Question

x + y + 2z =5
2 x-3y+5z =5
x + 2y- z. =6
Solve by matrix or adjestment metrix
And also calculate x,y & z.

Answer 1

Answer on Question #49789 – Math – Matrix | Tensor Analysis

$$\begin{array}{r c c c c} x & + & y & + & 2 z = 5 \\ 2 x & - & 3 y & + & 5 z = 5 \\ x & + & 2 y & - & z = 6 \end{array}$$

Solve by matrix. And also calculate $x, y, z$.

Solution:

The first step is to turn three variable systems of equations into a 3x4 Augmented matrix.

$$\left( \begin{array}{c c c} 1 & 1 & 2 \\ 2 & - 3 & 5 \\ 1 & 2 & - 1 \end{array} \right) \left( \begin{array}{c} 5 \\ 5 \\ 6 \end{array} \right)$$

Next we label the rows of the matrix:

$$\left( \begin{array}{c c c} 1 & 1 & 2 \\ 2 & - 3 & 5 \\ 1 & 2 & - 1 \end{array} \right) \left( \begin{array}{c} 5 \\ 5 \\ 6 \end{array} \right) R _ {1} R _ {2} R _ {3}$$

Next we need to change all the entries below the leading coefficient of the first row to zeros.

For the second row, we can achieve this by multiplying by $-\frac{1}{2}$ and then adding the result to row 1. For the third row, we can multiply by $-1$ and then also add the result to row 1.

$$\begin{array}{l} - \frac {1}{2} R _ {2} \left( \begin{array}{c c c} 1 & 1 & 2 \\ - 1 & \frac {3}{2} & - \frac {5}{2} \\ - 1 & - 2 & 1 \end{array} \right) \left( \begin{array}{c} 5 \\ - \frac {5}{2} \\ - 6 \end{array} \right) R _ {1} ^ {\prime} R _ {2} \\ - \frac {1}{2} R _ {2} + R _ {1} \left( \begin{array}{c c c} 1 & 1 & 2 \\ 0 & \frac {5}{2} & - \frac {1}{2} \\ 0 & - 1 & 3 \end{array} \right) \left( \begin{array}{c} 5 \\ \frac {5}{2} \\ - 1 \end{array} \right) R _ {1} ^ {\prime} R _ {2} ^ {\prime} R _ {3} ^ {\prime} \\ - 1 R _ {3} + R _ {1} \\ \end{array}$$

We need the leading element in the second row to also be one. We obtain this result by multiplying the second row by $\frac{2}{5}$.

$$\frac {2}{5} R _ {2} ^ {\prime} \left( \begin{array}{c c c} 1 & 1 & 2 \\ 0 & 1 & - \frac {1}{5} \\ 0 & - 1 & 3 \end{array} \right) \left( \begin{array}{c} 5 \\ 1 \\ - 1 \end{array} \right) R _ {2} ^ {\prime \prime} R _ {3} ^ {\prime \prime}$$

Next we zero out the element in row three beneath the leading coefficient in row two. For achieve these adding row two and three.

$$R _ {2} ^ {\prime \prime} + R _ {3} ^ {\prime} \left( \begin{array}{c c c} 1 & 1 & 2 \\ 0 & 1 & - \frac {1}{5} \\ 0 & 0 & \frac {14}{5} \end{array} \right) \left( \begin{array}{c} 5 \\ 1 \\ 0 \end{array} \right) R _ {3} ^ {\prime \prime}$$

From the above matrix, we solve for the variables starting with $z$ in the last row

$$\frac{14}{5} z = 0$$

So $z = 0$

Next we solve for $y$ by substituting for $z$ in the equation formed by the second row:

$$y - \frac{1}{5} z = 1$$ $$y = 1 + 0$$ $$y = 1$$

Finally we solve for $x$ by substituting the values of $y$ and $z$ into the equation formed by the first row:

$$x + y + 2z = 5$$ $$x = 5 - 1 - 0$$ $$x = 4$$

Therefore, the solution to the system of equations is $\{x,y,z\} = \{4,1,0\}$

Answer: $\{4,1,0\}$

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