7.6 Solving systems with gaussian elimination  (Page 3/13)

First, we write this as an augmented matrix.

We want a 1 in row 1, column 1. This can be accomplished by interchanging row 1 and row 2.

We now have a 1 as the first entry in row 1, column 1. Now let’s obtain a 0 in row 2, column 1. This can be accomplished by multiplying row 1 by $\mathrm{-2},$ and then adding the result to row 2.

We only have one more step, to multiply row 2 by $\frac{1}{5}.$

Use back-substitution. The second row of the matrix represents $y=1.$ Back-substitute $y=1$ into the first equation.

The solution is the point $\left(\frac{3}{2},1\right).$

Write the system as an augmented matrix    .

Obtain a 1 in row 1, column 1. This can be accomplished by multiplying the first row by $\frac{1}{2}.$

Next, we want a 0 in row 2, column 1. Multiply row 1 by $\mathrm{-4}$ and add row 1 to row 2.

The second row represents the equation $0=4.$ Therefore, the system is inconsistent and has no solution.

Perform row operations    on the augmented matrix to try and achieve row-echelon form    .

The matrix ends up with all zeros in the last row: $0y=0.$ Thus, there are an infinite number of solutions and the system is classified as dependent. To find the generic solution, return to one of the original equations and solve for $y.$

So the solution to this system is $\left(x,3-\frac{3}{4}x\right).$

The first row already has a 1 in row 1, column 1. The next step is to multiply row 1 by $\mathrm{-2}$ and add it to row 2. Then replace row 2 with the result.

Next, obtain a zero in row 3, column 1.

Next, obtain a zero in row 3, column 2.

The last step is to obtain a 1 in row 3, column 3.

First, we write the augmented matrix.

Next, we perform row operations to obtain row-echelon form.

The easiest way to obtain a 1 in row 2 of column 1 is to interchange $R_2$ and $R_3.$

Then

The last matrix represents the equivalent system.

Using back-substitution, we obtain the solution as $\left(4,\mathrm{-3},1\right).$