Question
Obtain the inverse of the following matrix, by the Gauss - Jordan method 2 -1 0 1 3 2 4 0 1
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Answer to a math question Obtain the inverse of the following matrix, by the Gauss - Jordan method 2 -1 0 1 3 2 4 0 1
Sigrid
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1. Write the augmented matrix by appending the identity matrix:
\left( \begin{array}{ccc|ccc}2 & -1 & 0 & 1 & 0 & 0 \\1 & 3 & 2 & 0 & 1 & 0 \\4 & 0 & 1 & 0 & 0 & 1\end{array} \right)
2. Convert the first row to have a leading 1 by multiplying by\frac{1}{2}
\left( \begin{array}{ccc|ccc}1 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\1 & 3 & 2 & 0 & 1 & 0 \\4 & 0 & 1 & 0 & 0 & 1\end{array} \right)
3. Eliminate the first column entries below the leading 1 in the first row:
Subtract row 1 from row 2:R2 \rightarrow R2 - R1
Subtract 4 times row 1 from row 3:R3 \rightarrow R3 - 4R1
\left( \begin{array}{ccc|ccc}1 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\0 & \frac{7}{2} & 2 & -\frac{1}{2} & 1 & 0 \\0 & 2 & 1 & -2 & 0 & 1\end{array} \right)
4. Convert the second row second element to 1 by multiplying by\frac{2}{7}
\left( \begin{array}{ccc|ccc}1 & -\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\0 & 1 & \frac{4}{7} & -\frac{1}{7} & \frac{2}{7} & 0 \\0 & 2 & 1 & -2 & 0 & 1\end{array} \right)
5. Eliminate the second column entries above and below the leading 1 in the second row:
Add\frac{1}{2} R1 \rightarrow R1 + \frac{1}{2}R2
Subtract 2 times row 2 from row 3:R3 \rightarrow R3 - 2R2
\left( \begin{array}{ccc|ccc}1 & 0 & \frac{2}{7} & \frac{3}{7} & \frac{1}{7} & 0 \\0 & 1 & \frac{4}{7} & -\frac{1}{7} & \frac{2}{7} & 0 \\0 & 0 & \frac{-1}{7} & -\frac{12}{7} & -\frac{4}{7} & 1\end{array} \right)
6. Convert the third row third element to 1 by multiplying by-7
\left( \begin{array}{ccc|ccc}1 & 0 & \frac{2}{7} & \frac{3}{7} & \frac{1}{7} & 0 \\0 & 1 & \frac{4}{7} & -\frac{1}{7} & \frac{2}{7} & 0 \\0 & 0 & 1 & \frac{12}{7} & \frac{4}{7} & -7\end{array} \right)
7. Eliminate the third column entries above the leading 1 in the third row:
Subtract\frac{2}{7} R1 \rightarrow R1 - \frac{2}{7}R3
Subtract\frac{4}{7} R2 \rightarrow R2 - \frac{4}{7}R3
\left( \begin{array}{ccc|ccc}1 & 0 & 0 & \frac{2}{7} & \frac{1}{7} & 2 \\0 & 1 & 0 & -\frac{2}{7} & \frac{2}{7} & 1 \\0 & 0 & 1 & \frac{12}{7} & \frac{4}{7} & -7\end{array} \right)
Thus, the inverse of the given matrix is:
\begin{pmatrix}\frac{2}{7} & \frac{1}{7} & 2 \\-\frac{2}{7} & \frac{2}{7} & 1 \\\frac{12}{7} & \frac{4}{7} & -7\end{pmatrix}
2. Convert the first row to have a leading 1 by multiplying by
3. Eliminate the first column entries below the leading 1 in the first row:
Subtract row 1 from row 2:
Subtract 4 times row 1 from row 3:
4. Convert the second row second element to 1 by multiplying by
5. Eliminate the second column entries above and below the leading 1 in the second row:
Add
Subtract 2 times row 2 from row 3:
6. Convert the third row third element to 1 by multiplying by
7. Eliminate the third column entries above the leading 1 in the third row:
Subtract
Subtract
Thus, the inverse of the given matrix is:
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