**Part
10:** **Row operations on matrices**

- Unassign
your variables again, and ask
*Mathematica*to solve

**x + 4y + 3z = 10****2x + y - z = -1****3x - y - 4z = 11**

- What
happens in Step 1? What do you think it means? To check this out,
we need
*Mathematica*'s linear algebra commands. Enter the augmented matrix for our current problem as

**A={{1,4,3,10},{2,1,-1,-1},{3,-1,-4,11}};**This tells

A // MatrixForm

*Mathematica*to construct a matrix and fill its entries from the following vectors of numbers, reading across the first row, then the second, then the third. The "MatrixForm" command tells*Mathematica*to print out A as a matrix rather than as a list of vectors.

- First
let's see how to row reduce a step at a time. Then we will see
that a single command will do the whole job. Try the following
sequence of commands:

**A1 = A;**Got the idea? The double square brackets allow you to work with one row of the matrix at a time. Finish the row reduction, or at least take it far enough to decide what went wrong when you asked

A1 // MatrixForm

A1[[2]] = -2*A1[[1]] + A1[[2]];

A1 // MatrixForm

A1[[3]] = -3*A1[[1]] + A1[[3]];

A1 // MatrixForm

A1[[2]] = (-1/7)*A1[[2]];

A1 // MatrixForm

*Mathematica*to solve the system of equations.

(The third type of row operation is swapping two rows. You may not need it for this problem, but if you do, it works like this:

**A2 = A;**See if you can figure out what this does!)

A2 // MatrixForm

temp = A2[[2]];

A2[[2]] = A2[[3]];

A2[[3]]=temp;

A2 // MatrixForm

- Finally,
let's go directly to the reduced row echelon form of in one
step:

**RowReduce[A];**This should show you in another way that our linear system is inconsistent.

% // MatrixForm