Question

Solve each of the following systems by Gauss-Jordan elimination. (b) X1-2x2+ x3- 4x4=1 X1+3x2 + 7x3 + 2x4=2 -12x2-11x3- 16x4 5 (a) 5x1+2x2 +6x3= 0 -2x1 +x2+3x3 = 0

Answer 1

a) The set of solutions is
`\{(0,-3x_3,x_3): x_3\; \text{es un real}\}` y b) the set of solutions is
`\{(-6,(-41)/(17)-(30)/(17)x_4 , (37)/(17)+(8)/(17) x_4 ,x_4): x_4\;\text{es un real}\}`.

Explanation:

a) Let's first find the echelon form of the matrix
`\left[\begin{array}{ccc}5&2&6\\-2&1&3\end{array}\right]`.

• We add
  `(2)/(5)` from row 1 to row 2 and we obtain the matrix
  `\left[\begin{array}{ccc}5&2&6\\0&(9)/(5) &(27)/(5)\end{array}\right]`

• From the previous matrix, we multiply row 1 by
  `(1)/(5)` and the row 2 by
  `(5)/(9)` and we obtain the matrix
  `\left[\begin{array}{ccc}1&(2)/(5) &(6)/(5) \\0&1&3\end{array}\right]`. This matrix is the echelon form of the initial matrix.

The system has a free variable (x3).

• x2+3x3=0, then x2=-3x3

• 0=x1+
  `(2)/(5)`x2+
  `(6)/(5)`x3=

x1+
`(2)/(5)`(-3x3)+
`(6)/(5)`x3=

x1-
`(6)/(5)`x3+
`(6)/(5)`x3

then x1=0.

The system has infinite solutions of the form (x1,x2,x3)=(0,-3x3,x3), where x3 is a real number.

b) Let's first find the echelon form of the aumented matrix
`\left[\begin{array}{ccccc}1&-2&1&-4&1\\1&3&7&2&2\\0&-12&-11&-16&5\end{array}\right]`.

• To row 2 we subtract row 1 and we obtain the matrix
  `\left[\begin{array}{ccccc}1&-2&1&-4&1\\0&5&6&6&1\\0&-12&-11&-16&5\end{array}\right]`

• From the previous matrix, we add to row 3,
  `(12)/(5)` of row 2 and we obtain the matrix
  `\left[\begin{array}{ccccc}1&-2&1&-4&1\\0&5&6&6&1\\0&0&(17)/(5)&(-8)/(5)&(37)/(5)   \end{array}\right]`.

• From the previous matrix, we multiply row 2 by
  `(1)/(5)` and the row 3 by
  `(5)/(17)` and we obtain the matrix
  `\left[\begin{array}{ccccc}1&-2&1&-4&1\\0&1&(6)/(5) &(6)/(5)&(1)/(5)\\0&0&1&(-8)/(17)&(37)/(17) \end{array}\right]`. This matrix is the echelon form of the initial matrix.

The system has a free variable (x4).

• x3-
  `(8)/(17)`x4=
  `(37)/(17)`, then x3=
  `(37)/(17)`+
  `(8)/(17)x4.</li></ul><ul><li>x2+[tex](6)/(5)`x3+
  `(6)/(5)`x4=
  `(1)/(5)`, x2+
  `(6)/(5)`(
  `(37)/(17)`+
  `(8)/(17)x4)+[tex](6)/(5)`x4=
  `(1)/(5)`, then

x2=
`(-41)/(17)-(30)/(17)`x4.

• x1-2x2+x3-4x4=1, x1+
  `(82)/(17)`+
  `(60)/(17)`x4+
  `(37)/(17)`+
  `(8)/(17)`x4-4x4=1, then x1=
  `1-(119)/(17)=-6`

The system has infinite solutions of the form (x1,x2,x3,x4)=(-6,
`(-41)/(17)-(30)/(17)`x4,
`(37)/(17)`+
`(8)/(17)`x4,x4), where x4 is a real number.