Question

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

Answer 1

Answer:

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 $x$ 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.

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

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

Step Two: Using row 1, eliminate the first unknown of row 1 $y$ 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,

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