91Ó°ÊÓ

Perform Gaussian Elimination for the first column. We can update the elements of the L and U matrices as follows. From \(A_{11} = 1*u_{1,1}\), we get \(u_{1,1} = 1\). Next, we find the elements of the first column of the L matrix. For \(A_{21} = 2 = l_{2,1}*u_{1,1}\), we have \(l_{2,1} = 2\). For \(A_{31} = -3 = l_{3,1}*u_{1,1}\), we have \(l_{3,1} = -3\). Now let's use the Gaussian Elimination for the second column: Subtract 2 times the first row from the second row to get the second row of A, $$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & -1\\ -3 & 1 & -2 \end{bmatrix}$$ Subtract -3 times the first row from the third row to get the third row of A, $$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & -1\\ 0 & 4 & -1 \end{bmatrix}$$ Now, using the updated values of A, we can find the elements of the L and U matrices: For \(A_{22} = 2 = l_{2,1}*u_{1,2} +u_{2,2}\), we have \(u_{2,2} = 2 -2*1 = 0\). Similarly, for \(A_{32} = 4\), we get \(l_{3,2} = (4 +(-3*1)) /0 = 1\). (Note: We divide by \(u_{2,2} = 0\) here) Finally, for \(A_{33} = -1\), we have \(u_{3,3} = -1 -(2*-1) = -1 + 2 = 1\). So, the matrices L and U are now given as: $$L = \left(\begin{array}{ccc} 1 & 0 & 0 \\\ 2 & 1 & 0 \\\ -3 & 1 & 1 \end{array}\right)$$ and $$U = \left(\begin{array}{ccc} 1 & 1 & 1 \\\ 0 & 0 & -1 \\\ 0 & 0 &1 \end{array}\right)$$.

We need to verify if L.U = A. $$L.U = \left(\begin{array}{ccc} 1 & 0 & 0 \\\ 2 & 1 & 0 \\\ -3 & 1 & 1 \end{array}\right)\cdot\left(\begin{array}{ccc} 1 & 1 & 1 \\\ 0 & 0 & -1 \\\ 0 & 0 &1 \end{array}\right)$$ $$=\left(\begin{array}{ccc} 1 & 1 & 1 \\\ 2 & 2 & 1 \\\ -3 & 1 & -2 \end{array}\right)$$ Which matches the given matrix A. Hence, the LU Decomposition is correct, and the LU factorization of A is: $$A = LU$$ $$A=\left(\begin{array}{ccc} 1 & 0 & 0 \\\ 2 & 1 & 0 \\\ -3 & 1 & 1 \end{array}\right)\cdot\left(\begin{array}{ccc} 1 & 1 & 1 \\\ 0 & 0 & -1 \\\ 0 & 0 &1 \end{array}\right)$$