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2 votes
(8 pts)

7. Solve the following system using a matrix and converting the matrix to reduced row
echelon form, without introducing any fractions.
3x - 2y + 5z = -7
x – y + 3z = -5
4x + y + z = 6

User Merve
by
4.1k points

2 Answers

5 votes

Write the system as the following augmented matrix:


\left[ \begin{array}ccc 3 & -2 & 5 & -7 \\ 1 & -1 & 3 & -5 \\ 4 & 1 & 1 & 6 \end{array} \right]

Swap rows 1 and 2 :


\left[ \begin{array}c 1 & -1 & 3 & -5 \\ 3 & -2 & 5 & -7 \\ 4 & 1 & 1 & 6 \end{array} \right]

Eliminate the x coefficient from the last two rows by add -3 (row 1) to row 2, and -4 (row 1) to row 3 :


\left[ \begin{array}ccc 1 & -1 & 3 & -5 \\ 0 & 1 & -4 & 8 \\ 0 & 5 & -11 & 26 \end{array} \right]

Eliminate the y coefficient from the first and last rows by adding row 2 to row 1, and -5 (row 2) to row 3 :


\left[ \begin{array}ccc 1 & 0 & -1 & 3 \\ 0 & 1 & -4 & 8 \\ 0 & 0 & 9 & -14 \end{array} \right]

Eliminate the z coefficient from the second row by adding 4 (row 3) to 9 (row 2) :


\left[ \begin{array}ccc 1 & 0 & -1 & 3 \\ 0 & 9 & 0 & 16 \\ 0 & 0 & 9 & -14 \end{array} \right]

Eliminate the z coefficient from the first row by adding row 3 to 9 (row 1) :


\left[ \begin{array}ccc 9 & 0 & 0 & 13 \\ 0 & 9 & 0 & 16 \\ 0 & 0 & 9 & -14 \end{array} \right]

Then the solution to the system is (x, y, z) such that

9x = 13 and 9y = 16 and 9z = -14

or

x = 13/9 and y = 16/9 and z = -14/9

User SUMguy
by
4.2k points
2 votes

Answer:

{x,y,z} = {-22/13,29/13,6/13}

User Akarsh SEGGEMU
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5.2k points