# Recitation 6A

Problem 1: When a deep space probe is launched, corrections may be necessary to place the probe on a precisely calculated trajectory. Radio telemetry provides a stream of vectors, \mathbf{x_1}, \ldots, \mathbf{x_k}, giving information at different times about how the probe’s position compares with its planned trajectory. Let X_k be the matrix [\mathbf{x_1} \ldots \mathbf{x_k}]. The matrix G_k=X_kX_k^T is computed as the radar data are analyzed. When \mathbf{x_{k+1}} arrives, a new G_{k+1} must be computed. Since the data vectors arrive at high speed, the computational burden could be severe. But partitioned matrix multiplication helps tremendously. Compute the column-row expansions of G_k and G_{k+1}, and describe what must be computed in order to update G_k to form G_{k+1}.

Solution: By the column-row expansion rule, G_k=X_kX_k^T=\mathbf{x_1}\mathbf{x_1}^T+\ldots+\mathbf{x_k}\mathbf{x_k}^T. Similarly, G_{k+1}=X_kX_k^T=\mathbf{x_1}\mathbf{x_1}^T+\ldots+\mathbf{x_{k+1}}\mathbf{x_{k+1}}^T. Therefore G_{k+1}=G_k+\mathbf{x_{k+1}}\mathbf{x_{k+1}}^T.

Problem 2: Solve the equation A\mathbf{x}=\mathbf{b} by using the LU factorization for A=\begin{bmatrix}2 & -6 & 4 \\ -4 & 8 & 0\\0 & -4 & 6\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ -2 & 1 & 0\\0 & 1 & 1\end{bmatrix}\begin{bmatrix}2 & -6 & 4 \\ 0 & -4 & 8\\0 & 0 & -2\end{bmatrix} and \mathbf{b}=\begin{bmatrix}2 \\ -4 \\ 6\end{bmatrix}.

Solution: Given the LU factorization of A=LU, it is enough to solve L\mathbf{y}=\mathbf{b} and U\mathbf{x}=\mathbf{y} consecutively. By row reduction, we have \begin{bmatrix}1 & 0 & 0 & 2 \\ -2 & 1 & 0 & -4\\0 & 1 & 1 & 6\end{bmatrix}\sim \begin{bmatrix}1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0\\0 & 1 & 1 & 6\end{bmatrix}\sim \begin{bmatrix}1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0\\0 & 0 & 1 & 6\end{bmatrix}. Therefore \mathbf{y}=\begin{bmatrix}2 \\ 0 \\ 6\end{bmatrix}. Again by row reduction, we have \begin{bmatrix}2 & -6 & 4 & 2 \\ 0 & -4 & 8 & 0\\0 & 0 & -2 & 6\end{bmatrix}\sim\begin{bmatrix}1 & -3 & 2 & 1 \\ 0 & 1 & -2 & 0\\0 & 0 & 1 & -3\end{bmatrix}\sim\begin{bmatrix}1 & -3 & 0 & 7 \\ 0 & 1 & 0 & -6\\0 & 0 & 1 & -3\end{bmatrix}\sim\begin{bmatrix}1 & 0 & 0 & -11 \\ 0 & 1 & 0 & -6\\0 & 0 & 1 & -3\end{bmatrix}. So the solution of the original equation is \mathbf{x}=\begin{bmatrix}-11 \\ -6 \\ -3\end{bmatrix}.