Question

7x1+2x2+3x3=15
5x1-3x2+2x3=15
10X1-11X2+5x3=36

Answer 1

9514 1404 393

Answer:

(x1, x2, x3) = (2, -1, 1)

Explanation:

Your graphing calculator or an on-line solver can reduce the augmented matrix for you.

`\left[\begin{array}c7&2&3&15\\5&-3&2&15\\10&-11&5&36\end{array}\right]`

The result is the reduced row-echelon form ...

`\left[\begin{array}c1&0&0&2\\0&1&0&-1\\0&0&1&1\end{array}\right]`

(x1, x2, x3) = (2, -1, 1)

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Manual solution

As you probably know, the idea is to eliminate variables by using combinations of rows of coefficients. It is useful to start by creating a row with 1 in the first column. That can be done many ways, including ...

• divide the first row by 7
• subtract 4 times the second row from 3 times the first row
• subtract 2 times the first row from 3 times the second row

The first of these choices has the disadvantage of introducing fractions right away. The result of the 3rd choice is the row ...

[1, -13, 0 | 15]

Now, the rest of column 1 can be made zeros by adding the row to the product of the first item in the row and this row. The row operations are ...

r2 ← (r2 -5×r1) ÷ 2

r3 ← r3 -10×r1

This gives ...

`\left[\begin{array}c1&-13&0&15\\0&31&1&-30\\0&119&5&-114\end{array}\right]`

The row operation ...

r2 ← 6×r3 -23×r2

will put a 1 on the diagonal of that row, making it be [0, 1, 7 | 6]. The rest of the second column can be made 0 by ...

r1 ← 13×r2 +r1

r3 ← (-119×r2 +r3)/(-828)

This gives ...

`\left[\begin{array}ccc1&0&91&93\\0&1&7&6\\0&0&1&1\end{array}\right]`

Finally, the replacements ...

r1 ← r1 -91×r3

r2 ← r2 -7×r3

will give the reduced row-echelon form shown in the top section of this explanation.

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Additional comment

For a 3-variable system of equations, it can work reasonably well to use available methods to solve the 2-variable system that results from substitution for the third variable. One of my favorite methods is to use a graphing calculator.