Answer:
D. (x, y, z) = (1, 1, 2). That is:
[tex]\left\{\begin{aligned}&x = 1\\ & y = 1 \\&z = 2\end{aligned}\right.[/tex].
Step-by-step explanation:
Step One: Make sure that the first coefficient of the first row is 1. In this case, the coefficient of [tex]x[/tex] in the first row is already 1.
Step Two: Using row 1, eliminate the first unknown of row 1 [tex]x[/tex] in the rest of the rows. For example, to eliminate [tex]x[/tex] from row 2, multiply row 1 by the opposite of the coefficient of [tex]x[/tex] in row 2 and add that multiple to row 2. The coefficient of [tex]x[/tex] in row 2 is [tex]3[/tex]. Thus, multiply row 1 by [tex]-3[/tex] to get its multiple:
[tex]-3x - 3y + 3z = 0[/tex].
Add this multiple to row 2 to eliminate [tex]x[/tex] in that row:
[tex]\begin{array}{lrrrcr}&-3x &-3y&+3z& =&0 \\ + & 3x& -y& +z& =& 4\\\cline{1-6}\\[-1.0em]\implies&&-4y &+ 4z&=&4\end{array}[/tex].
Similarly, for the third row, multiply row 1 by [tex]-5[/tex] to get:
[tex]-5x - 5y + 5z = 0[/tex].
Do not replace the initial row 1 with this multiple.
Add that multiple to row 3 to get:
[tex]\begin{array}{lrrrcr}&-5x &-5y&+5z& =&0 \\ + & 5x& & +z& =& 7\\\cline{1-6}\\[-1.0em]\implies&&-5y &+6z&=&7\end{array}[/tex].
After applying step one and two to all three rows, the system now resembles the following:
[tex]\left\{\begin{array}{rrrcr}x& + y & -z& = &0\\&-4y &+4z&=&4\\ &-5y &+6z&=&7\end{array}\right.[/tex].
Ignore the first row and apply step one and two to the second and third row of this new system.
[tex]\left\{\begin{array}{rrcr}-4y &+4z&=&4\\ -5y &+6z&=&7\end{array}\right.[/tex].
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.
[tex]\displaystyle (-\frac{1}{4})\cdot (-4y) + (-\frac{1}{4})\cdot 4z = (-\frac{1}{4})\times 4[/tex].
Row 1 is now [tex]y - z = -1[/tex].
Step Two: Using row 1, eliminate the first unknown of row 1 [tex]y[/tex] in the rest of the rows.
The coefficient of [tex]y[/tex] in row 2 is currently [tex]-5[/tex]. Multiply row 1 by [tex]5[/tex] to get:
[tex]5y - 5z = -5[/tex].
Do not replace the initial row 1 with this multiple.
Add this multiple to row 2:
[tex]\begin{array}{lrrcr}&-5y&+6z& =&7 \\ + & 5y& -5z& =& -5\\\cline{1-5}\\[-1.0em]\implies&&z &= &2\end{array}[/tex].
The system is now:
[tex]\left\{\begin{array}{rrcr}y & - z&=&-1\\& z &=&2\end{array}\right.[/tex].
Include the row that was previously ignored:
[tex]\left\{\begin{array}{rrrcr}x& + y & -z& = &0\\&y &-z&=&-1 \\ & &z&=&2\end{array}\right.[/tex].
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 [tex]1[/tex]. Find the value of each unknown by solving the row on the bottom and substituting back into previous rows.
From the third row: [tex]z = 2[/tex].
Substitute back into row 2:
[tex]y -2 = -1[/tex].
[tex]y = 1[/tex].
Substitute [tex]y = 1[/tex] and [tex]z = 2[/tex] to row 1:
[tex]x + 1 - 2 = 0[/tex].
[tex]x = 1[/tex].
In other words,
[tex]\left\{\begin{aligned}&x = 1\\ & y = 1 \\&z = 2\end{aligned}\right.[/tex].
