Calculating determinants from matrices

Question

1. $[10$ points $]$ Let $P$ be a $n \times n$ matrix, where the entries of of $P=\left(p_{i j}\right)$ are as follows: $$ p_{i j}=\left\{\begin{array}{l} 1 \text { if } i=j+1 \\ 1 \text { if } i=1, j=n \\ 0 \text { otherwise. } \end{array}\right. $$ Namely, $$ P=\left[\begin{array}{ccccc} 0 & 0 & \cdots & 0 & 1 \\ 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & \cdots & \cdots & 1 & 0 \end{array}\right] $$ Calculate the determinants $\operatorname{det}(P), \operatorname{det}(I+5 P),$ and $\operatorname{det}\left(I+3 P+4 P^{2}\right)$ where $I$ is the identity matrix.

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I solved this and got $\det(P) = 0+0+0+....+ 1(1) = 1$. So, $\det(P) = 1$. Now, we already know that $\det(I) = 1$ and $\det(I+5P) = \det(I) + 5 \det(P) = 1 + 5\times 1 = 1 + 5 = 6$, and $\det(I+3P+4P^2) = \det(I) + 3\det(P) + 4\det(P^2) = 1 + 3\times1 + 4\times0 = 1 + 3 = 4$ since $\det(P^2) = 0$.

Is this correct? Kindly check and let me know if there are any problems. Thanks a lot in advance for your help.

Answer 1

For $\det(I+5P)$. Consider the following: $$ \det(I+5P)= \begin{vmatrix} 1&0&0& \ldots &0 &5\\ 5&1&0& \ldots &0 &0\\ 0&5&1& \ldots &0 &0\\ \vdots & \vdots &\vdots & \ldots & \vdots & \vdots\\ 0&0&0& \ldots &5 &1\\ \end{vmatrix}. $$ Expand (using Laplace's cofactor expansion) about the first row to get \begin{align*} \det(I+5P)&=1 \, \underbrace{\begin{vmatrix} 1&0&0& \ldots &0 &0\\ 5&1&0& \ldots &0 &0\\ 0&5&1& \ldots &0 &0\\ \vdots & \vdots &\vdots & \ldots & \vdots & \vdots\\ 0&0&0& \ldots &5 &1\\ \end{vmatrix}}_{\text{lower triangular}}+(-1)^{n+1}5 \, \underbrace{\begin{vmatrix} 5&1&0& \ldots &0 &0\\ 0&5&1& \ldots &0 &0\\ 0&0&5& \ldots &0 &0\\ \vdots & \vdots &\vdots & \ldots & \vdots & \vdots\\ 0&0&0& \ldots &0 &5\\ \end{vmatrix}}_{\text{upper triangular}}\\ &=1(1)+(-1)^{n+1}5^{n}\\ &=\color{red}{1-(-5)^n}. \end{align*} Note here we have used the fact that for upper and lower triangular matrices the determinant is the product of the diagonal entries.

In case, you have studied eigenvalues and/or if you realize that $P$ is a permutation matrix, then it can help in approaching the remaining determinant.

Answer 2

Your argument isn't correct. Unless the matrix is $1\times1$, the determinant function is not an additive. You cannot argue that $\det(I+5P)=\det(I)+\det(5P)$ without any justification (and in fact, you cannot justify that because the two sides are not equal). Also, $\det(5P)$ is equal to $5^n\det(P)$, not $5\det(P)$. The way you handled $\det(I+3P+4P^2)$ is wrong too for the same reason.

For any nonzero polynomial $g$, by inspecting the entries on the first row of $g(P)$, one may argue that $g(P)\ne0$ when $\deg(g)<n$. Hence the minimal polynomial of $P$ has degree $n$. As $P^n=I$, the minimal polynomial of $P$ must be $f(x)=x^n-1$. Hence the characteristic polynomial of $P$ is also $f$, and \begin{aligned} \det(P)&=(-1)^n\det\left(0I-P\right)=(-1)^nf(0),\\ \det(I+5P)&=(-5)^n\det\left(-\frac15I-P\right)=(-5)^nf\left(-\frac15\right). \end{aligned} By factorising $1+3x+4x^2$ into a product of two linear factors, one can also obtain $\det(I+3P+4P^2)$ in a similar way.