Question

What is the solution to the system? x+y-z=0
3x-y+z=4
5x+z=7

Answer 1

D. (x, y, z) = (1, 1, 2). That is:

`\left\{\begin{aligned}&x = 1\\ & y = 1 \\&z = 2\end{aligned}\right.`.

Explanation:

Step One: Make sure that the first coefficient of the first row is 1. In this case, the coefficient of
`x` in the first row is already 1.

Step Two: Using row 1, eliminate the first unknown of row 1
`x` in the rest of the rows. For example, to eliminate
`x` from row 2, multiply row 1 by the opposite of the coefficient of
`x` in row 2 and add that multiple to row 2. The coefficient of
`x` in row 2 is
`3`. Thus, multiply row 1 by
`-3` to get its multiple:

`-3x - 3y + 3z = 0`.

Add this multiple to row 2 to eliminate
`x` in that row:

`\begin{array}{lrrrcr}&-3x &-3y&+3z& =&0 \\ + & 3x& -y& +z& =& 4\\\cline{1-6}\\[-1.0em]\implies&&-4y &+ 4z&=&4\end{array}`.

Similarly, for the third row, multiply row 1 by
`-5` to get:

`-5x - 5y + 5z = 0`.

Do not replace the initial row 1 with this multiple.

Add that multiple to row 3 to get:

`\begin{array}{lrrrcr}&-5x &-5y&+5z& =&0 \\ + & 5x& & +z& =& 7\\\cline{1-6}\\[-1.0em]\implies&&-5y &+6z&=&7\end{array}`.

After applying step one and two to all three rows, the system now resembles the following:

`\left\{\begin{array}{rrrcr}x& + y & -z& = &0\\&-4y &+4z&=&4\\ &-5y &+6z&=&7\end{array}\right.`.

Ignore the first row and apply step one and two to the second and third row of this new system.

`\left\{\begin{array}{rrcr}-4y &+4z&=&4\\ -5y &+6z&=&7\end{array}\right.`.

Step One: Make sure that the first coefficient of the first row is 1.

Multiply the first row by the opposite reciprocal of its first coefficient.

`\displaystyle (-(1)/(4))\cdot (-4y) + (-(1)/(4))\cdot 4z = (-(1)/(4))* 4`.

Row 1 is now
`y - z = -1`.

Step Two: Using row 1, eliminate the first unknown of row 1
`y` in the rest of the rows.

The coefficient of
`y` in row 2 is currently
`-5`. Multiply row 1 by
`5` to get:

`5y - 5z = -5`.

Do not replace the initial row 1 with this multiple.

Add this multiple to row 2:

`\begin{array}{lrrcr}&-5y&+6z& =&7 \\ + & 5y& -5z& =& -5\\\cline{1-5}\\[-1.0em]\implies&&z &= &2\end{array}`.

The system is now:

`\left\{\begin{array}{rrcr}y & - z&=&-1\\& z &=&2\end{array}\right.`.

Include the row that was previously ignored:

`\left\{\begin{array}{rrrcr}x& + y & -z& = &0\\&y &-z&=&-1 \\ & &z&=&2\end{array}\right.`.

This system is now in a staircase form called Row-Echelon Form. The length of the rows decreases from the top to the bottom. The first coefficient in each row is all
`1`. Find the value of each unknown by solving the row on the bottom and substituting back into previous rows.

From the third row:
`z = 2`.

Substitute back into row 2:

`y -2 = -1`.

`y = 1`.

Substitute
`y = 1` and
`z = 2` to row 1:

`x + 1 - 2 = 0`.

`x = 1`.

In other words,

`\left\{\begin{aligned}&x = 1\\ & y = 1 \\&z = 2\end{aligned}\right.`.