Question

Given the linear systems (a) x1+2x2 = 2 and 3x1+7x2 = 8 (b) x1+2x2 = 1 and 3x1+7x2 = 7 solve both systems by incorporating the right-hand sides into a 2x2 matrix B and computing the reduced row echelon form of (A|B) = [ 1 2 | 2 1 | 3 7 | 8 7 ]

Answer 1

Answer:

• (x1, x2) = (-2, 2)
• (x1, x2) = (-7, 4)

Step-by-step explanation:

The augmented matrix is shown after the indicated row operations. The last line shows the solutions.

  $\left[\begin{array}{cc|cc}1&2&2&1\\3&7&8&7\end{array}\right] \quad\text{given}\\\\\left[\begin{array}{cc|cc}1&2&2&1\\0&1&2&4\end{array}\right] \quad\text{subtract 3r1 from r2}\\\\\left[\begin{array}{cc|cc}1&0&-2&-7\\0&1&2&4\end{array}\right] \quad\text{subtract 2r2 from r1}$

The solution to the first system is (x1, x2) = (-2, 2).

The solution to the second system is (x1, x2) = (-7, 4).