Question

Use row operations to solve the system

x + y – z = -2
4x – y + z = 7
X – 3y + 2z =-5

Answer 1

(x, y, z) = (1, 12, 15)

Explanation:

As with any set of linear equations, there are many possible routes to a solution. We might simplify the notation a bit by writing the coefficients in an augmented matrix. The columns, left to right, represent the coefficients of x, y, and z, in order, and the constant term.

The row operations we'll use are multiplying a row by a value and adding that result to another row, replacing the other row by the sum.

We can make things a little simpler by writing the second equation first. Then the augmented matrix we're starting with is ...

`\left[\begin{array}c4&-1&1&7\\1&1&-1&-2\\1&-3&2&-5\end{array}\right]`

Adding the second row to the first, we get ...

`\left[\begin{array}ccc5&0&0&5\\1&1&-1&-2\\1&-3&2&-5\end{array}\right]`

Dividing the first row by 5 gives ...

`\left[\begin{array}c1&0&0&1\\1&1&-1&-2\\1&-3&2&-5\end{array}\right]`

Subtracting this from the second row, and again from the third row, we are left with ...

`\left[\begin{array}ccc1&0&0&1\\0&1&-1&-3\\0&-3&2&-6\end{array}\right]`

Multiplying the second row by 3 and adding that to the third row, we get ...

`\left[\begin{array}c1&0&0&1\\0&1&-1&-3\\0&0&-1&-15\end{array}\right]`

Subtracting the third row from the second gives ...

`\left[\begin{array}ccc1&0&0&1\\0&1&0&12\\0&0&-1&-15\end{array}\right]`

Finally, multiplying the last row by -1, we have the solution:

`\left[\begin{array}ccc1&0&0&1\\0&1&0&12\\0&0&1&15\end{array}\right]`

This matrix corresponds to the equations ...

• x = 1
• y = 12
• z = 15

_____

The purpose of our choice of row operations is to make the diagonal elements 1 and the off-diagonal elements 0. That is how we end up with the final equations shown.

As we said, there are many ways to go about this. In general, one can ...

• if necessary, swap rows until the diagonal term of interest is non-zero. If you are doing this using a computer program, generally you want the diagonal term to have the coefficient with the largest magnitude. When doing this by hand, you may want to arrange the rows to avoid fractions when you do the normalizing.
• divide the row by the coefficient of the diagonal element to "normalize" the diagonal element to a value of 1
• zero the other elements in that column by multiplying the row just normalized by the element in another row, then subtracting the product. (The 4th matrix shown above shows this for the first column.)
• proceed to the next diagonal element and repeat the process until all diagonal elements are 1. If you cannot make all diagonal elements 1, then the system of equations does not have a unique solution. If any row becomes all zeros, the system is "dependent" and has infinite solutions. If a row is zeros except for the rightmost column, the system is "inconsistent" and has no solutions.