Question 8.5.2: Solve the system {x-2y+3z= 4 2x+y-4z=3 -3x+4y-z= -2...
Solve the system
\left\{\begin{aligned}x-2 y+3 z= & 4 \\2 x+y-4 z= & 3 \\-3 x+4 y-z= & -2\end{aligned}\right.
Step 1:
We start with the augmented matrix of the system, which is a matrix that includes both the coefficients of the variables and the constants on the right side of the equations.
Step 2:
We apply elementary row transformations to obtain a simpler matrix that represents an equivalent system of equations. These transformations involve manipulating the rows of the matrix.
Step 3:
The first transformation we apply is -2 times the first row added to the second row, which gives us a new second row. Then, we apply 3 times the first row added to the third row, which gives us a new third row.
Step 4:
Next, we perform the operation of adding -2 times the first row to the second row and 3 times the first row to the third row.
Step 5:
We continue by multiplying the second row by 1/5 and the third row by -1/2.
Step 6:
After that, we subtract the second row from the third row.
Step 7:
Finally, we multiply the third row by -1/2.
Step 8:
We now have a new matrix that represents an equivalent system of equations. We can interpret this matrix as the system of equations:
x - 2y + 3z = 4 y - 2z = -1 z = 2
x - 2y + 3z = 4 y - 2z = -1 z = 2
Step 9:
This system is easier to solve than the original system. We can use back substitution to find the values of x, y, and z. In this case, we find that x = 4, y = 3, and z = 2.
Step 10:
Therefore, the solution to the original system of equations is x = 4, y = 3, and z = 2.
Final Answer
We begin with the matrix of the system—that is, with the augmented matrix:
\left[\begin{array}{rrr|r}1 & -2 & 3 & 4 \\2 & 1 & -4 & 3 \\-3 & 4 & -1 & -2\end{array}\right]We next apply elementary row transformations to obtain another (simpler) matrix of an equivalent system of equations. These transformations correspond to the manipulations used for equations in Example 1. We will place appropriate symbols between equivalent matrices.
\left[\begin{array}{rrr|r}1 & -2 & 3 & 4 \\2 & 1 & -4 & 3 \\-3 & 4 & -1 & -2\end{array}\right] \begin{aligned}& -2 {R}_1+{R}_2 \rightarrow {R}_2 \\& 3 {R}_1+{R}_3 \rightarrow {R}_3\end{aligned}\left[\begin{array}{rrr|r}1 & -2 & 3 & 4 \\0 & 5 & -10 & -5 \\0 & -2 & 8 & 10\end{array}\right] \quad \begin{aligned}& \text { add }-2 R_1 \text { to } R_2 \\& \text { add } 3 R_1 \text { to } R_3\end{aligned}
\begin{aligned}& \begin{array}{r}\frac{1}{5} {R}_2 \rightarrow {R}_2 \\-\frac{1}{2} {R}_3 \rightarrow {R}_3\end{array} \left[\begin{array}{rrr|r}1 & -2 & 3 & 4 \\0 & 1 & -2 & -1 \\0 & 1 & -4 & -5\end{array}\right] \quad \begin{array}{l}\text { multiply } \mathrm{R}_2 \text { by } \frac{1}{5} \\\text { multiply } \mathrm{R}_3 \text { by }-\frac{1}{2}\end{array} \\& -{R}_2+{R}_3 \rightarrow{R}_3\left[\begin{array}{rrr|r}1 & -2 & 3 & 4 \\0 & 1 & -2 & -1 \\0 & 0 & -2 & -4\end{array}\right] \quad \text { add }-\mathrm{R}_2 \text { to } \mathrm{R}_3 \\& -\frac{1}{2} {R}_3 \rightarrow {R}_3\left[\begin{array}{rrr|r}1 & -2 & 3 & 4 \\0 & 1 & -2 & -1 \\0 & 0 & 1 & 2\end{array}\right] \quad \text { multiply } \mathrm{R}_3 \text { by }-\frac{1}{2} \\&\end{aligned}
We use the last matrix to return to the system of equations
\left[\begin{array}{rrr|r}1 & -2 & 3 & 4 \\0 & 1 & -2 & -1 \\0 & 0 & 1 & 2\end{array}\right]\Longleftrightarrow\left\{\begin{aligned}x-2 y+3 z= & 4 \\y-2 z= & -1 \\z= & 2\end{aligned}\right.
which is equivalent to the original system. The solution x=4, y=3, z=2 may now be found by back substitution, as in Example 1.
Related Answered Questions
Question: 8.3.9
Verified Answer:
We must first solve the associated equality for y:...
Question: 8.9.4
Verified Answer:
The determinant of the coefficient matrix is
[late...
Question: 8.9.5
Verified Answer:
We shall merely list the various determinants. You...
Question: 8.3.3
Verified Answer:
We replace each ≤ with = and then sketch the resul...
Question: 8.3.4
Verified Answer:
The first two inequalities are the same as those c...
Question: 8.3.5
Verified Answer:
Using properties of absolute values (listed on pag...
Question: 8.3.6
Verified Answer:
The graphs of x²+y²=16 and x+y=2 are the circle an...
Question: 8.3.7
Verified Answer:
An equation of the circle is x²+y²=5². Since the i...
Question: 8.9.1
Verified Answer:
We plan to use property 3 of the theorem on row an...
Question: 8.9.3
Verified Answer:
\left|\begin{array}{ccc}1 & 1 & 1 \...