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Reduce to reduced row echelon form. \( \left[\begin{array}{ccc}4 & 1 & 21 \\ 2 & 5 & 15 \\ -2 & -1 & -11\end{array}\right] \)

Ask by Coleman Hill. in the United States
Jan 23,2025

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The reduced row echelon form of the matrix is: \[ \begin{bmatrix} 1 & 0 & 5 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \]

Solución

To reduce the given matrix to its **Reduced Row Echelon Form (RREF)**, we perform a series of row operations to achieve the following: 1. **Leading 1s** in each pivot position. 2. **Zeros** in all positions above and below each leading 1. 3. **Order** the leading 1s from left to right as you move down the rows. Let's start with the matrix: \[ \begin{bmatrix} 4 & 1 & 21 \\ 2 & 5 & 15 \\ -2 & -1 & -11 \end{bmatrix} \] ### Step 1: Create a Leading 1 in the First Row Divide the first row by 4 to make the first element a leading 1. \[ R1 = \frac{1}{4}R1 \Rightarrow \begin{bmatrix} 1 & \frac{1}{4} & \frac{21}{4} \\ 2 & 5 & 15 \\ -2 & -1 & -11 \end{bmatrix} \] ### Step 2: Eliminate Below the Leading 1 in the First Column - **Row 2:** Subtract 2 times Row 1 from Row 2. \[ R2 = R2 - 2R1 \Rightarrow \begin{bmatrix} 1 & \frac{1}{4} & \frac{21}{4} \\ 0 & \frac{9}{2} & \frac{9}{2} \\ -2 & -1 & -11 \end{bmatrix} \] - **Row 3:** Add 2 times Row 1 to Row 3. \[ R3 = R3 + 2R1 \Rightarrow \begin{bmatrix} 1 & \frac{1}{4} & \frac{21}{4} \\ 0 & \frac{9}{2} & \frac{9}{2} \\ 0 & -\frac{1}{2} & -\frac{1}{2} \end{bmatrix} \] ### Step 3: Create a Leading 1 in the Second Row Divide the second row by \(\frac{9}{2}\) to make the second element a leading 1. \[ R2 = \frac{2}{9}R2 \Rightarrow \begin{bmatrix} 1 & \frac{1}{4} & \frac{21}{4} \\ 0 & 1 & 1 \\ 0 & -\frac{1}{2} & -\frac{1}{2} \end{bmatrix} \] ### Step 4: Eliminate Above and Below the Leading 1 in the Second Column - **Row 1:** Subtract \(\frac{1}{4}\) times Row 2 from Row 1. \[ R1 = R1 - \frac{1}{4}R2 \Rightarrow \begin{bmatrix} 1 & 0 & 5 \\ 0 & 1 & 1 \\ 0 & -\frac{1}{2} & -\frac{1}{2} \end{bmatrix} \] - **Row 3:** Add \(\frac{1}{2}\) times Row 2 to Row 3. \[ R3 = R3 + \frac{1}{2}R2 \Rightarrow \begin{bmatrix} 1 & 0 & 5 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \] ### Final Reduced Row Echelon Form \[ \begin{bmatrix} 1 & 0 & 5 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \]

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To reduce the matrix to reduced row echelon form (RREF), we can use row operations. 1. Start with the original matrix: \[ \left[\begin{array}{ccc}4 & 1 & 21 \\ 2 & 5 & 15 \\ -2 & -1 & -11\end{array}\right] \] 2. Make the leading coefficient of the first row a 1 by dividing the entire row by 4: \[ R_1 = \frac{1}{4}R_1 \implies R_1 = \left[\begin{array}{ccc}1 & \frac{1}{4} & \frac{21}{4}\end{array}\right] \] Matrix becomes: \[ \left[\begin{array}{ccc}1 & \frac{1}{4} & \frac{21}{4} \\ 2 & 5 & 15 \\ -2 & -1 & -11\end{array}\right] \] 3. Eliminate the first column values below the leading 1 in row 1. Use row 1 to update rows 2 and 3: - For row 2: \(R_2 = R_2 - 2R_1 \) \[ R_2 = \left[\begin{array}{ccc}0 & \frac{19}{4} & \frac{-3}{4}\end{array}\right] \] - For row 3: \(R_3 = R_3 + 2R_1\) \[ R_3 = \left[\begin{array}{ccc}0 & \frac{1}{2} & \frac{1}{2}\end{array}\right] \] Matrix becomes: \[ \left[\begin{array}{ccc}1 & \frac{1}{4} & \frac{21}{4} \\ 0 & \frac{19}{4} & -\frac{3}{4} \\ 0 & \frac{1}{2} & \frac{1}{2}\end{array}\right] \] 4. Next, make the leading coefficient in row 2 a 1 by multiplying by \(\frac{4}{19}\): \[ R_2 = \frac{4}{19}R_2 \implies R_2 = \left[\begin{array}{ccc}0 & 1 & -\frac{3}{19}\end{array}\right] \] Matrix becomes: \[ \left[\begin{array}{ccc}1 & \frac{1}{4} & \frac{21}{4} \\ 0 & 1 & -\frac{3}{19} \\ 0 & \frac{1}{2} & \frac{1}{2}\end{array}\right] \] 5. Eliminate the values above and below 1 in column 2. Use row 2 to update rows 1 and 3: - For row 1: \(R_1 = R_1 - \frac{1}{4}R_2\) \[ R_1 = \left[\begin{array}{ccc}1 & 0 & \frac{27}{19}\end{array}\right] \] - For row 3: \(R_3 = R_3 - \frac{1}{2}R_2\) \[ R_3 = \left[\begin{array}{ccc}0 & 0 & \frac{11}{38}\end{array}\right] \] Matrix becomes: \[ \left[\begin{array}{ccc}1 & 0 & \frac{27}{19} \\ 0 & 1 & -\frac{3}{19} \\ 0 & 0 & \frac{11}{38}\end{array}\right] \] 6. Now, normalize row 3 by multiplying by \(\frac{38}{11}\): \[ R_3 = \frac{38}{11}R_3 \implies R_3 = \left[\begin{array}{ccc}0 & 0 & 1\end{array}\right] \] Matrix becomes: \[ \left[\begin{array}{ccc}1 & 0 & \frac{27}{19} \\ 0 & 1 & -\frac{3}{19} \\ 0 & 0 & 1\end{array}\right] \] 7. Lastly, eliminate the values in column 3 from

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