Algebra Know-It-ALL

656 Worked-Out Solutions to Exercises: Chapters 11 to 19

We morph the first equation as follows:

x=y−z− 7

x−y=−z− 7

x−y+z=− 7

Next, the second equation:

y= 2 x+ 2 z+ 2

− 2 x+y= 2 z+ 2

− 2 x+y− 2 z= 2

Finally, the third:

z= 3 x− 5 y+ 4

− 3 x+z=− 5 y+ 4

− 3 x+ 5 y+z= 4

Now we have this set of equations representing our three-by-three linear system:

x−y+z=− 7

− 2 x+y− 2 z= 2

− 3 x+ 5 y+z= 4

2. Before we write down the matrix, we must be sure we have the correct signs for the
   coefficients. Subtraction of a positive is equivalent to addition of a negative. With that
   in mind, we can “pigeonhole” the coefficients into the matrix form:

1 − 1 1 − 7

− 2 1 − 2 2

− 3 514

3. It’s easy to convert a matrix into a set of linear equations, but we must pay attention to
   the signs. Here’s the matrix:

(^04) − 1 − 2
(^5) −3/2 81
1111