Read the following paragraph carefully and answer the following questions. $$ \mathrm{A}=\left|\begin{array}{lll} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array}\right| $$ if, \(\mathrm{U}_{1}, \mathrm{U}_{2}\) and \(\mathrm{U}_{3}\) are column matrices satisfying $$ \mathrm{AU}_{1}=\left|\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right|, \mathrm{AU}_{2}=\left|\begin{array}{l} 2 \\ 3 \\ 0 \end{array}\right|, \mathrm{AU}_{3}=\left|\begin{array}{l} 2 \\ 3 \\ 1 \end{array}\right| $$ and \(\mathrm{U}\) is \(3 \times 3\) matrix whose columns are \(\mathrm{U}_{1}, \mathrm{U}_{2}, \mathrm{U}_{3}\), then The sum of the elements of \(\mathrm{U}^{-1}\) is (a) \(-1\) (b) 0 (c) 1 (d) 3

A lower triangular matrix is a type of square matrix where all the entries above the main diagonal are zero. In the matrix A given in the exercise, it is represented as:
\[\begin{pmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 3 & 2 & 1 \end{pmatrix}\]
Each element \(a_{ij} = 0\) for all \(i < j\), which signifies that the matrix is indeed a lower triangular matrix.

An intriguing property of lower triangular matrices, like the one in the exercise, is that their determinant is the product of the diagonal elements. This ease in computing determinants is advantageous since the determinant plays a crucial role in finding a matrix's inverse. For the case of a lower triangular matrix, the inverse—if it exists—will also be a lower triangular matrix, and we can calculate it by inverting each diagonal entry and adjusting the off-diagonal entries accordingly.

An additional point to note is that a matrix composed entirely of zeros below the main diagonal is termed an upper triangular matrix, which is conceptually the flip of a lower triangular matrix.

The identity matrix is the multiplicative identity in the world of matrices, denoted as I. In a matrix context, \(I\) signifies a matrix that doesn't change another matrix it's multiplied with. The defining structure of an identity matrix is having ones on the main diagonal and zeroes everywhere else. This is how it looks for a 3x3 matrix:
\[I = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}\]
The step-by-step solution demonstrated the significance of the identity matrix while finding the inverse of matrix U. The equation \(UU^{-1} = I\) shows us that when U multiplies its inverse, the result is the identity matrix. This understanding is pivotal since it confirms whether the obtained inverse is indeed correct. And in the context of exercises like the one given, this property becomes very handy for verification purposes.

Matrix multiplication is not like our regular number multiplication. It involves a row-by-column rule, where each entry \(c_{ij}\) of the product matrix C is obtained by multiplying the corresponding elements of row i of the first matrix with column j of the second matrix and adding up those products. It is key to remember that the number of columns in the first matrix must equal the number of rows in the second matrix for the multiplication to be defined. The step-by-step solution used matrix multiplication to calculate AU and then leveraged the resulting matrix to find the inverse of U using the fundamental equation \(AU = UU^{-1}\).

In the context of the problem, we multiply matrix A with the identity and the various U matrices. The elementary but critical point here is that matrix multiplication is not commutative, meaning \(AB\) does not necessarily equal \(BA\); this unique nature influences how we approach problems involving matrix equations. Moreover, the associative and distributive properties are often utilized to simplify and solve these matrix multiplication scenarios.