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A lower triangular matrix is a special type of square matrix where all the entries above the diagonal are zero. This means that any element \( a_{ij} \), where \( i < j \), must be zero. For example, in a 2x2 matrix, the structure of the lower triangular matrix \( L \) would be:

• First row: \( a_{11}, 0 \)
• Second row: \( a_{21}, a_{22} \)

In LU factorization,
the lower triangular matrix \( L \) is utilized to simplify solving systems of linear equations. In the given step-by-step exercise, our \( L \) matrix is represented as: \[ L = \begin{pmatrix} 1 & 0 \ l_{21} & 1 \end{pmatrix} \] where the diagonal elements are 1, reflecting an important attribute of some lower triangular matrices called unit lower triangular matrices.
This ensures stability and simplifies the computation.

An upper triangular matrix is another specialized square matrix. In this matrix configuration, all the elements below the main diagonal are zero. This means any element \( a_{ij} \), where \( i > j \), must be zero. Consider a 2x2 matrix,
the structure would be as follows:

• First row: \( a_{11}, a_{12} \)
• Second row: \( 0, a_{22} \)

In the context of LU factorization,
the upper triangular matrix \( U \) serves to complete one stage of breaking down the original matrix \( A \) into solvable parts. In our example, \( U \) is written as: \[ U = \begin{pmatrix} u_{11} & u_{12} \ 0 & u_{22} \end{pmatrix} \] This simplification into triangle shapes — both lower and upper —
makes the step of backward substitution straightforward when solving linear matrix equations.

Matrix multiplication is a process of multiplying two matrices by taking a row from one matrix and multiplying it with a column from another matrix. This operation is fundamental in mathematics, especially in linear algebra, when dealing with matrices like in LU factorization.
When multiplying two matrices, the element \( c_{ij} \) in the resulting matrix \( C = AB \) is obtained by this rule:

• Multiply each element of the \( i \)-th row of matrix \( A \) by the corresponding element of the \( j \)-th column of matrix \( B \).
• Sum the products up.

In our example, to verify the correctness of an LU factorization, the original matrix \( A \) must be re-obtained via
the multiplication of \( L \) and \( U \). Let's take a snippet from our exercise: \( A = \begin{pmatrix} 1 & 2 \ -3 & -1 \end{pmatrix} \) was broken down as \( LU \) = \[ \begin{pmatrix} 1 & 0 \ -3 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \ 0 & 5 \end{pmatrix} = \begin{pmatrix} 1 & 2 \ -3 & -1 \end{pmatrix} \]
Demonstrating the validity of matrix multiplication in recomposing the original matrix \( A \).