91Ó°ÊÓ

A determinant is a unique number associatedwith a square matrix.

A determinant is a scalar value that is a function of the entries of a square matrix.

It is the signed factor by which areas are scaled by this matrix. If the sign is negative the matrix reverses orientation.

Given linear transformation,

$T\left(f\right(t\left)\right)=f(-t)from{P}_{\mathrm{n}}to{P}_{\mathrm{n}}$

Consider the linear transformation$T:P_n→P_n$defined as$T\left(f\right)=f(-t)$.

Consider the basisfor$B=\left\{1,t,t^2,....t^{n-1},t^n\right\}\mathrm{for}{P}_n$.

We have that

$$\begin{gathered} T\left(1\right)=1 \\ T\left(1\right)=1\left(1\right)+0t+0t^2+.....+o{t}^{n-1}+0t^n \\ T\left(t\right)=-t \\ T\left(t\right)=0\left(1\right)+(-1)t+0t^2+.....+o{t}^{n-1}+0t^n \\ T\left(t^2\right)={(-t)}^{2}T\left(t^2\right)=t^2=0\left(1\right)+0t+1t^2+...+0t^{n-1}+0t^n \\ T\left(t^{n-1}\right)={(-t)}^{n-1}T\left(t^{n-1}\right)=0\left(1\right)+0t+1t^2+...+{(-1)}^{n-1}{t}^{n-1}+0t^n \\ T\left(t^n\right)={(-t)}^n \\ T\left(t^n\right)={(-1)}^n{t}^n=0\left(1\right)+0t+0t^2+...+0t^{n-1}+{(-1)}^n{t}^n \end{gathered}$$

Thus the matrix ofwith respect tois

$B=\left[\begin{array}{cccccc}1& 0& 0& ...& 0& 0\\ 0& -1& 0& ...& 0& 0\\ 0& 0& 1& ...& 0& 0\\ â‹\circledR & â‹\circledR & â‹\circledR & …& â‹\circledR & â‹\circledR \\ 0& 0& 0& …& {(-1)}^{n-1}& 0\\ 0& 0& 0& …& 0& {(-1)}^{n-1}\end{array}\right]$

Observe that

$$\begin{gathered} \det B=\left[\begin{array}{cccccc}1& 0& 0& ...& 0& 0\\ 0& -1& 0& ...& 0& 0\\ 0& 0& 1& ...& 0& 0\\ â‹\circledR & â‹\circledR & â‹\circledR & …& â‹\circledR & â‹\circledR \\ 0& 0& 0& …& {(-1)}^{n-1}& 0\\ 0& 0& 0& …& 0& {(-1)}^{n-1}\end{array}\right] \\ \det B=\left[\begin{array}{cccccc}{(-1)}^{°}& 0& 0& ...& 0& 0\\ 0& {(-1)}^{°}& 0& ...& 0& 0\\ 0& 0& {(-1)}^{°}& ...& 0& 0\\ â‹\circledR & â‹\circledR & â‹\circledR & …& â‹\circledR & â‹\circledR \\ 0& 0& 0& …& {(-1)}^{n-1}& 0\\ 0& 0& 0& …& 0& {(-1)}^n\end{array}\right] \\ \det B={(-1)}^{°}-{(-1)}^1.{(-1)}^2....{(-1)}^{n-1}.{(-1)}^n\det B={(-1)}^{0+1+2+...+n-1+n} \\ \det B={(-1)}^{\frac{n(n-1)}{2}} \end{gathered}$$

Therefore

role="math" localid="1659515144419" $\det T=\det B={(-1)}^{\frac{n(n-1)}{2}}$