Question

Solve the system of equations by finding the reduced row-echelon form of the augmented matrix for the system of equations. 9x-4y+z=-4. -x+2y-3z=20. 4x+4y-z=43

Answer 1

The answer is d. (3, 7, -3)

Explanation:

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Answer 2

d. (3, 7, -3)

Explanation:

The reduced row-echelon form of the augmented matrix has 1s on the diagonal and zeros off-diagonal, except for the answer in the last column. It can be convenient to start by writing the augmented matrix using the second equation first:

`\left[\begin{array}c-1&2&-3&20\\9&-4&1&-4\\4&4&-1&43\end{array}\right] \\\\\text{Add 9 times the first row to the second, and 4 times the first row to the third}\\\\\left[\begin{array}ccc-1&2&-3&20\\0&14&-26&176\\0&12&-13&123\end{array}\right]`

`\text{Subtract 2 times the third row from the second}\\\\\left[\begin{array}ccc-1&2&-3&20\\0&-10&0&-70\\0&12&-13&123\end{array}\right] \\\\\text{Divide the second row by -10}\\\\\left[\begin{array}ccc-1&2&-3&20\\0&1&0&7\\0&12&-13&123\end{array}\right]`

`\text{Subtract 2 times the second row from the first, and}\\\text{subtract 12 times the second row from the third}\\\\\left[\begin{array}ccc-1&0&-3&6\\0&1&0&7\\0&0&-13&39\end{array}\right] \\\\\text{Divide the third row by -13}\\\\\left[\begin{array}c-1&0&-3&6\\0&1&0&7\\0&0&1&-3\end{array}\right]`

`\text{Add 3 times the third row to the first, then multiply the result by -1}\\\\\left[\begin{array}ccc1&0&0&3\\0&1&0&7\\0&0&1&-3\end{array}\right]`

The method may not be strictly according to some algorithm, but it avoids fractions and gives the correct result: (x, y, z) = (3, 7, -3).