Testbook Test Series: Linear Algebra - Determinant

Question

If a matrix $A$ is given by $f(x)=a_{0}+a_{1}x+a_{2}x^{2}+....+a_{n-1}x^{n-1}+a_{n}x^{n},$then the determinant of $A$ is?

• A. $\frac{a_{0}}{a_{n}}$                 
• B. $\frac{a_{n}}{a_{0}}$
• C. $(-1)^{n}\frac{a_{0}}{a_{n}}$  
• D. $(-1)^{n}\frac{a_{n}}{a_{0}}$

Answer 1

$f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\dots +a_{n}x^{n}$

Hence   $f(A)=a_{0}+a_{1}A+a_{2}A^{2}+ \dots +a_{n}A^{n}$

Characteristic equation  $f(\lambda)=a_{0}+a_{1}\lambda+a_{2}\lambda^{2}+ \dots+a_{n}\lambda^{n}$

Lets take $ n =2$

$f(\lambda)=a_{0}+a_{1}\lambda+a_{2}\lambda^{2}$

$f(\lambda)= a_{2}\lambda^{2} + a_{1}\lambda + a_{0}$

Lets say roots are $\lambda_{1},\lambda_{2}$

$\mid A \mid  = \lambda_{1}\times \lambda_{2} = \dfrac{a_{0}}{a_{2}}$

Lets take $ n =3$

$f(\lambda)=a_{0}+a_{1}\lambda+a_{2}\lambda^{2} + a_{3}\lambda^{3}$

$f(\lambda)= a_{3}\lambda^{3} + a_{2}\lambda^{2} + a_{1}\lambda + a_{0}$

Lets say roots are $\lambda_{1},\lambda_{2},\lambda_{1}$

$\mid A \mid  = \lambda_{1}\times \lambda_{2}\times \lambda_{3} = -\:\dfrac{a_{0}}{a_{3}}$

$\text{det(A) = Product of eigen values} = \lambda_{1}\times \lambda_{2}\times \dots \times \lambda_{n}$

Hence $det(A)= (-1)^{n}\dfrac{a_{0}}{a_{n}}\:\: \text{(Product of roots of equation)}$

So, the correct answer is $(C).$

Comments

Now I corrected the solution.

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Any Proof???

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@Hradesh patel

Here we can apply the properties of eigen values

$\det(A) = \text{Product of Eigen values} = \text{Product of Characteristic roots}$