Find the LU Decomposition of the matrix \(A\) where \(A = \begin{bmatrix} 4 & 3 \ 6 & 3 \end{bmatrix}\)

Step 1 ：Step 1: We can write the matrix \(A\) as the product of a lower triangular matrix \(L\) and an upper triangular matrix \(U\). We first identify the upper triangular matrix \(U\) by reducing \(A\) to row echelon form. The matrix \(A\) is already in row echelon form, so \(U = \begin{bmatrix} 4 & 3 \ 0 & -3 \end{bmatrix}\)

Step 2 ：Step 2: We now find the lower triangular matrix \(L\). The first column of \(L\) is the first column of \(A\), and the second column of \(L\) is found by solving the equation \(LU = A\) for the second column of \(L\). So \(L = \begin{bmatrix} 4 & 0 \ 6 & 1 \end{bmatrix}\)

Step 3 ：Step 3: Therefore, the LU Decomposition of the matrix \(A\) is \(A = LU\), where \(L = \begin{bmatrix} 4 & 0 \ 6 & 1 \end{bmatrix}\) and \(U = \begin{bmatrix} 4 & 3 \ 0 & -3 \end{bmatrix}\)