Question #169347

Find the row echelon form of [

1 2 − 2 − 2 −3

1 3 − 2 0 −4

3 8 − 7 − 2 −11

2 1 − 9 − 10 − 3

]. 


1
Expert's answer
2021-03-09T06:58:55-0500

Solution:

Given: [1222313204387211219103]\begin{bmatrix}1&2&-2&-2&-3\\ 1&3&-2&0&-4\\ 3&8&-7&-2&-11\\ 2&1&-9&-10&-3\end{bmatrix}

R1R3R_1\:\leftrightarrow \:R_3

=[3872111320412223219103]=\begin{bmatrix}3&8&-7&-2&-11\\ 1&3&-2&0&-4\\ 1&2&-2&-2&-3\\ 2&1&-9&-10&-3\end{bmatrix}

R2R213R1R_2\:\rightarrow \:R_2-\frac{1}{3}\cdot \:R_1

=[38721101313231312223219103]=\begin{bmatrix}3&8&-7&-2&-11\\ 0&\frac{1}{3}&\frac{1}{3}&\frac{2}{3}&-\frac{1}{3}\\ 1&2&-2&-2&-3\\ 2&1&-9&-10&-3\end{bmatrix}

R3R313R1R_3\:\rightarrow \:R_3-\frac{1}{3}\cdot \:R_1

=[387211013132313023134323219103]=\begin{bmatrix}3&8&-7&-2&-11\\ 0&\frac{1}{3}&\frac{1}{3}&\frac{2}{3}&-\frac{1}{3}\\ 0&-\frac{2}{3}&\frac{1}{3}&-\frac{4}{3}&\frac{2}{3}\\ 2&1&-9&-10&-3\end{bmatrix}

R4R423R1R_4\:\rightarrow \:R_4-\frac{2}{3}\cdot \:R_1

=[3872110131323130231343230133133263133]=\begin{bmatrix}3&8&-7&-2&-11\\ 0&\frac{1}{3}&\frac{1}{3}&\frac{2}{3}&-\frac{1}{3}\\ 0&-\frac{2}{3}&\frac{1}{3}&-\frac{4}{3}&\frac{2}{3}\\ 0&-\frac{13}{3}&-\frac{13}{3}&-\frac{26}{3}&\frac{13}{3}\end{bmatrix}

R2R4R_2\:\leftrightarrow \:R_4

=[3872110133133263133023134323013132313]=\begin{bmatrix}3&8&-7&-2&-11\\ 0&-\frac{13}{3}&-\frac{13}{3}&-\frac{26}{3}&\frac{13}{3}\\ 0&-\frac{2}{3}&\frac{1}{3}&-\frac{4}{3}&\frac{2}{3}\\ 0&\frac{1}{3}&\frac{1}{3}&\frac{2}{3}&-\frac{1}{3}\end{bmatrix}

R3R3213R2=[387211013313326313300100013132313]R_3\:\rightarrow \:R_3-\frac{2}{13}\cdot \:R_2 \\ =\begin{bmatrix}3&8&-7&-2&-11\\ 0&-\frac{13}{3}&-\frac{13}{3}&-\frac{26}{3}&\frac{13}{3}\\ 0&0&1&0&0\\ 0&\frac{1}{3}&\frac{1}{3}&\frac{2}{3}&-\frac{1}{3}\end{bmatrix}

R4R4+113R2=[38721101331332631330010000000]R_4\:\rightarrow \:R_4+\frac{1}{13}\cdot \:R_2 \\ =\begin{bmatrix}3&8&-7&-2&-11\\ 0&-\frac{13}{3}&-\frac{13}{3}&-\frac{26}{3}&\frac{13}{3}\\ 0&0&1&0&0\\ 0&0&0&0&0\end{bmatrix}

This is required reduce matrix to row echelon form.


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