Questions: L= ( 1 0 0 0 1 1 0 0 -3 -1/2 1 0 2 0 -2 1 ) U= ( -1 2 3 1 0 4 -3 5 0 0 1/2 -9/2 0 0 0 12 )

L= ( 1 0 0 0 
      1 1 0 0 
      -3 -1/2 1 0 
      2 0 -2 1 ) 
U= ( -1 2 3 1 
      0 4 -3 5 
      0 0 1/2 -9/2 
      0 0 0 12 )
Transcript text: L=\left(\begin{array}{cccc}1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ -3 & \frac{-1}{2} & 1 & 0 \\ 2 & 0 & -2 & 1\end{array}\right) \\ U=\left(\begin{array}{cccc}-1 & 2 & 3 & 1 \\ 0 & 4 & -3 & 5 \\ 0 & 0 & \frac{1}{2} & \frac{-9}{2} \\ 0 & 0 & 0 & 12\end{array}\right)
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Solution

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Solution Steps

To solve for the LU decomposition of a given matrix, we need to find two matrices, L (lower triangular) and U (upper triangular), such that their product equals the original matrix. Given the matrices L and U, we can verify if their product results in the original matrix.

Step 1: Define the Matrices \(L\) and \(U\)

Given the matrices: \[ L = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ -3 & -0.5 & 1 & 0 \\ 2 & 0 & -2 & 1 \end{pmatrix} \] \[ U = \begin{pmatrix} -1 & 2 & 3 & 1 \\ 0 & 4 & -3 & 5 \\ 0 & 0 & 0.5 & -4.5 \\ 0 & 0 & 0 & 12 \end{pmatrix} \]

Step 2: Calculate the Product \(LU\)

To verify the LU decomposition, we calculate the product \(LU\): \[ LU = L \cdot U \] Performing the matrix multiplication: \[ LU = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ -3 & -0.5 & 1 & 0 \\ 2 & 0 & -2 & 1 \end{pmatrix} \cdot \begin{pmatrix} -1 & 2 & 3 & 1 \\ 0 & 4 & -3 & 5 \\ 0 & 0 & 0.5 & -4.5 \\ 0 & 0 & 0 & 12 \end{pmatrix} \]

Step 3: Perform the Matrix Multiplication

Calculating each element of the resulting matrix \(LU\): \[ LU = \begin{pmatrix} -1 & 2 & 3 & 1 \\ -1 & 6 & 0 & 6 \\ 3 & -8 & -7 & -10 \\ -2 & 4 & 5 & 23 \end{pmatrix} \]

Final Answer

\[ L = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ -3 & -\frac{1}{2} & 1 & 0 \\ 2 & 0 & -2 & 1 \end{pmatrix} \] \[ U = \begin{pmatrix} -1 & 2 & 3 & 1 \\ 0 & 4 & -3 & 5 \\ 0 & 0 & \frac{1}{2} & -\frac{9}{2} \\ 0 & 0 & 0 & 12 \end{pmatrix} \]

\[ LU = \begin{pmatrix} -1 & 2 & 3 & 1 \\ -1 & 6 & 0 & 6 \\ 3 & -8 & -7 & -10 \\ -2 & 4 & 5 & 23 \end{pmatrix} \]

\boxed{LU = \begin{pmatrix} -1 & 2 & 3 & 1 \\ -1 & 6 & 0 & 6 \\ 3 & -8 & -7 & -10 \\ -2 & 4 & 5 & 23 \end{pmatrix}}

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