Use back substitution to solve each of the following systems of equations:a) x 1 − 3 x 2 = 2 2 x 2 = 6b) x 1 + x 2 + x 3 = 8 2 x 2 + x 3 = 5 3 x 3 = 9c) x 1 + 2 x 2 + 2 x 3 + x 4 = 5 3 x 2 + x 3 − 2 x 4 = 1 − x 3 + 2 x 4 = − 1 4 x 4 = 4d) x 1 + x 2 + x 3 + x 4 + x 5 = 5 2 x 2 + x 3 − 2 x 4 + x 5 = 1 4 x 3 + x 4 − 2 x 5 = 1 x 4 − 3 x 5 = 0 2 x 5 = 2Write out the coefficient matrix for each of the systems in first exercise.

Question

Use back substitution to solve each of the following systems of equations:a)       x    1    −  3      x    2    =  2    2      x    2    =  6b)       x    1    +      x    2    +      x    3    =  8    2      x    2    +      x    3    =  5    3      x    3    =  9c)       x    1    +  2      x    2    +  2      x    3    +      x    4    =  5    3      x    2    +      x    3    −  2      x    4    =  1    −      x    3    +  2      x    4    =  −  1    4      x    4    =  4d)       x    1    +      x    2    +      x    3    +      x    4    +      x    5    =  5    2      x    2    +      x    3    −  2      x    4    +      x    5    =  1    4      x    3    +      x    4    −  2      x    5    =  1        x    4    −  3      x    5    =  0    2      x    5    =  2Write out the coefficient matrix for each of the systems in first exercise.

Aileen Powers · Accepted Answer

a  )         x    1    −  3      x    2    =  2    2      x    1    =  6  ⇒      x    2    =  3        x    1    −  9  =  2  ⇒      x    1    =  11    b  )     3      x    3    =  9  ⇒      x    3    =  3    2      x    2    +      x    3    =  5  ⇒  2      x    2    =  2  ⇒      x    1    =  1        x    1    +      x    2    +      x    3    =  8  ⇒      x    1    ⇒  4    c  )     4      x    4    =  4  ⇒      x    4    =  1    −      x    3    +  2      x    4    =  −  1  ⇒      x    3    =  3    3      x    2    +      x    3    −  2      x    4    =  1  ⇒      x    2    =  0        x    1    +  2      x    2    +  2      x    3    +      x    4    =  5  ⇒      x    1    =  −  2    d  )     2      x    5    =  2  ⇒      x    5    =  1        x    4    −  3      x    5    =  0  ⇒      x    4    =  3    4      x    3    +      x    4    −  2      x    5    =  1  ⇒      x    3    =  0    2      x    1    +      x    3    −  2      x    4    +      x    5    =  1  ⇒      x    1    =  3        x    1    +      x    2    +      x    3    +      x    4    +      x    5    =  5  ⇒      x    1    =  −  72) Coefficient matrix   a  )      [                            1                          −          3                                      0                          2                      ]      b  )      [                            1                          1                          1                                      0                          2                          1                                      0                          0                          3                      ]      c  )      [                            1                          2                          2                          1                                      0                          3                          1                          −          2                                      0                          0                          −          1                          2                                      0                          0                          0                          4                      ]      d  )      [                            1                          1                          1                          1                          1                                      0                          2                          1                          −          2                          1                                      0                          0                          4                          1                          −          2                                      0                          0                          0                          1                          −          3                                      0                          0                          0                          0                          2                      ]

Use back substitution to solve each of the following systems of equations: a) x_1-3x_2=2, 2x^2=6

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New Questions in Algebra I