Mathematics, 02.07.2019 13:10, ellarsteers

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

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

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answered: brittanysanders

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

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

Step-by-step explanation:

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

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

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

Add this multiple to row 2 to eliminate 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 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.

$\displaystyle (-\frac{1}{4})\cdot (-4y) + (-\frac{1}{4})\cdot 4z = (-\frac{1}{4})\times 4$ .

Row 1 is now y - z = -1 .

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

The coefficient of in row 2 is currently . Multiply row 1 by 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 . Find the value of each unknown by solving the row on the bottom and substituting back into previous rows.