Forward elimination changes:

|1 1|
|   | x = b
|1 2|

to a triangular:

|1 1|
|   | x = c
|0 1|

In other words,

x +  y = 5        x +  y = 5      |1 1 5|      |1 1 5|
             =>                   |     |  =>  |     |
x + 2y = 7             y = 2      |1 2 7|      |0 1 2|

That last subtracted l₂₁ = ___ times row 1 from row 2. The reverse step adds l₂₁ times row 1 to row 2. The matrix for that reverse step is L = ___. Multiply this L times the triangular system:

|1 1|      |5|
|   | x1 = | |
|0 1|      |2|

to get ___ = ___. In letters, L multiplies Ux = c to give ___.

(Fill in the blanks above.)

== Solution ==

That last subtracted l₂₁ = 1 times row 1 from row 2. The reverse step adds l₂₁ times row 1 to row 2. The matrix for that reverse step is

    |1 0|
L = |   |
    |1 1|

Multiply this L times the triangular system:

|1 1|       |5|
|   | x1  = | |
|0 1|       |2|

to get:

|1 0| |1 1|      |1 0| |5|   |5|
|   | |   | x1 = |   | | | = | |
|1 1| |0 1|      |1 1| |2|   |7|

In letters, L multiplies Ux = c to give b.