Given
The matrix,
[tex]\begin{bmatrix}{1} & 1{} & {1} & {|\text{ }4} \\ {-2} & {3} & {3} & {|\text{ }2} \\ {1} & {2} & {3} & {|\text{ }5} \\ {} & {} & {} & {}\end{bmatrix}[/tex]To find:
The solution to the system of equation.
Explanation:
It is given that,
[tex]\begin{bmatrix}{1} & 1{} & {1} & {|\text{ }4} \\ {-2} & {3} & {3} & {|\text{ }2} \\ {1} & {2} & {3} & {|\text{ }5} \\ {} & {} & {} & {}\end{bmatrix}[/tex]That implies,
Consider,
[tex]\begin{gathered} [A|B]=\begin{bmatrix}{1} & 1{} & {1} & {|\text{ }4} \\ {-2} & {3} & {3} & {|\text{ }2} \\ {1} & {2} & {3} & {|\text{ }5} \\ {} & {} & {} & {}\end{bmatrix} \\ \approx\begin{bmatrix}{1} & 1{} & {1} & {|\text{ }4} \\ {0} & 5 & {5} & |\text{ }10 \\ 0 & 1 & 2 & {|\text{ }1} \\ {} & {} & {} & {}\end{bmatrix}[R_2=R_2+2R_1,\text{ }R_3=R_3-R_1] \\ \approx\begin{bmatrix}{1} & 1{} & {1} & {|\text{ }4} \\ {0} & 5 & {5} & |\text{ }10 \\ 0 & 0 & 5 & {|\text{ }-5} \\ {} & {} & {} & {}\end{bmatrix}[R_3=5R_3-R_2] \end{gathered}[/tex]Therefore,
[tex]\begin{gathered} x+y+z=4\text{ \_\_\_\_\_\lparen1\rparen} \\ 5y+5z=10\text{ \_\_\_\_\_\lparen2\rparen} \\ 5z=-5\text{ \_\_\_\_\_\_\lparen3\rparen} \end{gathered}[/tex]Then, (3) implies,
[tex]\begin{gathered} (3)\Rightarrow z=\frac{-5}{5} \\ z=-1 \end{gathered}[/tex]Substitute z = -1 in (2).
Then,
[tex]\begin{gathered} (2)\Rightarrow5y+5(-1)=10 \\ \Rightarrow5y-5=10 \\ \Rightarrow5y=10+5 \\ \Rightarrow5y=15 \\ \Rightarrow y=\frac{15}{5} \\ \Rightarrow y=3 \end{gathered}[/tex]Substitute y = 3, z = -1 in (1).
Then,
[tex]\begin{gathered} (1)\Rightarrow x+3+(-1)=4 \\ \Rightarrow x+3-1=4 \\ \Rightarrow x=4-2 \\ \Rightarrow x=2 \end{gathered}[/tex]Hence, the solution is, option B) x = 2, y = 3, z = -1.
