Question

Use the row operations tool to solve the following system of equations, obtaining the solutions in fraction form.

15x + 13y + 4z = 8
11x + 13y + 9z = 1
3x + 5y + 7z = -5
8x + 8y + 2z = 6
7x + 5y + 2z = 2
Give the values for x, y, and z with the fractions reduced to lowest terms (for example 1/2 rather than 3/6).
x =
y =
z =

Answer 1

`x=-(7)/(26), y=(37)/(26), z=-(21)/(13)`

Explanation:

The matrix representation of the system of linear equations is:

`\left(\begin{array}{ccccc}15&13&4&\vdots&8\\11&13&9&\vdots&1\\3&5&7&\vdots&-5\\8&8&2&\vdots&6\\7&5&2&\vdots&2\end{array}\right)`

First apply the following row operations:

`\begin{array}{c}R_(2)\to R_(2)+(-(11)/(15))R_(1)\\R_(3)\to R_(3)+(-(1)/(5))R_(1)\\R_(4)\to R_(4)+(-(8)/(15))R_(1)\\R_(5)\to R_(5)+(-(7)/(15))R_(1)\end{array}`

The resulting matrix is:

`\left(\begin{array}{ccccc}15&13&4&\vdots&8\\0&(52)/(15)&(91)/(15)&\vdots&-(73)/(15)\\0&(12)/(5)&(31)/(5)&\vdots&-(33)/(5)\\0&(16)/(15)&-(2)/(15)&\vdots&(26)/(15)\\0&-(16)/(15)&(2)/(15)&\vdots&-(26)/(15)\end{array}\right)`

Then apply the row operations:

`\begin{array}{c}R_(3)\to R_(3)+(-(12)/(5)\cdot (15)/(52))R_(2)\\R_(4)\to R_(4)+(-(-16)/(15)\cdot (15)/(52))R_(2)\\R_(5)\to R_(5)+((16)/(15)\cdot (15)/(52))R_(2)\end{array}`

The resulting matrix is:

`\left(\begin{array}{ccccc}15&13&4&\vdots&8\\0&(52)/(15)&(91)/(15)&\vdots&-(73)/(15)\\0&0&2&\vdots&-(42)/(13)\\0&0&-2&\vdots&(42)/(13)\\0&0&2&\vdots&-(42)/(13)\end{array}\right)`

Now apply the row operations:

`\begin{array}{c} R_(4)\to R_(4)+R_(3)\\R_(5)\to R_(5)+(-1)R_(3)\end{array}`

The resulting matrix is:

`\left(\begin{array}{ccccc}15&13&4&\vdots&8\\0&(52)/(15)&(91)/(15)&\vdots&-(73)/(15)\\0&0&2&\vdots&-(42)/(13)\\0&0&0&\vdots&0\\0&0&0&\vdots&0\end{array} \right)`

The equivalent linear system associated to this matrix is

`\begin{cases}15x+13y+4z=8\\(52)/(15)y+(91)/(15)z=-(73)/(15)\\2z=-(42)/(13)\end{cases}`

To Solve this last system is very simple by substitution. The solutions are:

`x=-(7)/(26), y=(37)/(26), z=-(21)/(13)`