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1. Find sequence of elementary matrices whose product is \( A=\left[\begin{array}{cc}1 / 2 & -3 \\ 2 & 3 / 15\end{array}\right] \)

Ask by Ruiz Campbell. in the Philippines
Mar 14,2025

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The sequence of elementary matrices is: 1. \( \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & 1 \end{bmatrix} \) 2. \( \begin{bmatrix} 1 & 0 \\ -3 & 1 \end{bmatrix} \) 3. \( \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{5} \end{bmatrix} \) Their product equals matrix \( A \).

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To express the matrix \( A = \left[ \begin{array}{cc} 1/2 & -3 \\ 2 & 3/15 \end{array} \right] \) as a product of elementary matrices, we can perform row operations to transform the identity matrix into \( A \). Start with the identity matrix \( I = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] \). We’ll apply a series of row operations: 1. To get the first entry in the first row to be \( 1/2 \), we can multiply the first row of the identity matrix by \( 1/2 \): \[ E_1 = \left[ \begin{array}{cc} 1/2 & 0 \\ 0 & 1 \end{array} \right] \] 2. Next, we need to make the first column in the second row equal to \( 2 \). We can achieve this by replacing the second row with its own twice its first row, leading to: \[ E_2 = \left[ \begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array} \right] \] 3. Finally, we want to adjust the second row to have the second entry equal to \( 3/15 = 1/5 \). We can do this by replacing the second row using a combination involving the first row to adjust the \( y \) entry. We can use: \[ E_3 = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1/5 \end{array} \right] \] Now, we can compute the product of these elementary matrices: \[ A = E_3 E_2 E_1 I \] Thus, a sequence of elementary matrices that can reproduce \( A \) from the identity matrix is: 1. \( E_1 = \left[ \begin{array}{cc} 1/2 & 0 \\ 0 & 1 \end{array} \right] \), 2. \( E_2 = \left[ \begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array} \right] \), 3. \( E_3 = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1/5 \end{array} \right] \).

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