Your problem statement shows that you have correctly selected the matrices representing the initial problem setup (middle left) and the problem solution (middle right).

Of the remaining matrices, the upper left is an incorrect setup, and the lower left is an incorrect solution matrix.

We notice that in the remaining matrices on the right that the (2,3) term is 0, and the (3,2) and (3,3) terms are both 1.

The easiest way to get a 0 in the 3rd column of row 2 is to add the first row to the second. When you do that, you get ...

$\left[\begin{array}{ccc|c}1&1&1&29000\\1+2&1-3&1-1&1000(29+1)\\0&0.15&0.15&2100\end{array}\right] =\left[\begin{array}{ccc|c}1&1&1&29000\\3&-2&0&30000\\0&0.15&0.15&2100\end{array}\right]$

Already, we see that the second row matches that in the lower right matrix.

The easiest way to get 1's in the last row is to divide that row by 0.15. When we do that, the (3,4) entry becomes 2100/0.15 = 14000, matching exactly the lower right matrix.

The correct choices here are the two you have selected, and <em>the lower right matrix</em>.

3 0

3 years ago

Select the solution for the system of linear equations y = x + 6 and y = -2x. How do you solve it as well.

Maurinko [17]

You can just plug the coordinates in.

6 0

3 years ago