Question

What do I need to do to be able to solve this problem?

Answer 1

9514 1404 393

Answer:

(x, y, z) = (5, 11, 6)

Explanation:

To solve this problem, you need to understand "row operations". The ones we're concerned with are multiplying a row by a scalar, and adding rows together.

You also need to understand what the solution looks like, and the usual way that is achieved. The solution will have the 3×3 matrix left of the vertical line be a diagonal of 1s (the identity matrix). Then the numbers to the right of the vertical line represent the solution.

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In the following, we will use the notation [a]+b[c]⇒[d] to mean that row 'a' is added to the product of row 'c' and the scalar 'b' and that sum replaces row [d]. Multiplying a row by a scalar multiplies each element in the row by that scalar.

The first several operations look like this. Notice we have made the upper left 2×2 matrix an identity matrix. Steps will continue to take care of the 3rd column.

`\left[\begin{array}c1&1&1&22\\6&1&8&89\\-6&4&4&38\end{array}\right] \qquad\text{given}\\\\\left[\begin{array}ccc1&1&1&22\\0&-5&2&-43\\0&5&12&127\end{array}\right]\qquad\text{[2]+[3]$\rightarrow$[3];\ 6[1]+[2]$\rightarrow$[2]}\\\\\left[\begin{array}ccc1&1&1&22\\0&1&-0.4&8.6\\0&0&14&84\end{array}\right]\qquad\text{[2]+[3]$\rightarrow$[3];\ $-(1)/(5)$[2]$\rightarrow$[2]}\\\\\left[\begin{array}ccc1&0&1.4&13.4\\0&1&-0.4&8.6\\0&0&14&84\end{array}\right]\qquad\text{[1]-[2]$\rightarrow$[1]}`

Now, we normalize the third column.

`\left[\begin{array}c1&0&1.4&13.4\\0&1&-0.4&8.6\\0&0&1&6\end{array}\right] \qquad\text{$(1)/(14)$[3]$\rightarrow$[3]}\\\\\left[\begin{array}c1&0&0&5\\0&1&0&11\\0&0&1&6\end{array}\right] \qquad\text{$-(7)/(5)$[3]+1$\rightarrow$[1];\ $(2)/(5)$[3]+[2]$\rightarrow$[2]}`

This is the target of our row operations. It tells us the solution to the system of equations is ...

(x, y, z) = (5, 11, 6)

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A number of online calculators and phone or tablet apps are available for putting a coefficient matrix into this "reduced row-echelon form." Many graphing calculators will do this, too.

In the end, this is not terribly different from ad hoc solution using "elimination" methods. That is precisely what we did when we created the zeros in the first two columns of row 3.