Question

x-y-z=1, 2x+y+z=2, x-2y+z=4​

Answer 1

Answer:

the answer for this question is

X=1

y=-1

z=1

Answer 2

The solution of the system of linear equations is $x = 1$, $y = -1$, and $z = 1$.

Step-by-step explanation:

Given: Linear equations, $x-y-z=1, \ 2x+y+z=2, \ \text{and} \ x-2y+z=4$

To Find: Solution of the systems of linear equations

Solution:

• Solution of the system of linear equations

The linear equations are $x-y-z=1, \ 2x+y+z=2, \ \text{and} \ x-2y+z=4$ such that they can be written in the form of matrix AX = B. Therefore, we can write,

$\left[\begin{array}{ccc}1&-1&-1\\2&1&1\\1&-2&1\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\2\\4\end{array}\right]$

Using the matrix operation $R_3 \rightarrow R_3 - R_1$

$\Rightarrow \left[\begin{array}{ccc}1&-1&-1\\2&1&1\\0&-1&2\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\2\\3\end{array}\right]$

Using the matrix operation $R_2 \rightarrow R_2 - 2R_1$

$\Rightarrow \left[\begin{array}{ccc}1&-1&-1\\0&3&3\\0&-1&2\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\0\\3\end{array}\right]$

Using the matrix operation $R_2 \rightarrow R_2/3$

$\Rightarrow \left[\begin{array}{ccc}1&-1&-1\\0&1&1\\0&-1&2\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\0\\3\end{array}\right]$

Using the matrix operation $R_3 \rightarrow R_3 + R_2$

$\Rightarrow \left[\begin{array}{ccc}1&-1&-1\\0&1&1\\0&0&3\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\0\\3\end{array}\right]$

Using the matrix operation $R_3 \rightarrow R_3/3$

$\Rightarrow \left[\begin{array}{ccc}1&-1&-1\\0&1&1\\0&0&1\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\0\\1\end{array}\right]$

Using the matrix operation $R_1 \rightarrow R_1 + R_2$

$\Rightarrow \left[\begin{array}{ccc}1&0&0\\0&1&1\\0&0&1\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\0\\1\end{array}\right]$

Using the matrix operation $R_2 \rightarrow R_2 - R_3$

$\Rightarrow \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\-1\\1\end{array}\right]$

Since the matrix $\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$ is the identity matrix (I), therefore,

$\Rightarrow \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}1\\-1\\1\end{array}\right]$

Hence, the solution of the system of linear equations is $x = 1$, $y = -1$, and $z = 1$.