Question 10.1.4: Comparing Linear Systems and Matrices Solve the system of li...
Comparing Linear Systems and Matrices
Solve the system of linear equations.
\left\{\begin{array}{rr}x-y-z= & 1 & (1)\\2 x-3 y+z= & 10 & (2)\\x+y-2 z= & 0 & (3)\end{array}\right.
| Linear System | Augmented Matrix |
| \left\{\begin{array}{r}x-y-z=1 &(1)\\2 x-3 y+z=10 &(2)\\x+y-2 z=0 &(3)\end{array}\right. | A=\left[\begin{array}{rrr|r}1 & -1 & -1 & 1 \\2 & -3 & 1 & 10 \\1 & 1 & -2 & 0\end{array}\right] |
| \underbrace{\begin{aligned} \text{Add -2 times equation (1) to equation (2).} \\ \text {Add -1 times equation (1) to equation (3).} \end{aligned}}_{} | \underbrace{\begin{aligned} \text{Use the first row to produce zeros at} \\ \text {the (2, 1) and (3, 1) positions.} \end{aligned}}_{} |
| \left\{\begin{array}{r}x-z=1 &(1)\\-y+3 z=8 &(4)\\2 y-z=-1 &(5)\end{array}\right. | \begin{aligned} \underrightarrow{-2 R_{1}+R_{2} \rightarrow R_{2}} \\ \underrightarrow{(-1) R_{1}+R_{3} \rightarrow R_{3}} \end{aligned} \left[\begin{array}{rrr|r} 1 & -1 & -1 & 1 \\ 0 & -1 & 3 & 8 \\ 0 & 2 & -1 & -1 \end{array}\right] |
| \underbrace{\begin{aligned} \text{Multiply equation (4) by -1.} \end{aligned}}_{} | \underbrace{\text{Produce a 1 at the (2, 2) position.}}_{} |
| \left\{\begin{array}{rr}x-y-z= & 1 &(1)\\y-3 z= & -8 &(6)\\2 y-z= & -1 &(5)\end{array}\right. | \underrightarrow{(-1) R_{2}} \left[\begin{array}{rrr|r}1 & -1 & -1 & 1 \\0 & 1 & -3 & -8 \\0 & 2 & -1 & -1\end{array}\right] |
| \underbrace{\begin{aligned} \text{Add -2 times equation (6) } \\ \text {to equation (5).} \end{aligned}}_{} | \underbrace{\begin{aligned} \text{Use the second row to produce a zero} \\ \text {at the (3, 2) position.} \end{aligned}}_{} |
| \left\{\begin{array}{rr}x-y-z= & 1 &(1)\\y-3 z= & -8 &(6)\\5z= &15 &(7)\end{array}\right. | \underrightarrow{-2 R_{2}+R_{3} \rightarrow R_{3}} \left[\begin{array}{rrr|r}1 & -1 & -1 & 1 \\0 & 1 & -3 & -8 \\0 & 0 & 5 & 15\end{array}\right] |
Equation (7), 5z = 15, which corresponds to row 3 of the final matrix, gives the value z = 3. We find by back-substitution into equation (6).
\begin{aligned}y-3 z &=-8 & & \text { Equation }(6) ; \text { or row } 2 \text { of the final matrix } \\y-3(3) &=-8 & & \text { Replace } z \text { with } 3 \\y &=1 & & \text { Solve for } y\end{aligned}
We find x by back-substitution.
\begin{aligned}&x-y-z=1 \quad \text { Equation (1); or row } 1 \text { of the final matrix }\\&x-1-3=1 \quad \text { Replace } y \text { with } 1 \text { and } z \text { with } 3 \text {. }\\&x=5 \quad \text { Solve for } x .\end{aligned}
The solution of the system is x=5, y=1, \text { and } z=3. You should check this solution by substituting it into the original system of equations. The solution set is \{(5,1,3)\} \text {. }
