Consider the system of $d$ equations contained in the rows of $R\bar{x} = \bar{e_k}$ for the $d × d$ upper-triangular matrix $R$, where $\bar{e_k}$ is a $d$-dimensional column vector with a single value of $1$ in the $kth$ entry and $0$ in all other entries. Discuss why solving for $\bar{x} = [x_1 \cdots x_d]^T$ is simple in this case by solving for the variables in the order $x_d, x_{d−1}, \dots, x_1$. Furthermore, discuss why the solution for $R\bar{x} = \bar{e_k}$ must satisfy $x_i = 0$ for $i > k$. Why is the solution $x$ equal to the $kth$ column of the inverse of $R$? Discuss why the inverse of $R$ is also upper-triangular.
I am confused by the first part of the question and don't know where to start. For the last two questions, my answer for Why is the solution $x$ equal to the $kth$ column of the inverse of $R$? is that solving the equation $\bar{x}=R^{-1}\bar{e_k}$, all the columns of $R^{-1}$ would be multiplied by the scalars $ith$ entries $0$ of $\bar{e_k}$ except for the $kth$ entry of $\bar{e_k}$ where it is equal to $1$. For the last part Discuss why the inverse of R is also upper-triangular. The polynomials and inverses of triangular matrices are also triangular matrices of the same type.
It is all about definitions. Any matrix is equivalent to its reduced row echelon form. In order to guarantee that $R$ is invertible, none of its diagonal entries can be zero (See Exercise 1.6.9 in Hoffman and Kunze). Now performing elementary row operations (Only operation types 1 and 2. In other words, only scaling and shearing, no switching is needed) on $R$ will convert $R$ to $I$. Since each elementary row operation works on the upper part of $R$, it is equivalent to multiply an upper elementary matrix on the left side of $R$. Hence, $E_s \cdots E_1 R=I$, which implies $R^{-1}=E_s\cdots E_1$. The product of upper-triangular matrices is still upper-triangular.