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**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 ${{\lambda }_{1}},{{\lambda }_{2}}\ldots \ldots {{\lambda }_{n}}$ to be the characteristic roots of the A and then $\varnothing \left( \lambda \right)$ will be characteristic equation. Then we will use $\varnothing \left( \lambda \right)=\left| A-\lambda I \right|$ to find $\left| A-\lambda I \right|$ 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{matrix}

{{a}_{11}} & {{a}_{12}} & {{a}_{13}} & \ldots & {{a}_{1n}} \\

{{a}_{21}} & {{a}_{22}} & {} & {} & {} \\

{{a}_{31}} & {{a}_{32}} & {{a}_{33}} & {} & {} \\

\vdots & \vdots & {} & \ddots & {} \\

{{a}_{n1}} & {{a}_{n2}} & {{a}_{n3}} & \ldots & {{a}_{nn}} \\

\end{matrix} \right]\]

Now, since the matrix is of order n, so it will have n characteristic roots. Let us suppose these roots to be ${{\lambda }_{1}},{{\lambda }_{2}}\ldots \ldots {{\lambda }_{n}}$. Now, characteristic equation will be given by $\varnothing \left( \lambda \right)$ where we can also write $\varnothing \left( \lambda \right)=\left| A-\lambda I \right|$ so let us find $\left| A-\lambda I \right|$.

\[A=\left[ \begin{matrix}

{{a}_{11}}-\lambda & {{a}_{12}} & {{a}_{13}} & \ldots & {{a}_{1n}} \\

{{a}_{21}} & {{a}_{22}}-\lambda & \ldots & \ldots & {{a}_{2n}} \\

{{a}_{31}} & {{a}_{32}} & {{a}_{33}}-\lambda & {} & \vdots \\

\vdots & \vdots & {} & \ddots & \vdots \\

{{a}_{n1}} & {{a}_{n2}} & {{a}_{n3}} & \ldots & {{a}_{nn}} \\

\end{matrix} \right]\]

Solving the determinants in general form, we will get determinant as \[{{\left( -1 \right)}^{n}}\left[ {{\lambda }^{n}}+{{p}_{1}}{{\lambda }^{n-1}}+{{p}_{2}}{{\lambda }^{n-2}}+\ldots \ldots +{{p}_{n}} \right]\] where \[{{p}_{1}},{{p}_{2}},{{p}_{3}},\ldots \ldots {{p}_{n}}\] are general function of terms of matrix.

Since, ${{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}}\ldots \ldots {{\lambda }_{n}}$ are solution of this characteristic equation so equation must look like this, \[{{\left( -1 \right)}^{n}}\left( \lambda -{{\lambda }_{1}} \right)\left( \lambda -{{\lambda }_{2}} \right)\ldots \ldots \left( \lambda -{{\lambda }_{n}} \right)\]

(Putting them equal to zero will give us value of $\lambda $ as ${{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}}\ldots \ldots {{\lambda }_{n}}$).

Hence, value of $\varnothing \left( \lambda \right)=\left| A-\lambda I \right|={{\left( -1 \right)}^{n}}\left( \lambda -{{\lambda }_{1}} \right)\left( \lambda -{{\lambda }_{2}} \right)\ldots \ldots \left( \lambda -{{\lambda }_{n}} \right)$.

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{align}

& \varnothing \left( 0 \right)=\left| A-0 \right|={{\lambda }_{1}}\cdot {{\lambda }_{2}}\cdot {{\lambda }_{3}}\ldots \ldots {{\lambda }_{n}} \\

& \Rightarrow \left| A \right|={{\lambda }_{1}}\cdot {{\lambda }_{2}}\cdot {{\lambda }_{3}}\ldots \ldots {{\lambda }_{n}} \\

\end{align}$

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 ${{\left( -1 \right)}^{n}}$ after putting $\lambda =0$ because n number of (-1) will cancel out with negative signs of ${{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}}\ldots \ldots {{\lambda }_{n}}$. $\left[ \left( -1 \right)\left( -{{\lambda }_{1}} \right),\left( -1 \right)\left( -{{\lambda }_{2}} \right),\ldots \ldots \left( -1 \right)\left( -{{\lambda }_{n}} \right)={{\lambda }_{1}},{{\lambda }_{2}},\ldots \ldots {{\lambda }_{n}} \right]$

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