Question

Solve the system using the Gauss-Jordan method. 2x – 3y – 9z = 12 4x + 5y – 6z = –14 –5x + 3y – 9z = 21 A. (–3, 7, –5) B. C. No solution D.

Answer 1

Answer:

$(-3,-2,\frac{-4}{3})$

Step-by-step explanation:

We are given the equations,

2x - 3y - 9z = 12

4x + 5y - 6z = - 14

-5x + 3y - 9z = 21

Then, the augmented matrix is given by, $\begin{bmatrix}2&-3&-9&12\\4&5&-6&-14\\-5&3&-9&21\end{bmatrix}$

Now, we will apply some rules to get the row-echlon form of the matrix.

1. $R_{1}=>\frac {R_{1}}{2}$.

This gives, $\begin{bmatrix}1&\frac{-3}{2}&\frac{-9}{2}&6\\4&5&-6&-14\\-5&3&-9&21\end{bmatrix}$

2. $R_{2}=>R_{2}-4R_{1}$ and $R_{3}=>R_{3}+5R_{1}$

We get, $\begin{bmatrix}1&\frac{-3}{2}&\frac{-9}{2}&6\\0&11&12&-38\\0&\frac{-9}{2}&\frac{-63}{2}&51\end{bmatrix}$

3. $R_{2}=>\frac {R_{2}}{11}$

So, $\begin{bmatrix}1&\frac{-3}{2}&\frac{-9}{2}&6\\0&1&\frac{12}{11}&\frac{-38}{11}\\0&\frac{-9}{2}&\frac{-63}{2}&51\end{bmatrix}$

4. $R_{3}=>R_{3}+\frac{9}{2}R_{2}$ and $R_{1}=>R_{1}+\frac{3}{2}R_{2}$

We get, $\begin{bmatrix}1&0&\frac{-63}{22}&\frac{9}{11}\\0&1&\frac{12}{11}&\frac{-38}{11}\\0&0&\frac{-585}{22}&\frac{390}{11}\end{bmatrix}$

5. $R_{3}=>\frac{-22}{585}\times R_{3}$

We get, $\begin{bmatrix}1&0&\frac{-63}{22}&\frac{9}{11}\\0&1&\frac{12}{11}&\frac{-38}{11}\\0&0&1&\frac{-4}{3}\end{bmatrix}$

6. $R_{1}=>R_{1}+\frac{63}{22}R_{3}$ and $R_{2}=>R_{2}+\frac{-12}{11}R_{3}$

So, $\begin{bmatrix}1&0&0&-3\\0&1&0&-2\\0&0&1&\frac{-4}{3}\end{bmatrix}$

Thus, we get,

x = -3, y = -2 and z = $\frac{-4}{3}$.

Hence, the solution is $(-3,-2,\frac{-4}{3})$.