\dots \dots {\lambda }_n$. Now, characteristic equation will be given by $\varnothing \left(\lambda \right)$ where we can also write $\varnothing \left(\lambda \right)=|A-\lambda I|$ so let us find $|A-\lambda I|$.
$$A=\left[\begin{array}{ccccc}{a}_{11}-\lambda & a_{12}& a_{13}& \dots & a_{1n}\\ a_{21}& a_{22}-\lambda & \dots & \dots & a_{2n}\\ a_{31}& a_{32}& a_{33}-\lambda & & ⋮\\ ⋮& ⋮& & \ddots & ⋮\\ a_{n1}& a_{n2}& a_{n3}& \dots & a_{nn}\end{array}\right]$$
Solving the determinants in general form, we will get determinant as $${(-1)}^n[{\lambda }^n+p_1{\lambda }^{n-1}+p_2{\lambda }^{n-2}+\dots \dots +p_n]$$ where $$p_1,p_2,p_3,\dots \dots p_n$$ are general function of terms of matrix.
Since, ${\lambda }_1,{\lambda }_2,{\lambda }_3\dots \dots {\lambda }_n$ are solution of this characteristic equation so equation must look like this, $${(-1)}^n(\lambda -{\lambda }_1)(\lambda -{\lambda }_2)\dots \dots (\lambda -{\lambda }_n)$$
(Putting them equal to zero will give us value of $\lambda$ as ${\lambda }_1,{\lambda }_2,{\lambda }_3\dots \dots {\lambda }_n$).
Hence, value of $\varnothing \left(\lambda \right)=|A-\lambda I|={(-1)}^n(\lambda -{\lambda }_1)(\lambda -{\lambda }_2)\dots \dots (\lambda -{\lambda }_n)$.
Now as we can see from the equation, if we put $\lambda =0$ then the right hand side of the equation will become a product of the characteristic roots of matrix A which we need to find. Hence, putting $\lambda =0$ in above equation we get:
$\begin{array}{rl}& \varnothing \left(0\right)=|A-0|={\lambda }_1\cdot {\lambda }_2\cdot {\lambda }_3\dots \dots {\lambda }_n\\ & \Rightarrow \left|A\right|={\lambda }_1\cdot {\lambda }_2\cdot {\lambda }_3\dots \dots {\lambda }_n\end{array}$
Hence, we can see that, the product of characteristic roots of the matrix A is equal to the determinant of A $\left(\left|A\right|\right)$.

So, the correct answer is “Option B”.

Note: Students should note that, we have taken n characteristics roots because the order of the matrix was n and we would get characteristic equation of order n and hence have n roots of the equation which are n characteristic roots. We have not taken ${(-1)}^n$ after putting $\lambda =0$ because n number of (-1) will cancel out with negative signs of ${\lambda }_1,{\lambda }_2,{\lambda }_3\dots \dots {\lambda }_n$. $[(-1)(-{\lambda }_1),(-1)(-{\lambda }_2),\dots \dots (-1)(-{\lambda }_n)={\lambda }_1,{\lambda }_2,\dots \dots {\lambda }_n]$